problem solving grouped data

Mean, Median and Mode from Grouped Frequencies

Explained with Three Examples

The Race and the Naughty Puppy

This starts with some raw data ( not a grouped frequency yet ) ...

Alex timed 21 people in the sprint race, to the nearest second:

59, 65, 61, 62, 53, 55, 60, 70, 64, 56, 58, 58, 62, 62, 68, 65, 56, 59, 68, 61, 67

To find the Mean Alex adds up all the numbers, then divides by how many numbers:

Mean = 59 + 65 + 61 + 62 + 53 + 55 + 60 + 70 + 64 + 56 + 58 + 58 + 62 + 62 + 68 + 65 + 56 + 59 + 68 + 61 + 67 21 Mean = 61.38095...

To find the Median Alex places the numbers in value order and finds the middle number.

In this case the median is the 11 th number:

53, 55, 56, 56, 58, 58, 59, 59, 60, 61, 61 , 62, 62, 62, 64, 65, 65, 67, 68, 68, 70

Median = 61 

To find the Mode , or modal value, Alex places the numbers in value order then counts how many of each number. The Mode is the number which appears most often (there can be more than one mode):

53, 55, 56, 56, 58, 58, 59, 59, 60, 61, 61, 62, 62, 62 , 64, 65, 65, 67, 68, 68, 70

62 appears three times, more often than the other values, so Mode = 62

Grouped Frequency Table

Alex then makes a Grouped Frequency Table :

So 2 runners took between 51 and 55 seconds, 7 took between 56 and 60 seconds, etc

Suddenly all the original data gets lost (naughty pup!)

Only the Grouped Frequency Table survived ...

... can we help Alex calculate the Mean, Median and Mode from just that table?

The answer is ... no we can't. Not accurately anyway. But, we can make estimates .

Estimating the Mean from Grouped Data

So all we have left is:

The groups (51-55, 56-60, etc), also called class intervals , are of width 5

The midpoints are in the middle of each class: 53, 58, 63 and 68

We can estimate the Mean by using the midpoints .

So, how does this work?

Think about the 7 runners in the group 56 - 60 : all we know is that they ran somewhere between 56 and 60 seconds:

• Maybe all seven of them did 56 seconds,
• Maybe all seven of them did 60 seconds,
• But it is more likely that there is a spread of numbers: some at 56, some at 57, etc

So we take an average and assume that all seven of them took 58 seconds.

Let's now make the table using midpoints:

Our thinking is: "2 people took 53 sec, 7 people took 58 sec, 8 people took 63 sec and 4 took 68 sec". In other words we imagine the data looks like this:

53, 53, 58, 58, 58, 58, 58, 58, 58, 63, 63, 63, 63, 63, 63, 63, 63, 68, 68, 68, 68

Then we add them all up and divide by 21. The quick way to do it is to multiply each midpoint by each frequency:

And then our estimate of the mean time to complete the race is:

Estimated Mean =   1288 21   = 61.333...

Very close to the exact answer we got earlier.

Estimating the Median from Grouped Data

Let's look at our data again:

The median is the middle value, which in our case is the 11 th one, which is in the 61 - 65 group:

We can say "the median group is 61 - 65"

But if we want an estimated Median value we need to look more closely at the 61 - 65 group.

We call it "61 - 65", but it really includes values from 60.5 up to (but not including) 65.5.

Why? Well, the values are in whole seconds, so a real time of 60.5 is measured as 61. Likewise 65.4 is measured as 65.

At 60.5 we already have 9 runners, and by the next boundary at 65.5 we have 17 runners. By drawing a straight line in between we can pick out where the median frequency of n/2 runners is:

And this handy formula does the calculation:

Estimated Median = L +   (n/2) − B G  × w

• L is the lower class boundary of the group containing the median
• n is the total number of values
• B is the cumulative frequency of the groups before the median group
• G is the frequency of the median group
• w is the group width

For our example:

• B = 2 + 7 = 9

Estimating the Mode from Grouped Data

Again, looking at our data:

We can easily find the modal group (the group with the highest frequency), which is 61 - 65

We can say "the modal group is 61 - 65"

But the actual Mode may not even be in that group! Or there may be more than one mode. Without the raw data we don't really know.

But, we can estimate the Mode using the following formula:

Estimated Mode = L +   f m − f m-1 (f m − f m-1 ) + (f m − f m+1 )  × w

• L is the lower class boundary of the modal group
• f m-1 is the frequency of the group before the modal group
• f m is the frequency of the modal group
• f m+1 is the frequency of the group after the modal group

In this example:

Our final result is:

• Estimated Mean: 61.333...
• Estimated Median: 61.4375
• Estimated Mode: 61.5

(Compare that with the true Mean, Median and Mode of 61.38..., 61 and 62 that we got at the very start.)

And that is how it is done.

Now let us look at two more examples, and get some more practice along the way!

Baby Carrots Example

Example: You grew fifty baby carrots using special soil. You dig them up and measure their lengths (to the nearest mm) and group the results :

Estimated Mean =   8530 50   = 170.6 mm

The Median is the mean of the 25 th and the 26 th length, so is in the 170 - 174 group:

• L = 169.5 (the lower class boundary of the 170 - 174 group)
• B = 5 + 2 + 6 + 8 = 21

The Modal group is the one with the highest frequency, which is 175 - 179 :

• L = 174.5 (the lower class boundary of the 175 - 179 group)

Age Example

Age is a special case.

When we say "Sarah is 17" she stays "17" up until her eighteenth birthday. She might be 17 years and 364 days old and still be called "17".

This changes the midpoints and class boundaries.

Example: The ages of the 112 people who live on a tropical island are grouped as follows:

A child in the first group 0 - 9 could be almost 10 years old. So the midpoint for this group is 5 not 4.5

The midpoints are 5, 15, 25, 35, 45, 55, 65, 75 and 85

Similarly, in the calculations of Median and Mode, we will use the class boundaries 0, 10, 20 etc

Estimated Mean =   3360 112   = 30

The Median is the mean of the ages of the 56 th and the 57 th people, so is in the 20 - 29 group:

• L = 20 (the lower class boundary of the class interval containing the median)
• B = 20 + 21 = 41

The Modal group is the one with the highest frequency, which is 20 - 29:

• L = 20 (the lower class boundary of the modal class)
• For grouped data, we cannot find the exact Mean, Median and Mode, we can only give estimates.

Estimated Mean = Sum of (Midpoint × Frequency) Sum of Frequency

• n is the total number of data

Mean of Grouped Data

Mean of grouped data is the data set formed by aggregating individual observations of a variable into different groups. Grouped data is data that is grouped together in different categories. Mean is considered as the average of the data. For the mean of grouped data, it might be difficult to find the exact value however, we can always estimate it. Let us learn more about the mean of grouped data, the methods to find the mean of grouped data, and solve a few examples to understand this concept better.

What is Mean of Grouped Data?

Mean of grouped data is the process of finding the average of a set of data that are grouped together in different categories. To determine the mean of a grouped data, a frequency table is required to set across the frequencies of the data which makes it simple to calculate. There are three main methods of calculating the mean of grouped data, they are - direct method, assumed mean method, and step deviation method. Each of these methods has its own formulas and ways to calculate the mean.

Definition of Mean

The mean is the average or a calculated central value of a set of numbers that is used to measure the central tendency of the data. Central tendency is the statistical measure that recognizes the entire set of data or distribution through a single value. In statistics, the mean can also be defined as the sum of all observations to the total number of observations. Given a data set, \( X = x_{1},x_{2}, . . . ,x_{n}\), the mean (or arithmetic mean, or average), denoted x̄, is the mean of the n values \(x_{1},x_{2}, . . . ,x_{n}\). The mean is represented as x-bar, x̄.

Mean of Grouped Data Formula

The mean formula is defined as the sum of the observations divided by the total number of observations. There are two different formulas for calculating the mean for ungrouped data and the mean for grouped data. Let us look at the formula to calculate the mean of grouped data. The formula is: x̄ = Σf\(_i\)/N Where,

• x̄ = the mean value of the set of given data.
• f = frequency of the individual data
• N = sum of frequencies

Hence, the average of all the data points is termed as mean.

Methods of Calculating Mean of Grouped Data

To calculate the mean of grouped data we have three different methods - direct method, assumed mean method, and step deviation method. The mean of grouped data deals with the frequencies of different observations or variables that are grouped together. Let us look at each of these methods separately.

Direct Method

The direct method is the simplest method to find the mean of the grouped data. If the values of the observations are x\(_1\), x\(_2\), x\(_3\),.....x\(_n\) with their corresponding frequencies are f\(_1\), f\(_2\), f\(_3\),.....f\(_n\) then the mean of the data is given by,

x̄ = x\(_1\)f\(_1\) + x\(_2\)f\(_2\) + x\(_3\)f\(_3\) +.....x\(_n\)f\(_n\) / f\(_1\) + f\(_2\) + f\(_3\) +.....f\(_n\)

x̄ = ∑x\(_i\)f\(_i\) / ∑f\(_i\), where i = 1, 2, 3, 4,......n

Here are the steps that can be followed to find the mean for grouped data using the direct method,

• Create a table containing four columns such as class interval, class marks (corresponding), denoted by x\(_i\), frequencies f\(_i\) (corresponding), and x\(_i\)f\(_i\).
• Calculate Mean by the Formula Mean = ∑x\(_i\)f\(_i\) / ∑f\(_i\). Where f\(_i\) is the frequency and x\(_i\) is the midpoint of the class interval.
• Calculate the midpoint, x\(_i\), we use this formula x\(_i\) = (upper class limit + lower class limit)/2.

Let us look at an example.

Example: Find the mean of the following data.

Solution: The first step is to create the table with the midpoint or marks and the product of the frequency and midpoint. To calculate the midpoint we find the average between the class interval by using the formula mentioned above.

Midpoint x\(_i\) = 0 - 10 = 5 ([10 + 0]/2), 10 - 20 = 15 ([20 + 10]/2) and so on.

x\(_i\)f\(_i\) = For the class interval 0 - 10 = 5 × 9 = 45, For the class interval 10 - 20 = 13 × 15 = 195 and so on.

Once we have determined the totals, let us use the formula to calculate the estimated mean.

Estimated Mean = ∑x\(_i\)f\(_i\) / ∑f\(_i\) = 1415/55 = 25.73.

Assumed Mean Method

The next method to calculate the estimated mean for a group of data is the assumed mean method. In the direct method, we know the observations and the corresponding frequencies. In the assumed mean method, let us assume that a is any assumed number which the deviation of the observation is d\(_i\) = x\(_i\) - a. Substituting this in the direct method formula we get,

x̄ = ∑[a + d\(_i\)]f\(_i\) / ∑f\(_i\)

x̄ = ∑af\(_i\) + ∑d\(_i\)]f\(_i\) / ∑f\(_i\)

x̄ = a∑f\(_i\) + ∑d\(_i\)]f\(_i\) / ∑f\(_i\)

x̄ = a + ∑d\(_i\)f\(_i\) / ∑f\(_i\)

Therefore, the assumed mean formula = a + ∑d\(_i\)]f\(_i\) / ∑f\(_i\), where d\(_i\) = x\(_i\) - a.

To calculate the mean of grouped data using the assumed mean method, here are the steps:

• Calculate the midpoint or x\(_i\) for the class interval as we did in the direct method.
• Take the central value from the class marks as the assumed mean and denote it as A.
• Calculate the deviation d\(_i\) = x\(_i\) - A for each i.
• Calculate the product of d\(_i\)f\(_i\) for each i.
• Find the total of f\(_i\)
• Calculate the mean by using the assumed mean method = a + ∑d\(_i\)f\(_i\) / ∑f\(_i\).

Let us look at an example to understand this better.

Solution: The first step is to create the table with the midpoint or marks and the product of the frequency and midpoint. To calculate the midpoint we find the average between the class intervals.

Midpoint x\(_i\) = 0 - 10 = 5 ([10 + 0]/2), 10 - 20 = 15 ([20 + 10]/2) and so on. From the midpoint let us select the assumed mean, so A = 25.

Find the deviation, d\(_i\) = x\(_i\) - A for each i. So, d\(_i\) = x\(_i\) - 25. Let find for each i, 2 - 25 = - 20 , 15 - 25 = -10 and so on.

f\(_i\) × d\(_i\) = - 20 × 12 = - 240 , - 10 × 15 = - 150 and so on.

Once we have determined the totals, let us use the assume mean formula to calculate the estimated mean.

Assumed Mean Formula = a + ∑d\(_i\)f\(_i\) / ∑f\(_i\)

= 25 + 260/82

= 25 + 3.170

Therefore, the assumed mean for the data is 28.17.

Step Deviation Method

The step deviation method is used when the deviations of the class marks from the assumed mean are large and they all have a common factor. We already know the direct method mean formula, let us derive the step deviation formula by using the direct method formula and the deviation process in the assumed mean method. Let us consider the class size to be h and the assumed mean to be A. Using the same formula used in assumed mean, d\(_i\) = x\(_i\) - A and calculating the value of u\(_i\) with the equation u\(_i\) = d\(_i\)/h, where h is the class width and d\(_i\) = x\(_i\) - A. Hence, the formula for step deviation method for grouped data is:

Step Deviation of Mean = a + h [∑u\(_i\)f\(_i\) / ∑f\(_i\)],

• a is the assumed mean
• h is the class size
• u\(_i\) = d\(_i\)/h

Midpoint x\(_i\) = 50 - 70 = 60 ([70 + 50]/2), 70 - 90 = 80 ([90 + 70]/2) and so on. From the midpoint let us select the assumed mean, so A = 100 and the value of h = 20 which is the class size.

Find the value, u\(_i\) = d\(_i\)/h, where d\(_i\) = x\(_i\) - A. Hence, we can write it as u\(_i\) = x\(_i\) - A / h. Let us find the value for each class interval. So, 50 - 70 = (60 - 100) / 20 = -2 , 70 - 90 = (80 - 100) / 20 = -1 and so on.

Step Deviation of Mean = a + h [∑u\(_i\)f\(_i\) / ∑f\(_i\)]

= 100 + 20 [65/100]

Therefore, the mean of the data is 113.

Related Topics

Listed below are a few interesting topics that are related to mean of grouped data, take a look.

• Mean, Median, Mode
• Categorical Data

Examples on Mean of Grouped Data

Example 1: There are 40 students in Grade 8. The marks obtained by the students in mathematics are tabulated below. Calculate the mean marks.

The total number of students in Grade 8 = 40

x\(_1\) = 100, x\(_2\) = 95, x\(_3\) = 88, x\(_4\) = 76, x\(_5\) = 69, f\(_1\) = 6, f\(_2\) = 8, f\(_3\) = 10, f\(_4\) = 9, f\(_5\) = 7

\(x_1f_1\) = 100 × 6 = 600 \(x_2f_2\) = 95 × 8 = 760 \(x_3f_3\) = 88 × 10 = 880 \(x_1f_1\) = 76 × 9 = 684 \( x_1f_1\) = 69 × 7 = 483

\(f_1x_1 + f_2x_2 + f_3x_3 + f_4x_4 + f_5x_5\) = 600 + 760 + 880 + 684 + 483 = 3,407

\(f_1 + f_2 + f_3 + f_4 + f_5\) = 6 + 8 + 10 + 9 + 7 = 40.

We will use the formula given below.

x̄ = Σf\(_i\)x\(_i\)/Σf\(_i\)

Mean marks = 3407/40 = 85.175

Therefore, the mean marks = 85.175.

Example 2: Find the mean percentage of the work completed for a project in a country where the assumed mean is 50, the class size is 20, frequency is 100, and the product of the frequency and deviation is - 42. Solve this by using the step-deviation method.

Solution: Given,

a = 50, h = 20, f\(_i\) = 100, f\(_i\)u\(_i\) = - 42

Using the step deviation method formula,

= 50 + 20 [-42/100]

= 50 - 42/5

Therefore, the mean percentage is 41.6.

Example 3: The marks obtained by 8 students in a class test are 12, 14, 16, 18, 20, 10, 11, and19. Use the mean formula and find out what is the mean of the marks obtained by the students?

To find: Mean of marks obtained by 8 students Marks obtained by 8 students in class test = 12, 14, 16, 18, 20, 10, 11, and19 (given) Total marks obtained by 8 students in class test = (12 + 14 + 16 + 18 + 20 + 10 + 11 + 19) = 120 Using the mean formula, Mean = (Sum of Observation) ÷ (Total numbers of Observations) = 120/8 = 15

Therefore, the mean of marks obtained by 8 students is 15.

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Practice Questions on Mean of Grouped Data

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FAQs on Mean of Grouped Data

Mean of grouped data is expressed as a data set formed by aggregating individual observations of a variable into different groups. To determine the mean of a grouped data, a frequency table is required to set across the frequencies of the data which makes it simple to calculate. There are three main methods of calculating the mean of grouped data, they are - direct method, assumed mean method, and step deviation method.

What is Mean Formula for Grouped Data?

The mean formula to find the mean of a grouped set of data can be given as, x̄ = Σfx/Σf, where, x̄ is the mean value of the set of given data, f is the frequency of each class and x is the mid-interval value of each class

What is the Mean Formula for Ungrouped Data?

The mean formula to find the mean for an ungrouped set of data can be given as, Mean = (Sum of Observations) ÷ (Total Numbers of Observations)

How Do You Find the Mean of Grouped Data From a Frequency Table?

While calculating mean of a grouped data, we always created a frequency table with the midpoint, derivation, and the product of the frequency and midpoint or frequency and derivation. This criteria depends on the type of method used. To calculate the mean from the frequency table we add all the numbers and then divide it by the numbers there are.

What are Different Types of Mean?

The different types of means in mathematics are,

• Arithmetic Mean
• Weighed Mean
• Geometric Mean
• Harmonic Mean

Statistics Made Easy

How to Find Mean & Standard Deviation of Grouped Data

Often we may want to calculate the mean and standard deviation of data that is grouped in some way. For example, suppose we have the following grouped data:

While it’s not possible to calculate the exact mean and standard deviation since we don’t know the raw data values , it is possible to estimate the mean and standard deviation.

The following steps explain how to do so.

Related: How to Find the Mode of Grouped Data

Calculate the Mean of Grouped Data

We can use the following formula to estimate the mean of grouped data:

Mean:  Σm i n i  / N

• m i :  The midpoint of the i th group
• n i :  The frequency of the i th group
• N:  The total sample size

Here’s how we would apply this formula to our dataset from earlier:

The mean of the dataset turns out to be  22.89 . 

Note: The midpoint for each group can be found by taking the average of the lower and upper value in the range. For example, the midpoint for the first group is calculated as: (1+10) / 2 = 5.5.

Calculate the Standard Deviation of Grouped Data

We can use the following formula to estimate the standard deviation of grouped data:

Standard Deviation: √ Σn i (m i -μ) 2 / (N-1)

• μ : The mean

Here’s how we would apply this formula to our dataset:

Additional Resources

How to Estimate the Mean and Median of Any Histogram How to Calculate Percentile Rank for Grouped Data How to Find the Median of Grouped Data

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3.3: Measures of Position

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• Page ID 10928

The common measures of location are quartiles and percentiles. Quartiles are special percentiles. The first quartile, Q 1 , is the same as the 25 th percentile, and the third quartile, Q 3 , is the same as the 75 th percentile. The median, M , is called both the second quartile and the 50 th percentile.

To calculate quartiles and percentiles, the data must be ordered from smallest to largest. Quartiles divide ordered data into quarters. Percentiles divide ordered data into hundredths. To score in the 90 th percentile of an exam does not mean, necessarily, that you received 90% on a test. It means that 90% of test scores are the same or less than your score and 10% of the test scores are the same or greater than your test score.

Percentiles are useful for comparing values. For this reason, universities and colleges use percentiles extensively. One instance in which colleges and universities use percentiles is when SAT results are used to determine a minimum testing score that will be used as an acceptance factor. For example, suppose Duke accepts SAT scores at or above the 75 th percentile. That translates into a score of at least 1220.

Percentiles are mostly used with very large populations. Therefore, if you were to say that 90% of the test scores are less (and not the same or less) than your score, it would be acceptable because removing one particular data value is not significant.

The median is a number that measures the "center" of the data. You can think of the median as the "middle value," but it does not actually have to be one of the observed values. It is a number that separates ordered data into halves. Half the values are the same number or smaller than the median, and half the values are the same number or larger. For example, consider the following data.

1; 11.5; 6; 7.2; 4; 8; 9; 10; 6.8; 8.3; 2; 2; 10; 1

Ordered from smallest to largest:

1; 1; 2; 2; 4; 6; 6.8; 7.2; 8; 8.3; 9; 10; 10; 11.5

Since there are 14 observations, the median is between the seventh value, 6.8, and the eighth value, 7.2. To find the median, add the two values together and divide by two.

The median is seven. Half of the values are smaller than seven and half of the values are larger than seven.

Quartiles are numbers that separate the data into quarters. Quartiles may or may not be part of the data. To find the quartiles, first find the median or second quartile. The first quartile, Q 1 , is the middle value of the lower half of the data, and the third quartile, Q 3 , is the middle value, or median, of the upper half of the data. To get the idea, consider the same data set:

The median or second quartile is seven. The lower half of the data are 1, 1, 2, 2, 4, 6, 6.8. The middle value of the lower half is two.

1; 1; 2; 2; 4; 6; 6.8

The number two, which is part of the data, is the first quartile . One-fourth of the entire sets of values are the same as or less than two and three-fourths of the values are more than two.

The upper half of the data is 7.2, 8, 8.3, 9, 10, 10, 11.5. The middle value of the upper half is nine.

The third quartile , Q 3, is nine. Three-fourths (75%) of the ordered data set are less than nine. One-fourth (25%) of the ordered data set are greater than nine. The third quartile is part of the data set in this example.

The interquartile range is a number that indicates the spread of the middle half or the middle 50% of the data. It is the difference between the third quartile ( Q 3 ) and the first quartile ( Q 1 ).

\[IQR = Q_3 – Q_1 \tag{2.4.1}\]

The IQR can help to determine potential outliers . A value is suspected to be a potential outlier if it is less than (1.5)( IQR ) below the first quartile or more than (1.5)( IQR ) above the third quartile . Potential outliers always require further investigation.

Definition: Outliers

A potential outlier is a data point that is significantly different from the other data points. These special data points may be errors or some kind of abnormality or they may be a key to understanding the data.

Example 2.4.1

For the following 13 real estate prices, calculate the IQR and determine if any prices are potential outliers. Prices are in dollars.

389,950; 230,500; 158,000; 479,000; 639,000; 114,950; 5,500,000; 387,000; 659,000; 529,000; 575,000; 488,800; 1,095,000

• Order the data from smallest to largest.

114,950; 158,000; 230,500; 387,000; 389,950; 479,000; 488,800; 529,000; 575,000; 639,000; 659,000; 1,095,000; 5,500,000

\[M = 488,800 \nonumber\]

\[Q_{1} = \dfrac{230,500 + 387,000}{2} = 308,750\nonumber\]

\[Q_{3} = \dfrac{639,000 + 659,000}{2} = 649,000\nonumber\]

\[IQR = 649,000 - 308,750 = 340,250\nonumber\]

\[(1.5)(IQR) = (1.5)(340,250) = 510,375\nonumber\]

\[Q_{1} - (1.5)(IQR) = 308,750 - 510,375 = –201,625\nonumber\]

\[Q_{3} + (1.5)(IQR) = 649,000 + 510,375 = 1,159,375\nonumber\]

No house price is less than –201,625. However, 5,500,000 is more than 1,159,375. Therefore, 5,500,000 is a potential outlier .

Exercise \(\PageIndex{1}\)

For the following 11 salaries, calculate the IQR and determine if any salaries are outliers. The salaries are in dollars.

$33,000; $64,500; $28,000; $54,000; $72,000; $68,500; $69,000; $42,000; $54,000; $120,000; $40,500

$28,000; $33,000; $40,500; $42,000; $54,000; $54,000; $64,500; $68,500; $69,000; $72,000; $120,000

Median = $54,000

\[Q_{1} = $40,500\nonumber\]

\[Q_{3} = $69,000\nonumber\]

\[IQR = $69,000 - $40,500 = $28,500\nonumber\]

\[(1.5)(IQR) = (1.5)($28,500) = $42,750\nonumber\]

\[Q_{1} - (1.5)(IQR) = $40,500 - $42,750 = -$2,250\nonumber\]

\[Q_{3} + (1.5)(IQR) = $69,000 + $42,750 = $111,750\nonumber\]

No salary is less than –$2,250. However, $120,000 is more than $11,750, so $120,000 is a potential outlier.

Example 2.4.2

For the two data sets in the test scores example , find the following:

• The interquartile range. Compare the two interquartile ranges.
• Any outliers in either set.

The five number summary for the day and night classes is

The IQR for the night group is \(Q_{3} - Q_{1} = 89 - 78 = 11\)

The interquartile range (the spread or variability) for the day class is larger than the night class IQR . This suggests more variation will be found in the day class’s class test scores.

• \(Q_{1} - IQR(1.5) = 56 – 26.5(1.5) = 16.25\)
• \(Q_{3} + IQR(1.5) = 82.5 + 26.5(1.5) = 122.25\)

Since the minimum and maximum values for the day class are greater than 16.25 and less than 122.25, there are no outliers.

Night class outliers are calculated as:

• \(Q_{1} - IQR (1.5) = 78 – 11(1.5) = 61.5\)
• \(Q_{3} + IQR(1.5) = 89 + 11(1.5) = 105.5\)

For this class, any test score less than 61.5 is an outlier. Therefore, the scores of 45 and 25.5 are outliers. Since no test score is greater than 105.5, there is no upper end outlier.

Exercise \(\PageIndex{2}\)

Find the interquartile range for the following two data sets and compare them.

Test Scores for Class A

69; 96; 81; 79; 65; 76; 83; 99; 89; 67; 90; 77; 85; 98; 66; 91; 77; 69; 80; 94

Test Scores for Class B

90; 72; 80; 92; 90; 97; 92; 75; 79; 68; 70; 80; 99; 95; 78; 73; 71; 68; 95; 100

65; 66; 67; 69; 69; 76; 77; 77; 79; 80; 81; 83; 85; 89; 90; 91; 94; 96; 98; 99

\(Median = \dfrac{80 + 81}{2}\) = 80.5

\(Q_{1} = \dfrac{69 + 76}{2} = 72.5\)

\(Q_{3} = \dfrac{90 + 91}{2} = 90.5\)

\(IQR = 90.5 - 72.5 = 18\)

68; 68; 70; 71; 72; 73; 75; 78; 79; 80; 80; 90; 90; 92; 92; 95; 95; 97; 99; 100

\(Median = \dfrac{80 + 80}{2} = 80\)

\(Q_{1} = \dfrac{72 + 73}{2} = 72.5\)

\(Q_{3} = \dfrac{92 + 95}{2} = 93.5\)

\(IQR = 93.5 - 72.5 = 21\)

The data for Class B has a larger IQR , so the scores between Q 3 and Q 1 (middle 50%) for the data for Class B are more spread out and not clustered about the median.

Example \(\PageIndex{3}\)

Fifty statistics students were asked how much sleep they get per school night (rounded to the nearest hour). The results were:

Find the 28 th percentile . Notice the 0.28 in the "cumulative relative frequency" column. Twenty-eight percent of 50 data values is 14 values. There are 14 values less than the 28 th percentile. They include the two 4s, the five 5s, and the seven 6s. The 28 th percentile is between the last six and the first seven. The 28 th percentile is 6.5.

Find the median . Look again at the "cumulative relative frequency" column and find 0.52. The median is the 50 th percentile or the second quartile. 50% of 50 is 25. There are 25 values less than the median. They include the two 4s, the five 5s, the seven 6s, and eleven of the 7s. The median or 50 th percentile is between the 25 th , or seven, and 26 th , or seven, values. The median is seven.

Find the third quartile . The third quartile is the same as the 75 th percentile. You can "eyeball" this answer. If you look at the "cumulative relative frequency" column, you find 0.52 and 0.80. When you have all the fours, fives, sixes and sevens, you have 52% of the data. When you include all the 8s, you have 80% of the data. The 75 th percentile, then, must be an eight . Another way to look at the problem is to find 75% of 50, which is 37.5, and round up to 38. The third quartile, Q 3 , is the 38 th value, which is an eight. You can check this answer by counting the values. (There are 37 values below the third quartile and 12 values above.)

Exercise \(\PageIndex{3}\)

Forty bus drivers were asked how many hours they spend each day running their routes (rounded to the nearest hour). Find the 65 th percentile.

The 65 th percentile is between the last three and the first four.

The 65 th percentile is 3.5.

Example 2.4.4

Using the table above in Example \(\PageIndex{3}\)

• Find the 80 th percentile.
• Find the 90 th percentile.
• Find the first quartile. What is another name for the first quartile?

Using the data from the frequency table, we have:

• The 80 th percentile is between the last eight and the first nine in the table (between the 40 th and 41 st values). Therefore, we need to take the mean of the 40 th an 41 st values. The 80 th percentile \(= \dfrac{8+9}{2} = 8.5\)
• The 90 th percentile will be the 45 th data value (location is \(0.90(50) = 45\)) and the 45 th data value is nine.
• Q 1 is also the 25 th percentile. The 25 th percentile location calculation: \(P_{25} = 0.25(50) = 12.5 \approx 13\) the 13 th data value. Thus, the 25 th percentile is six.

Exercise \(\PageIndex{4}\)

Refer to the table above in Exercise \(\PageIndex{3}\) . Find the third quartile. What is another name for the third quartile?

The third quartile is the 75 th percentile, which is four. The 65 th percentile is between three and four, and the 90 th percentile is between four and 5.75. The third quartile is between 65 and 90, so it must be four.

COLLABORATIVE STATISTICS

Your instructor or a member of the class will ask everyone in class how many sweaters they own. Answer the following questions:

• How many students were surveyed?
• What kind of sampling did you do?
• Construct two different histograms. For each, starting value = _____ ending value = ____.
• Find the median, first quartile, and third quartile.
• the 10 th percentile
• the 70 th percentile
• the percent of students who own less than four sweaters

A Formula for Finding the k th Percentile

If you were to do a little research, you would find several formulas for calculating the kth percentile. Here is one of them.

• \(k =\) the kth percentile. It may or may not be part of the data.
• \(i =\) the index (ranking or position of a data value)
• \(n =\) the total number of data

Calculate \(i = \dfrac{k}{100}(n + 1)\)

If \(i\) is an integer, then the \(k^{th}\) percentile is the data value in the \(i^{th}\) position in the ordered set of data.

If \(i\) is not an integer, then round \(i\) up and round \(i\) down to the nearest integers. Average the two data values in these two positions in the ordered data set. This is easier to understand in an example.

Example 2.4.5

Listed are 29 ages for Academy Award winning best actors in order from smallest to largest.

• Find the 70 th percentile.
• Find the 83 rd percentile.
• \(i\) = the index
• \(k\) = 83 rd percentile
• \(i \)= the index

Exercise \(\PageIndex{5}\)

18; 21; 22; 25; 26; 27; 29; 30; 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77

Calculate the 20 th percentile and the 55 th percentile.

\(k = 20\). Index \(= i = \dfrac{k}{100}(n+1) = \dfrac{20}{100}(29 + 1) = 6\). The age in the sixth position is 27. The 20 th percentile is 27 years.

\(k = 55\). Index \(= i = \dfrac{k}{100}(n+1) = \dfrac{55}{100}(29 + 1) = 16.5\). Round down to 16 and up to 17. The age in the 16 th position is 52 and the age in the 17 th position is 55. The average of 52 and 55 is 53.5. The 55 th percentile is 53.5 years.

You can calculate percentiles using calculators and computers. There are a variety of online calculators.

A Formula for Finding the Percentile of a Value in a Data Set

• \(x =\) the number of data values counting from the bottom of the data list up to but not including the data value for which you want to find the percentile.
• \(y =\) the number of data values equal to the data value for which you want to find the percentile.
• \(n =\) the total number of data.
• Calculate \(\dfrac{x + 0.5y}{n}(100)\). Then round to the nearest integer.

Example 2.4.6

• Find the percentile for 58.
• Find the percentile for 25.

\(x = 18\) and \(y = 1\). \(\dfrac{x + 0.5y}{n}(100) = \dfrac{18 + 0.5(1)}{29}(100) = 63.80\). 58 is the 64 th percentile.

\(x = 3\) and \(y = 1\). \(\dfrac{x + 0.5y}{n}(100) = \dfrac{3 + 0.5(1)}{29}(100) = 12.07\). Twenty-five is the 12 th percentile.

Exercise \(\PageIndex{6}\)

Listed are 30 ages for Academy Award winning best actors in order from smallest to largest.

18; 21; 22; 25; 26; 27; 29; 30; 31, 31; 33; 36; 37; 41; 42; 47; 52; 55; 57; 58; 62; 64; 67; 69; 71; 72; 73; 74; 76; 77

Find the percentiles for 47 and 31.

Percentile for 47: Counting from the bottom of the list, there are 15 data values less than 47. There is one value of 47.

\(x = 15\) and \(y = 1\). \(\dfrac{x + 0.5y}{n}(100) = \dfrac{15 + 0.5(1)}{30}(100) = 51.67\). 47 is the 52 nd percentile.

Percentile for 31: Counting from the bottom of the list, there are eight data values less than 31. There are two values of 31.

\(x = 8\) and \(y = 2\). \(\dfrac{x + 0.5y}{n}(100) = \dfrac{8 + 0.5(2)}{30}(100) = 30\). 31 is the 30 th percentile.

Interpreting Percentiles, Quartiles, and Median

A percentile indicates the relative standing of a data value when data are sorted into numerical order from smallest to largest. Percentages of data values are less than or equal to the p th percentile. For example, 15% of data values are less than or equal to the 15 th percentile.

• Low percentiles always correspond to lower data values.
• High percentiles always correspond to higher data values.

A percentile may or may not correspond to a value judgment about whether it is "good" or "bad." The interpretation of whether a certain percentile is "good" or "bad" depends on the context of the situation to which the data applies. In some situations, a low percentile would be considered "good;" in other contexts a high percentile might be considered "good". In many situations, there is no value judgment that applies.

Understanding how to interpret percentiles properly is important not only when describing data, but also when calculating probabilities in later chapters of this text.

When writing the interpretation of a percentile in the context of the given data, the sentence should contain the following information.

• information about the context of the situation being considered
• the data value (value of the variable) that represents the percentile
• the percent of individuals or items with data values below the percentile
• the percent of individuals or items with data values above the percentile.

Example 2.4.7

On a timed math test, the first quartile for time it took to finish the exam was 35 minutes. Interpret the first quartile in the context of this situation.

• Twenty-five percent of students finished the exam in 35 minutes or less.
• Seventy-five percent of students finished the exam in 35 minutes or more.
• A low percentile could be considered good, as finishing more quickly on a timed exam is desirable. (If you take too long, you might not be able to finish.)

Exercise \(\PageIndex{7}\)

For the 100-meter dash, the third quartile for times for finishing the race was 11.5 seconds. Interpret the third quartile in the context of the situation.

Twenty-five percent of runners finished the race in 11.5 seconds or more. Seventy-five percent of runners finished the race in 11.5 seconds or less. A lower percentile is good because finishing a race more quickly is desirable.

Example 2.4.8

On a 20 question math test, the 70 th percentile for number of correct answers was 16. Interpret the 70 th percentile in the context of this situation.

• Seventy percent of students answered 16 or fewer questions correctly.
• Thirty percent of students answered 16 or more questions correctly.
• A higher percentile could be considered good, as answering more questions correctly is desirable.

Exercise \(\PageIndex{8}\)

On a 60 point written assignment, the 80 th percentile for the number of points earned was 49. Interpret the 80 th percentile in the context of this situation.

Eighty percent of students earned 49 points or fewer. Twenty percent of students earned 49 or more points. A higher percentile is good because getting more points on an assignment is desirable.

Example 2.4.9

At a community college, it was found that the 30 th percentile of credit units that students are enrolled for is seven units. Interpret the 30 th percentile in the context of this situation.

• Thirty percent of students are enrolled in seven or fewer credit units.
• Seventy percent of students are enrolled in seven or more credit units.
• In this example, there is no "good" or "bad" value judgment associated with a higher or lower percentile. Students attend community college for varied reasons and needs, and their course load varies according to their needs.

Exercise \(\PageIndex{9}\)

During a season, the 40 th percentile for points scored per player in a game is eight. Interpret the 40 th percentile in the context of this situation.

Forty percent of players scored eight points or fewer. Sixty percent of players scored eight points or more. A higher percentile is good because getting more points in a basketball game is desirable.

Example 2.4.10

Sharpe Middle School is applying for a grant that will be used to add fitness equipment to the gym. The principal surveyed 15 anonymous students to determine how many minutes a day the students spend exercising. The results from the 15 anonymous students are shown.

0 minutes; 40 minutes; 60 minutes; 30 minutes; 60 minutes

10 minutes; 45 minutes; 30 minutes; 300 minutes; 90 minutes;

30 minutes; 120 minutes; 60 minutes; 0 minutes; 20 minutes

Determine the following five values.

If you were the principal, would you be justified in purchasing new fitness equipment? Since 75% of the students exercise for 60 minutes or less daily, and since the IQR is 40 minutes (60 – 20 = 40), we know that half of the students surveyed exercise between 20 minutes and 60 minutes daily. This seems a reasonable amount of time spent exercising, so the principal would be justified in purchasing the new equipment.

However, the principal needs to be careful. The value 300 appears to be a potential outlier.

\[Q_{3} + 1.5(IQR) = 60 + (1.5)(40) = 120\].

The value 300 is greater than 120 so it is a potential outlier. If we delete it and calculate the five values, we get the following values:

We still have 75% of the students exercising for 60 minutes or less daily and half of the students exercising between 20 and 60 minutes a day. However, 15 students is a small sample and the principal should survey more students to be sure of his survey results.

• Cauchon, Dennis, Paul Overberg. “Census data shows minorities now a majority of U.S. births.” USA Today, 2012. Available online at usatoday30.usatoday.com/news/...sus/55029100/1 (accessed April 3, 2013).
• Data from the United States Department of Commerce: United States Census Bureau. Available online at http://www.census.gov/ (accessed April 3, 2013).
• “1990 Census.” United States Department of Commerce: United States Census Bureau. Available online at http://www.census.gov/main/www/cen1990.html (accessed April 3, 2013).
• Data from San Jose Mercury News .
• Data from Time Magazine ; survey by Yankelovich Partners, Inc.

The values that divide a rank-ordered set of data into 100 equal parts are called percentiles. Percentiles are used to compare and interpret data. For example, an observation at the 50 th percentile would be greater than 50 percent of the other obeservations in the set. Quartiles divide data into quarters. The first quartile ( Q 1 ) is the 25 th percentile,the second quartile ( Q 2 or median) is 50 th percentile, and the third quartile ( Q 3 ) is the the 75 th percentile. The interquartile range, or IQR , is the range of the middle 50 percent of the data values. The IQR is found by subtracting Q 1 from Q 3 , and can help determine outliers by using the following two expressions.

• \(Q_{3} + IQR(1.5)\)
• \(Q_{1} - IQR(1.5)\)

Formula Review

\[i = \dfrac{k}{100}(n+1) \nonumber\]

where \(i\) = the ranking or position of a data value,

• \(k\) = the k th percentile,
• \(n\) = total number of data.

Expression for finding the percentile of a data value: \(\left(\dfrac{x + 0.5y}{n}\right)(100)\)

where \(x =\) the number of values counting from the bottom of the data list up to but not including the data value for which you want to find the percentile,

\(y =\) the number of data values equal to the data value for which you want to find the percentile,

\(n =\) total number of data

Excel Tutorial: How To Calculate Mode For Grouped Data In Excel

Introduction.

Welcome to our Excel tutorial on how to calculate mode for grouped data in Excel. Understanding mode is crucial for data analysis as it helps identify the most frequently occurring value or values in a dataset. This knowledge is vital for making informed decisions and drawing accurate conclusions from your data.

Key Takeaways

• Understanding mode is crucial for identifying the most frequently occurring values in a dataset
• Calculating mode for grouped data in Excel helps in making informed decisions and drawing accurate conclusions
• The mode for grouped data is calculated using a specific formula, which involves organizing the data into a frequency distribution table
• Mode assists in understanding the central tendency of the data and can be used in real-life scenarios for decision making and problem-solving
• Effective data analysis using mode in Excel involves utilizing functions for sorting and organizing data, and cross-verifying the calculated mode with other measures of central tendency for validation

Understanding Mode in Statistics

In the field of statistics, mode is a measure of central tendency that represents the most frequently occurring value in a data set. It is a valuable tool for understanding the distribution and characteristics of a particular data set.

Define what mode is in statistics

The mode of a data set is the value that appears most frequently. In a set of numbers, the mode is the number that occurs the highest number of times. It is a useful measure of central tendency, along with mean and median, and provides insight into the common or typical values in a data set.

Explain the difference between mode for individual data and mode for grouped data

When dealing with individual data, calculating the mode is straightforward - it is simply the number that appears most frequently in the data set. However, when working with grouped data, where the values are organized into intervals or classes, the process of calculating the mode becomes slightly more complex. In this scenario, the mode represents the interval with the highest frequency, rather than a specific value.

Steps for Calculating Mode for Grouped Data in Excel

When dealing with grouped data in Excel, calculating the mode can be a bit more complex. Follow these steps to efficiently calculate the mode for grouped data in Excel.

A. Organize the grouped data into a frequency distribution table

• B egin by creating a frequency distribution table to organize the grouped data. This table will list the class intervals and their corresponding frequencies.

B. Identify the class interval with the highest frequency

• I dentify the class interval with the highest frequency. This class interval will represent the modal class.

C. Calculate the mode using the formula: Mode = L + (f1 - f0) / (2f1 - f0 - f2) * w

• C alculate the mode using the formula where L represents the lower boundary of the modal class, f0 is the frequency of the class interval before the modal class, f1 is the frequency of the modal class, f2 is the frequency of the class interval after the modal class, and w is the class width.

D. Discuss the steps to input the formula in Excel for efficient calculation

• D iscuss the steps to input the mode calculation formula in Excel. This may involve using cell references to input the necessary values from the frequency distribution table and applying the formula to calculate the mode efficiently.

Examples of Calculating Mode for Grouped Data in Excel

Calculating mode for grouped data in Excel can be a useful skill for anyone working with statistical data. Let's take a look at some examples to understand how to do this effectively.

A. Provide a sample frequency distribution table

First, we need to have a sample frequency distribution table to work with. This table should include the categories or intervals of the data, as well as the frequency of each category.

• Category Frequency

B. Demonstrate the calculation of mode using the given data

Once we have our frequency distribution table, we can proceed to calculate the mode. In Excel, we can use the MODE.SNGL function to find the mode for grouped data. This function returns the most frequently occurring value in a data set.

For the given frequency distribution table, we would input the category ranges and their corresponding frequencies into an Excel worksheet. Then, we would use the MODE.SNGL function to calculate the mode for the grouped data.

C. Discuss any potential challenges or common mistakes to avoid

One potential challenge when calculating mode for grouped data in Excel is ensuring that the data is properly grouped and the frequencies are accurately represented. It's important to double-check the input data to avoid any errors in the calculation.

Another common mistake to avoid is misinterpreting the results. The mode for grouped data represents the category with the highest frequency, not the individual data point. It's important to understand the distinction and communicate the findings accurately.

Advantages of Calculating Mode for Grouped Data

Calculating the mode for grouped data in Excel can provide valuable insights into the central tendency of the data and can be used in various real-life scenarios for decision making and problem-solving.

When dealing with grouped data, calculating the mode can help in understanding the most frequently occurring value or class interval. This can give a clear indication of the central tendency of the data and can be useful in identifying the most common value or range within the dataset.

Understanding the mode for grouped data can be beneficial in various real-life scenarios. For example, in business and finance, it can be used to identify the most common sales figure, customer preference, or market trend. In healthcare, it can help in identifying the most common age group for a certain illness or condition. In education, it can be used to identify the most common test scores or grade distribution. Overall, it can aid in making informed decisions and solving problems based on the most prevalent values within the data.

Tips for Effective Data Analysis Using Mode in Excel

When working with grouped data in Excel, calculating the mode can provide valuable insights into the most frequent value or category. Here are some tips for effectively analyzing data using the mode in Excel:

Use the SORT function:

Utilize the filter function:, calculate the mean and median:, use data visualization tools:.

Calculating the mode for grouped data in Excel is important for identifying the most frequently occurring value in a dataset, especially when the data is grouped into categories or ranges. This can provide valuable insights for decision-making, trend analysis, and forecasting.

As you continue to enhance your data analysis skills, I encourage you to practice the steps outlined in this tutorial and apply mode calculation in your own data analysis endeavors. By doing so, you will be better equipped to make informed decisions and draw meaningful conclusions from your data.

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Median of Grouped Data

Median of Grouped Data is the value of the middlemost data point in any dataset when dataset is grouped. For ungrouped data, it is easy to find the median but finding the Median of Grouped Data is slightly complex. When we have any data in statistics , we try to find some basic parameters related to it which provide ease in data interpretation and making further predictions related to data. To measure the central tendency of data , i.e., a single value that can be used to represent an entire distribution, we generally have three parameters: mean, median, and mode.

In this article, we will discuss what is meant by Grouped Data and Median of Grouped Data. We will also learn about the Formula for Median of Grouped Data and steps to calculate the Median of Grouped Data, solved examples, and some frequently asked questions as well.

Table of Content

What is Grouped Data?

What is median of grouped data, median of grouped data formula, how to calculate median of grouped data, mean, median and mode of grouped data.

Data can be classified mainly into two types on basis of how it is organized, i.e. Grouped Data and Ungrouped Data. Ungrouped data is the raw data which consists of a simple list of values where each value corresponds to a distinct observation or measurement.

For example, a list of marks of students in a classroom (e.g. 95, 94, 96, 92, 98, 99, …). This kind of representation is used when we have a smaller dataset and we need to deal with individual data points.

But, when we have a larger dataset, it is preferable to group similar data values in form of an interval and assign a value to that interval corresponding to the frequency of data points in that range. The interval should have a uniform length defined by its upper limits and lower limits.

The interval so formed is also referred to as class or class interval, and the table containing classes (intervals) and their frequencies is known as the frequency distribution table. For example, a frequency distribution table showing marks of students in a class in various ranges,

Grouped data is useful when we have large datasets and when our analysis is concerned to study specific patterns and relationships with respect to ranges of values rather than individual data points.

The genric meaning of median, i.e. the middle value corresponding to a given distribution, remains same in this case too. As we have data in form of intervals (classes) in this case, we have a corresponding median class to find the value of median.

Also, we need to define cumulative frequencies for each class, which is a kind of prefix sum of frequencies of classes taken in order. The median value lies between the lower limit and upper limit of the median class. This value can be used by using a specified formula discussed as follows.

What is Median?

Median is a value corresponding to the middlemost data point in a dataset, when arranged in ascending order. The value of median helps one to know about center of a dataset. On comparing the value of median with that of mean, one can get idea of distribution of values in a dataset.

To find median of ungrouped data, one can simply sort the data points in ascending order. In case of odd number of observations, the middle value would be the median. On the other hand , for even number of observations, one can take mean of the two middle values to find the median. But there is a different method to find median of grouped data discussed later in this article.

We can use the following formula to calculate median of grouped data:

Median = l + ((n/2-cf)/f)×h Where, l is the lower limit of the median class, n is the total number of observations, cf is the cumulative frequency of the class preceding median class, f is the frequency of the median class, and h is the class size (upper limit – lower limit).

The steps listed below illustrate the procedure to find median of grouped data.

The steps followed to calculate median of grouped data are discussed as follows:

Step 1: First, we find out the total number of observations by summing up all the frequencies. Step 2: Then, we need to find the median class, i.e. the class having cumulative frequency just greater than half of total number of observations. Step 3: Now, we note the values of lower limit of median class (l), frequency of the median class (f), cumulative frequency of the class preceding median class (cf), and class size (h). Step 4: Next, we can substitute these values in the formula to calculate median of grouped data, i.e. Median = l + ((n/2-cf)/f)×h

Below are some examples that will help you understand above mentioned steps to find the Median of Grouped Data in a better way.

A comparison between Mean, Median and Mode of Grouped Data has been discussed in the table below:

Mean, Median and Mode of Grouped Data Mean of Grouped Data Frequency Distribution

Solved Examples on Median of Grouped Data

Example 1: calculate the value of median for the following data distribution:.

To find the median of given data, we build a table containing cumulative frequencies for each class interval along with the frequencies.

Here, the total number of observations are 40, i.e. n = 40. We have, n/2 = 20, now the class having cumulative frequency just greater than or equal to 20 is the class interval 20-30 (cf = 24).

Thus, the median class is 20-30 . Also, here the value of class size (h) is 10 (upper limit – lower limit). The lower limit (l) and frequency (f) of the median class are 20 and 12 respectively. And, the cumulative frequency (cf) of class preceding the median class is 12. Now, we can substitute these values in the formula to calculate value of median,

Median = l + ((n/2-cf)/f)×h = 20 + ((20-12)/12)×10 = 20 + (8/12)×10 = 20 + 6.67 Median = 26.67

Thus, the value of median corresponding to the given grouped data comes out to be 26.67 .

Example 2: Find the median age of employees working at XYZ organisation, based on the following data:

Solution: To find median of the given grouped data, first of all we form a frequency distribution table as follows:

Here, we have total number of employees, n = 40. So, the median class is the class having cumulative frequency just greater than or equal to 20 (i.e. n/2). Thus, median class is 35-40 .

Now, we have,

Lower limit of median class, l = 35. Class size, h = 5. Cumulative frequency of the class preceding the median class, cf = 20 Frequency of median class, f = 10 On substituitng these values in the formula, i.e. Median = l + ((n/2-cf)/f)×h we get, Median = 35 + ((20-20)/10)×5 Median = 35 Thus, median age of employees based upon given distribution comes out to be 35 years.

Example 3: Find the median score of a cricket team in past 20 matches based on the following data:

Solution: Let us create a frequency distribution table for the given data,

Here, total number of observations (n) are 20. Now, the class having cumulative frequency just greater than or equal to n/2, i.e. 10, is the class 120-140. Thus, it is the median class for the given distribution. Lower limit of the median class, l = 120, Frequency of the median class, f = 4, Cumulative frequency of the class preceding median class, cf = 9, Class size (upper limit – lower limit), h = 20, On substituting these values in formula to find median of grouped data, i.e. Median = l + ((n/2-cf)/f)×h we get, Median = 120 + ((10-9)/4)×20 Median = 120 + 5 = 125 Thus, the median score of team comes out to be 125.

Practice Problems on Median of Grouped Data

Q1. find the value of median for following grouped data distribution:, q2. find the median salary of employees working at an abc organisation:.

Q3. Find the median height of students in a class based upon following data:

Median of Grouped Data – FAQs

1. what do you mean by median in statistics.

Median refers to the middle value of the given dataset when arranged in ascending order.

2. What is difference between Grouped Data and Ungrouped Data?

Ungrouped data is the data presented in form of discrete individual points. Each data point corresponds to a single observation in this case. In grouped data, we represent the data in form of ranges or intervals, and the observations having values corresponding to that range are counted against them named as frequency of that interval.

3. What is Median Class in Grouped Data?

Median class is the class having cumulative frequency just greater than or equal to half of the total number of observations.

4. Why is Median is also called as Positional Average in Statistics?

Median is the middle value of the given data distribution when data points are arranged in ascending order. As it depends upon the position of data values when arranged in a specific order, so it is also called as positional average.

5. What is the formula to Calculate Median of Grouped Data in statistics?

We use the following formula to calculate median of grouped data in statistics, Median = l + ((n/2-cf)/f)×h Where, l is the lower limit of the median class, n is the total number of observations, cf is the cumulative frequency of the class preceding median class, f is the frequency of the median class, and h is the class size (upper limit – lower limit).

6. What are the steps involved in finding Median of Grouped Data?

Below are the steps involved in finding median of a grouped data: Arrange data in groups or classes. Find the midpoint of the data in each group. Calculate the cumulative frequency. Determine the group containing the median. Apply the formula  Median = l + ((n/2-cf)/f)×h Calculate the median using the formula.

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MIXED QUESTIONS ON MEAN MEDIAN AND MODE FOR UNGROUPED DATA

Question 1 :

The monthly salary (in $) of 10 employees in a factory are given below :

5000, 7000, 5000, 7000, 8000, 7000, 7000, 8000, 7000, 5000

Find the mean, median and mode.

Solution : 

= ( 5000 +  7000 +    5000 +   7000 +   8000 +   7000 +  7000 +  8000 +  7000 +   5000)/10

  =  66000/10

  =  6600

5000, 5000, 5000,  7000,  7000,  7000,  7000, 7000,   8000,   8000

Number of observations  =  10 (Even)

Median  =  {(10/2) th observation + [(10/2) + 1] th }/2

  =  (5 th observation + 6 th observation)/2

  =  (7000 + 7000)/2

  =  14000/2

  =  7000

7000 is repeating 5 times. Hence mode is 7000.

Question 2 :

Find the mode of the given data : 3.1, 3.2, 3.3, 2.1, 1.3, 3.3, 3.1

3.1 and 3.3 are repeating twice, so mode is 3.1 and 3.3

It is a bimodal data.

Question 3 :

For the data 11, 15, 17, x+1, 19, x–2, 3 if the mean is 14 , find the value of x. Also find the mode of the data.

Mean  =  (11 + 15 + 17 + x + 1 + 19 + x - 2 + 3)/7

14  =  (64 + 2x)/7

14(7)  =  64 + 2x

2x  =  98 - 64

2x  =  34

x  =  34/2  =  17

By applying the value of x in the given observation, we get 

11, 15, 17, 18, 19, 15, 3

Mode  =  15 (Repeating twice)

Question 4 :

The demand of track suit of different sizes as obtained by a survey is given below:

Demand for the size 40 is 37.

Hence the demand of size 40 is high.

Steps in Finding the Mode of Grouped Data

In case of a grouped frequency distribution, the exact values of the variables are not known and as such it is very difficult to locate mode accurately

The class interval with maximum frequency is called the modal class.

Where l - lower limit of the modal class;

f - frequency of the modal class

f1 - frequency of the class just preceding the modal class

f2 - frequency of the class succeeding the modal class

c - width of the class interval

Question 5 :

Find the mode of the following data:

The highest frequency is 46

modal class is 20 - 30

l = 20, f = 46, f1 = 38, f2 = 34, c = 10

  =  20 + [(46-38)/2(46) - 38 - 34] x 10

  =  20 + [8/(92 - 38 - 34)] x 10

  =  20 + [8/20] x 10

  =  20 + 4

  =  24

Hence the mode is 24.

Question 6 :

Find the mode of the following distribution:

The highest frequency is 14

modal class is 54.5 - 64.5

l = 54.5, f = 14, f1 = 10, f2 = 8, c = 10

  =  54.5 + [(14 - 10)/2(14) - 10 - 8] x 10

  =  54.5 + [4/(28 - 18)] x 10

  =  54.5 + [4/10] x 10

  =  54.5 + 4

  =  58.5

Hence the mode is 58.5.

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• Open access
• Published: 07 April 2024

Efficacy of a virtual nursing simulation-based education to provide psychological support for patients affected by infectious disease disasters: a randomized controlled trial

• Eunjung Ko 1 &
• Yun-Jung Choi 1  

BMC Nursing volume  23 , Article number:  230 ( 2024 ) Cite this article

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Virtual simulation-based education for healthcare professionals has emerged as a strategy for dealing with infectious disease disasters, particularly when training at clinical sites is restricted due to the risk of infection and a lack of personal protective equipment. This research evaluated a virtual simulation-based education program intended to increase nurses’ perceived competence in providing psychological support to patients affected by infectious disease disasters.

The efficacy of the program was evaluated via a randomized controlled trial. We recruited 104 nurses for participation in the study and allocated them randomly and evenly to an experimental group and a control group. The experimental group was given a web address through which they could access the program, whereas the control group was provided with a web address that directed them to text-based education materials. Data were then collected through an online survey of competence in addressing disaster mental health, after which the data were analyzed using the Statistical Package for the Social Sciences(version 23.0).

The analysis showed that the experimental group’s disaster mental health competence (F = 5.149, p  =.026), problem solving process (t = 3.024, p  =.003), self-leadership (t = 2.063, p  =.042), learning self-efficacy (t = 3.450, p  =.001), and transfer motivation (t = 2.095, p  =.039) significantly statistically differed from those of the control group.

Conclusions

A virtual nursing simulation-based education program for psychological support can overcome limitations of time and space. The program would also be an effective learning resource during infectious disease outbreaks.

Clinical trial registration

This Korean clinical trial was retrospectively registered (21/11/2023) in the Clinical Research Information Service ( https://cris.nih.go.kr ) with trial registration number KCT0008965.

Peer Review reports

The last two decades have confronted the world with a variety of infectious diseases, such as severe acute respiratory syndrome, which first occurred in Asia in 2003 before spreading worldwide, including Korea, in only a few months. Since then, infectious disease outbreaks began to be recognized as severe disasters. Other examples include the 2009 H1N1 influenza outbreak, which caused more than 10,000 deaths worldwide and 140 deaths in Korea; the proliferation of the Ebola virus, which resulted in a fatality rate of more than 90% in Africa in 2014; and the outbreak of Middle East respiratory syndrome in 2015, Zika virus disease in 2016, and coronavirus disease (COVID-19) in 2019 [ 1 ]. The COVID-19 pandemic, in particular, has caused infections among approximately 64 million people and the deaths of 1.5 million individuals as of December 2020 [ 2 ].

Direct victims of infectious disease disasters, infected patients, and quarantined individuals suffer from a fear of stigma or social blame and guilt, but even people who are unexposed to sources of infection experience psychological distress from anxiety and fear of disease or possible death [ 3 ]. They also blame infected people and harbor hatred toward them [ 3 ]. This assertion is supported by an examination of web search behaviors and infodemic attitudes toward COVID-19, which identified superficial and racist attitudes [ 4 ]. Additionally, in research using a health stigma and discrimination framework related to communicable diseases, the authors found that people exhibit negative stereotypes, biases, and discriminatory conduct toward infected groups owing to fears of contagion, concerns about potential harm, and perceptions that individuals violate central values [ 5 ]. Stigmatized individuals experience adverse effects on their health because of both the stress induced by stigma and the decreased use of available services [ 5 ].

Severe and prolonged anxiety, fear, blame, and aggression can lead to mental health problems, including depression, anxiety, panic attacks, somatic symptoms, post-traumatic stress disorder, psychosis, and even suicide and life-threatening behaviors [ 6 ]. Therefore, recovery from the psychological trauma caused by a disaster should be regarded as equally necessary as physical recovery, with emphasis placed on psychological support activities that prevent the deterioration of mental health [ 7 ].

Disasters pose a significant threat to mental health support systems, wherein the lack of healthcare professionals or psychologists trained to address these conditions exacerbates the psychological distress and psychopathological risk experienced by society [ 8 ]. When training at clinical sites is restricted due to infection risks and a lack of personal protective equipment (PPE), an emerging solution is virtual simulation [ 9 ].

A virtual simulation is a simulation modality developed on the basis of video or graphic recordings featuring virtual patients and delivered via either a static or mobile device. It replicates real-world clinical situations and affords learners an interactive experience [ 10 ]. Virtual simulation-based education provides an immersive clinical environment, as virtual patients respond to a learner’s assessments and interventions [ 11 , 12 ]. It enables two-way communication, and allows medical professionals to practice making clinical decisions [ 10 ]. Virtual patients are equipped with voice, intonation, and expressions that reinforce the educational narrative within the virtual environment, thereby enhancing the effectiveness of the learning experience [ 13 ]. One of the primary advantages of virtual simulation-based education is its provision of a safe and non-threatening environment in which learners can practice. It also offers flexible and reproducible learning experiences, thus catering to the diverse needs of learners [ 14 ].

Self-assessment is the most commonly used competence evaluation tool, as it is cost-effective and helps nurses improve their practice by identifying their strengths and weaknesses for development [ 15 ]. Self-assessed competence is also related to the quality of patient care because nurses promote continuous learning by determining educational needs through such evaluations [ 16 ]. The competence perceived by a nurse is inherently subjective given its self-reported nature and poses a challenge in establishing a direct correlation with the actual care of patients [ 17 , 18 ]. However, studies have indicated that increased levels of self-perceived competence are associated with a significant increase in core competencies related to patient care and frequent use of clinical skills [ 19 , 20 ]. Perceived competence likewise influences the job satisfaction and organizational citizenship behavior of nurses and is significantly related to absenteeism, one of the deterrents to the delivery of quality care [ 21 , 22 ].

Competence refers to the possession of qualifications and abilities to satisfy professional standards, as well as the capability to perform tasks and duties in a suitable and effective manner [ 23 ]. Competencies for disaster mental health are crucial for enhancing disaster response capabilities. These competencies encompass a range of skills, knowledge, and attitudes necessary for mental health professionals to effectively support individuals and communities affected by disasters [ 24 ]. Such competencies and how they are affected by simulation-based training have been explored in some studies, which reported a significant increase in competence after exposure to the aforementioned education [ 25 , 26 ].

The simulation education defined in mock training designs based on real situations provides opportunities to exercise problem-solving through various strategies. Problem-solving process is considered key competency through which learners are expected to enhance their relevant knowledge and clinical performance abilities [ 27 ]. In particular, problem-solving processes for identifying and assessing problems and finding solutions are psychological strategies that help people cope and recover after a disaster [ 28 ]. A scoping review on the effect of simulation-based education on the problem-solving process indicated that out of 32 studies reviewed, 21 demonstrated statistically significant improvement in people’s ability to resolve problems [ 29 ].

Simulation training can also address self-leadership, which is an essential self-learning quality that aids individuals in staying motivated and focused on their learning goals. It is also required as a basic qualification of professional nurses, who must be able to take initiative and make responsible decisions [ 30 , 31 ]. Previous studies have reported statistically significant improvements in self-leadership following simulation training [ 32 , 33 ].

Another aspect that benefits from simulation-driven education is learning self-efficacy, which plays a crucial role in predicting learners’ levels of engagement and academic success in online education. It reflects learners’ confidence in their ability to manage their own learning process. It is a significant predictor of both learners’ participation levels and their academic achievements in online education settings [ 34 , 35 ]. Several studies have demonstrated virtual simulation- or online education-induced significant improvements in learning self-efficacy [ 36 , 37 ]. Finally, virtual simulation-based education can also improve the motivation to transfer new knowledge and skills learned through education to clinical practice [ 38 ]. This motivation is considered an essential measure of effective learning for nurses working in the clinical field [ 38 ]. A previous study reported that psychiatric nursing simulation training combined with post-course debriefing significantly increases participants’ level of motivation to transfer [ 38 ].

On the basis of the discussion above, this study evaluated a virtual nursing simulation-based education program on disaster psychology designed to provide psychological support to patients affected by infectious disease disasters.

Study design

This study conducted a randomized controlled trial (RCT) to test the virtual nursing simulation-based education program of interest. The RCT protocol used was based on CONSORT guidelines.

Participants

We recruited nurses working at general hospitals in South Korea. With permission from the nurse managers of these hospitals, a participation notice was posted on the institutions’ internet bulletin boards for nurses for a week. The two-sided test criterion, with a significance level (α) of 0.05, a power (1-β) of 0.80, and a medium effect size of 0.6, dictates that the minimum number of participants per group be 90. The effect size was based on a virtual simulation intervention study conducted by Kim and Choi [ 36 ]. Taking the dropout rate into consideration, we recruited 104 nurses, who were assigned to an experimental group and a control group using the random sampling functionality of the Statistical Package for the Social Sciences (SPSS version 23.0). Out of the initial sample, 11 participants were excluded because they were on vacation, could not be contacted, or provided incomplete responses during data collection (Fig.  1 ).

Flowchart of the randomized controlled trial

The virtual nursing simulation-based education program

This study probed into the virtual nursing simulation-based education program developed by Ko [ 39 ]. The program is implemented using an e-learning development platform, Articulate Storyline, whose operating environment is compatible with all web browsers (Internet Explorer, Microsoft Edge, Firefox, Google Chrome, etc.). It is a mobile-friendly application that can run in devices with Android and iOS operating systems. When an individual uses their smartphone or personal computer to access the server via the web address corresponding to the education program, the content functions execute. Ko’s [ 39 ] program involves five stages of learning completed in 100 min: (1) preparatory learning (30 min), (2) pre-test (5 min), (3) pre-briefing (5 min), (4) simulation game (30 min), and (5) structured self-debriefing (30 min) (Fig.  2 ).

Preparatory learning comes with lecture materials on guidelines for providing psychological support to victims of infectious disease disasters, administering psychological first aid, donning and doffing PPE, and exercising mindfulness through videos and pictures. In the pretest stage, a learner answers five questions and can immediately check the correct responses, which come with detailed explanations. In the prebriefing stage, an overview of a nursing simulation scenario, patient information, learning objectives, and instructions on using the virtual simulation are provided. During the simulation game, a video of the simulation is presented. It starts with a 39-year-old female, a standardized patient who is age- and gender-matched to the scenario, confirmed to have contracted COVID-19 and transferred to a negative pressure isolation room. The patient presents with extreme anxiety and feeling of tightness in her chest. During the game, learners are expected to complete 12 quizzes. In the debriefing stage, a summary of the simulation quiz results and self-debriefing questions are provided, and the comments made by learners are saved in the Naver cloud platform.

The evaluated virtual nursing simulation-based education program (examples are our own work)

Measurements

Disaster mental health competence.

Disaster mental health competence was measured using the perceived competence scale for disaster mental health workforce (PCS-DMHW), which was developed by Yoon and Choi [ 40 ]. This tool consists of 24 questions related to knowledge (6 questions), attitudes (9 questions), and skills (9 questions). Each item is rated using a five-point Likert scale (0 = strongly disagree, 4 = strongly agree), and the responses are summed. The higher the score, the greater the perception of competence in a relevant area [ 40 ]. The Cronbach’s α values of the PCS-DMHW were 0.95 and 0.94 at the time of tool development and the present study, respectively.

Problem solving process

Problem solving process was determined using a tool modified and supplemented by Park and Woo [ 41 ] on the grounds of the problem solving process and behavior survey developed by Lee [ 42 ]. This tool is composed of 25 questions on five factors, namely, problem discovery, problem definition, problem solution design, problem solution execution, and problem solving review [ 41 ]. The reliability of the tool was 0.89 at the time of development [ 41 ], but the Cronbach’s α found in the current research was 0.94.

Self-leadership

Self-leadership was measured using a tool developed by Manz [ 43 ] and modified by Kim [ 44 ]. The tool consists of 18 questions distributed over six factors (three questions each): self-defense, rehearsal, goal setting, self-compensation, self-expense edition, and constructive thinking. The reliability of the tool at the time of development and the present research was (Cronbach’s α) 0.87 and 0.82, respectively.

Learning self-efficacy

To ascertain learning self-efficacy, we used the tool developed by Ayres [45] and translated by Park and Kweon [ 38 ]. This tool consists of 10 questions, and it had a reliability (Cronbach’s ⍺) of 0.94 and 0.93 at the time of development and the current study, respectively.

Motivation to transfer

We used Ayres’s [45] motivation to transfer scale, which was translated by Park and Kweon [ 38 ]. Its reliability at the time of development and the present research was (Cronbach’s ⍺) 0.80 and 0.93, respectively.

Data collection

The experimental and control groups were administered a pretest through an online survey. The web address through which the evaluated virtual simulation-based education program could be accessed was provided to the experimental group, whereas text-based education materials on psychological support for victims of infectious disease disasters were given to the control group. The groups were simultaneously sent the program’s instruction manual, and their inquiries were answered through chat. After the interventions, each participant was administered a posttest through another online survey.

Data analysis

The collected data were analyzed using SPSS version 23.0. The homogeneity test for general characteristics between the experimental and control groups was analyzed using a t-test, a chi-square test, and Fisher’s exact test. The normality of the dependent variables was analyzed using the Kolmogorov-Smirnov test. Changes in the dependent variables between the pretest and posttest were analyzed using a paired t-test. Differences in the dependent variables before and after the groups’ use of the interventions were examined via a t-test and ANCOVA.

Ethical considerations

We completed education in bioethics law prior to the research and obtained approval of the research proposal and questionnaire from the Institutional Review Board of the affiliated university (IRB approval number 1041078-202003-HRSB-070-01CC). A signed consent form was also obtained from each participant after the purpose and methods of the research, the confidentiality of personal information, and the voluntary nature of participation or their right to withdraw from the study were explained to them. All collected data were kept in a lockable cabinet, and electronic data were encrypted and stored. These data are to be discarded after three years.

A total of 93 participants (45 in the experimental group and 48 in the control group) were left after the exclusion of unsuitable respondents. of the between-group comparisons of the subjects indicated no significant differences between them (5% significance level) in terms of general characteristics, such as gender, age, work unit, and clinical experience (Table  1 ).

The score of the experimental group on disaster mental health competence increased from 48.13 in the pretest to 70.51 in the posttest (+ 22.38), whereas that of the control group increased from 53.33 in the pretest to 68.38 in the posttest (+ 15.04). These findings reflect a statistically significant difference in competence between the groups (F = 5.149, p  =.026). The scores of the experimental and control groups on problem solving process increased from 73.07 in the pretest to 88.24 in the posttest (+ 15.18) and from 75.75 in the pretest to 83.77 in the posttest (+ 8.02), respectively. As with the competence findings, these point to a significant difference between the groups in terms of the ability to resolve problems (t = 3.024, p  =.003) (Table  2 ).

The score of the experimental group on self-leadership increased from 54.87 in the pretest to 59.58 in the posttest (+ 4.71), and that of the control group increased from 57.48 in the pretest to 60.10 in the posttest (+ 2.63). These results denote a statistically significant difference in this ability between the groups (t = 2.063, p  =.042). The scores of the experimental and control participants on learning self-rose from 55.40 in the pretest to 58.84 in the posttest (+ 3.44) and from 56.81 in the pretest to 57.13 in the posttest (+ 0.31), respectively. Again, a statistically significant difference was found between the groups (t = 3.450, p  =.001). Their scores on motivation to transfer rose from 49.31 in the pretest to 54.29 in the posttest (+ 4.98) (experimental group) and the score increased from 50.50 in the pretest to 51.85 in the posttest (+ 1.35) (control group), pointing to a significant difference between the groups (t = 2.095, p  =.039).

As previously stated, this research was evaluated a virtual nursing simulation-based education program designed to provide psychological support to patients affected by infectious disease disasters. The results showed statistically significant increases in the experimental group’s pretest and posttest scores on disaster mental health competence, problem solving process, self-leadership, learning self-efficacy, and motivation to transfer.

The experimental group achieved more statistically significant improvements in disaster mental health competence than did the control group. This finding is similar to the statistically significant increase in the average disaster mental health competence shown by providers of disaster mental health services providers and non-expert groups after PFA training involving lecture and practice [ 46 ]. It is also consistent with the significant increase in the scores of school counselors on disaster mental health competence after a lecture and simulation on PFA [ 25 ]. In their study on disaster relief workers, Kang and Choi [ 26 ] measured the participants’ performance competence in PFA after the delivery of a lecture and simulation-based education using a standardized patient. The authors found a significant increase in PFA performance competence, consistent with the present research. Since there are currently no other virtual simulation-based education programs for disaster psychological support available, we compared the effectiveness of various PFA training methods with the program assessed in the present work.

In the current research, the posttest scores of the experimental group on problem solving process significantly increased, similar to the results of Kim et al.’s study on virtual simulation- and blended simulation-based education on asthmatic child nursing [ 47 ]. Both the control and experimental groups (virtual simulation only and blended simulation featuring high-fidelity and virtual simulations, respectively) showed an increase in their problem solving process scores. These results and those derived in the present work are similar because reading and pretest phases were incorporated into the design of the previous study. Given that researchers have used commercial virtual simulations featuring avatars rather than standardized patient videos available through English-based platforms, user experiences may differ, thus requiring a qualitative analysis to identify differences. However, Kim et al. [ 47 ] did not implement a debriefing after the virtual simulation program, rendering comparison impossible. Another research reported that a multimodality simulation education that combines such methods as virtual simulation, the use of mannequins, and part-task training increase increased the scores of hospital nurses’ on problem solving process [ 48 ].

In the present work, the experimental group’s self-leadership scores increased after they used the program, and these scores were higher [ 49 , 50 ]. This difference can be explained by the fact that our respondents voluntarily participated in our research given their interest in self-learning programs for disaster psychological support; even in the comparison studies, participants with stronger interest in leadership education typically exhibited heightened degrees of self-leadership [ 51 ]. The increase in self-leadership scores in the current research is consistent with a previous study involving a two-hour simulation education about PPE donning and doffing, medication administration, and medical specimen treatment in a scenario of patients suspected of having infectious diseases [ 32 ]. Another research showed that simulation education on high-risk pregnancy enhances nursing students’ problem-solving processes and self-leadership [ 52 ].

Learning self-efficacy is a key variable that enables the prediction of learners’ degrees of participation in online education and the prediction of their academic achievements, as it points to the ability to manage their learning processes [ 34 , 53 ]. The results of the current research in this regard are consistent with those of a study on the online practice of basic nursing skills, which increased participants’ learning self-efficacy [ 54 ]. The researchers included an online quiz about basic nursing skills and feedback sections for learners’ self-evaluations of their performance as avenues through which to encourage autonomy in learning. A similar approach was used in the present study, which involved both a pretest for self-evaluation, direct feedback on the virtual simulation, and a self-debriefing session, enabling the participants to reflect on their simulation experiences while reviewing other participants’ answers during self-debriefing. These functions of the evaluated program were expected to factor importantly in the significant increase in the participants’ learning self-efficacy scores.

Many studies on practice education have examined participants’ motivations to transfer knowledge and skills alongside their learning self-efficacies. In the current research, the motivation to transfer scores of the experimental increased, and the difference between the two groups was statistically meaningful. This result is consistent with the findings of Park and Kweon on the simulation education about psychiatric nursing, during which post-course debriefing increased the participants’ average scores on motivation to transfer and learning self-efficacy [ 38 ]. Conversely, Kang and Kim found that a six-week simulation program for alcoholic patient care did not generate a significant increase in the participants’ motivation to transfer and learning self-efficacy scores [ 55 ]. This finding was attributed to the unfamiliarity of the local community scenario used in the research to the participants, who were in their senior year of nursing school [ 55 ]. This limitation was overcome in the current research by administering a qualitative survey of nurses’ actual demand for education on psychological support for infectious disease patients. That is, the survey presented scenarios that the participants needed.

As with other studies, the present research was encumbered by several limitations. First, the self-assessment measures used in this study may be unreliable, because they are based on individuals’ subjective perceptions and interpretations of their abilities. There is also the possibility of respondent fatigue given that the participants were compelled to answer numerous questions. Future studies should incorporate both subjective and objective measures into data collection and consider as concise an evaluation method as possible to prevent respondent fatigue. Second, this study did not establish a direct link between the obtained results and actual changes in practice or improvements in patient outcomes. We propose a follow-up study to investigate the impact of the education program examined in this study on either the mental health of patients or the quality of patient care. Third, simulation-based education tends to be accompanied with more guidance than text-based program because the former has diverse components, including quiz games, and participants are predisposed to allocate more time to simulation-based education. These may potentially influence the results. In the future, we propose to conduct research by modifying education under the same time and guided condition.

This study proposed that a well-designed virtual nursing simulation-based education program can be an effective modality with which to satisfy the educational needs of nurses in the context of infectious disease outbreaks. Such programs can be easily used by nurses anywhere and anytime before they are deployed to provide psychological support to patients with infectious diseases. They are also expected to contribute to enhancing competence in addressing disaster mental health and improving the quality of care of patients afflicted with infectious diseases.

Data availability

The datasets used and/or analyzed in this study are available from the corresponding author upon reasonable request.

Abbreviations

Coronavirus disease 2019

Randomized controlled trial

Personal protective equipment

Statistical Package for the Social Sciences

Analysis of covariance

Psychological first aid

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Acknowledgements

The authors would like to thank Eun-Joo Choi and Dong-Hee Cho for their contributions to the development of the simulation program.

This work was supported by the National Research Foundation of Korea (NRF) through a grant funded by the Korean government (Ministry of Science and ICT) (NRF-2020R1A2B5B0100208).

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Ko, E., Choi, YJ. Efficacy of a virtual nursing simulation-based education to provide psychological support for patients affected by infectious disease disasters: a randomized controlled trial. BMC Nurs 23 , 230 (2024). https://doi.org/10.1186/s12912-024-01901-4

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• Math Article

Mode Formula

What is Mode?

The mode is the observation’s value, which occurs most frequently, i.e., an observation with the maximum frequency is called the mode. A data set can have more than one mode, which means more than one observation has the same maximum frequency. In this article, mode formulas for grouped and ungrouped data are explained with the solved examples.

Learn what is the  mode in detail here.

Mode Formula For Ungrouped Data

To find the mode for ungrouped data, it would be better to arrange the data values either in ascending or descending order, so that we can easily find the repeated values and their frequency. Hence, the observation with the highest frequency will be the mode of the given data. Alternatively, we can form a frequency distribution table to get the mode.

Thus, the mode formula for ungrouped data is:

The most frequently occurred value in the data set

Mode Formula for Grouped Data

Now, let us discuss the way of obtaining the mode of grouped data. As we know, more than one value may have the same maximum frequency. In such situations, the data is said to be multimodal. Though grouped data can also be multimodal, we will solve problems having only a single mode.

In a grouped frequency distribution, unlike ungrouped data, it is impossible to determine the mode by looking at the frequencies. Here, we can only locate a class with the maximum frequency, called the modal class. The mode is a value that lies in the modal class and is calculated using the formula given as:

This is the mode formula for grouped data in statistics.

l = Lower limit of the modal class

h = Size of the class interval (assuming all class sizes to be equal)

f 1 = Frequency of the modal class

f 0 = Frequency of the class preceding the modal class

f 2 = Frequency of the class succeeding the modal class

Mode Formula Class 10

In Class 10 maths, the modal formula is given for grouped data. However, the formula is suitable for the data having a single mode. Several solved examples and practice problems have been provided in Chapter 14 of the curriculum. All these problems will help to improve the knowledge of one of the measures of central tendency , i.e. mode.

Solved Examples

Go through the examples provided below for a better understanding of the concept and formulas explained above.

Example 1: Find the mode of the following marks obtained by 25 students in a mathematics test out of 50.

34, 46, 45, 39, 43, 22, 27, 37, 46, 35, 34, 39, 40, 30, 30, 41, 37, 46, 39, 29, 34, 39, 35, 43, 30

The ascending order of the data:

22, 27, 29, 30, 30, 30, 34, 34, 34, 35, 35, 37, 37, 39, 39, 39, 39, 40, 41, 43, 43, 45, 46, 46, 46

The most frequently occurred value is 39.

Hence, the mode of given marks is 39.

Alternatively, let us form the table with observations and their frequencies to get the mode.

The mode of the given data can be obtained by making the frequency table and choosing the highest frequency. Such as:

Here, the highest frequency is 4.

Therefore, the mode is 39.

Example 2: Calculate the mode of the following frequency distribution.

From the given table,

The highest frequency = 20

This value lies in the interval 50-60. Thus, it is the modal class.

Modal class = 50 – 60

l = Lower limit of the modal class = 50

h = Size of the class interval (assuming all class sizes to be equal) = 10

f 1 = Frequency of the modal class = 20

f 0 = Frequency of the class preceding the modal class = 12

f 2 = Frequency of the class succeeding the modal class = 11

= 50 + (80/17)

= 50 + 4.706

Therefore, the mode is 54.706.

Practice Problems

• Calculate the mode for the data: 8, 2, 3, 5, 4, 2, 8, 2, 5, 3, 8, 5, 6, 3, 2, 3, 8, 5, 5, 6

Frequently Asked Questions – FAQs

What is the mode of 2, 5, 7, 5, 2, 5, 5, 3, what is the definition of mode, what is the mode of 20, 21, 22, 22, 23, 22, 25, the mode of the data 12, 10, 11, 12, 12, 15, 16, 17 is, mode of the data 35, 38, 32, 35, 38, 39 is.

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What is causing MLB’s rash of pitching injuries? Analyzing the data on all the biggest questions

There might not be a solution to the pitching injury problem in baseball. If you sort the research and data on the subject to answer the questions most asked about the subject, you don’t end up in a place where there’s an easy way forward.

But that exercise, of answering what we think we can answer the best way possible, seems like a worthy enterprise either way.

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What is the main source of pitcher injury ?

Velocity. Throwing hard is a direct stressor on the elbow , and throwing hard has been shown to lead to injury by multiple studies over the years . One study found that fastball velocity was the most predictive factor of needing elbow surgery in pro pitchers. Every additional tick is more stress on the elbow ligament.

Are pitchers getting hurt because they are throwers, not pitchers now?

Although this has an element of shaking a fist at clouds, there’s also an element of truth in this. Glenn Fleisig, Jonathan Slowik, et al. found that the closer a pitcher pitches to their maximum velocity, the more stress they put on their elbow, and other studies have found similar answers . The bad news is that baseball, as a sport, is throwing closer to its maximum with every year. And yes, this is keeping the method of measuring that velocity constant, it has nothing to do with radar technology changes.

The hardest tracked throw remains Aroldis Chapman ’s 105.7 mph in 2016, and the league’s maximum has settled in around 105 mph most seasons. But the average fastball just keeps climbing, meaning as a league, pitchers are throwing closer to their maximum. And this is true on the individual level — according to STATS Perform, the difference between the average starting pitcher’s sitting and max fastball velocity (minimum 500 thrown in a season) was down to a pitch-tracking low of 3.2 mph in 2023, and that’s more than a full tick less than where it started when pitch tracking began. Today’s baseball is a max-effort game. Unfortunately, while varying the velocity on the fastball may help a pitcher “save bullets” and reduce stress, it does not help them perform better .

Are analytics to blame?

There’s an obvious bias here between the author and the answer, but this line of questioning does not reflect well on professional pitchers’ ability to understand the risks and rewards in their own sport. In other words, yes, there have been all sorts of studies that link fastball velocity to better outcomes, starting with Atlanta Braves executive Mike Fast’s seminal piece on the subject and culminating most recently in things like Stuff+ , but does a pitcher really need an analyst to tell them they need to get strikeouts, and that throwing hard will get them there? Listen to Justin Verlander on the subject. He’s smart, but he also captures the feeling of most pitchers when faced with the reality of getting outs in today’s game — especially given the rules changes that have favored offense.

Justin Verlander on the rash of pithcer injuries: "…I think the game has changed a lot, it would be easiest to blame the pitch clock, in reality everything has a little bit of influence, the biggest thing is the style of pitching has changed so much, everyone is throwing as… pic.twitter.com/rzmvwhB27R — Ari Alexander (@AriA1exander) April 7, 2024

Pitchers organize their talents to get outs, and pitchers know velocity is good for that. The role of the analyst, as it always has been, is to support the players in getting as many wins as possible. Don’t hate the players (or the analysts), they’re just trying to win games.

Are the injuries because of all the breaking balls?

One of the studies that looked at direct stress on the elbow for different pitch types did find that, although overall velocity was the biggest source of stress, once you adjusted for velocity, breaking balls provided more stress at any given mph . Eighty-five mph is a bit of a magic spot for breaking balls — they get better above that velocity, at least. In 2023, pitchers threw nearly 44,000 more breaking balls over 85 mph than they did when we started tracking pitches in 2008. But this is a distinction without a difference, probably: Whether it’s fastball velocity or breaking ball velocity, it’s still velocity.

What role does the pitch clock have in injury?

Theoretically, if you ask an athlete to do the same amount of work in less time, you’re increasing their fatigue. That’s something so basic it shouldn’t require supporting research, but before Dr. Mike Sonne went to work for the Cubs , he wrote for The Athletic about how that works. And how that fatigue should lead to more injuries .

The weird thing is… it hasn’t. Yet. Not at the major-league level.

Despite an early surge in injuries in the first year of the clock , once the year finished, there was no real discernible difference in injury rates . And because March and April are the biggest months for injury list placements, it wouldn’t make much sense to report this year’s injury rates as a big predictor, not until we sum it all up in the end again. The increase in injuries on the major-league level has been a slow burn, not a big spike. The one caveat is that minor-league UCL injuries have exploded since the pitch clock was first introduced in 2018, going from 152 in 2017 to consistently over 200 in each of the past three seasons. (Then again, there were only 86 UCL injuries in the minors in 2011, so there’s more happening here.)

It’s probably most accurate to say that the clock has some role, but that role is undefined as of now, and there’s a longer trend at play, so it can’t all be the clock.

Does sticky stuff (and its ban) have any role in the injury increase?

Tyler Glasnow famously felt that banning sticky stuff led to him having to grip the ball harder, which led to his injury.

Tyler Glasnow made it very clear why pitchers were getting injured 2 years ago. It’s not the pitch clock Nor even joking, this is all Trevor Bauer’s fault pic.twitter.com/Tr0XN8y4En — Nate (@notNate99) April 7, 2024

Research on grip strength isn’t conclusive. Grip strength probably doesn’t lead to more spin , but could lead to better health outcomes , and doesn’t seem to stress the elbow — but these are all studies about grip strength relative to other players. There may not be a study out there that looks at what happens when a pitcher grips the ball harder than he normally does.

But, again, there’s no real spike in injuries after enforcement. There were 243 pitcher injuries in 2021, 226 in 2022 and 233 last year. Sticky stuff enforcement happened in mid-2021.

Could year-round throwing be the problem?

“You have to build up your fitness,” said Sonne, who is a data scientist for the Cubs now. “If you’re running a marathon, you don’t not run for months so you can run on race day. The April spike in injuries happens because people STOP throwing and then try to build up.”

There might be a difference here between adult professionals and kids, though.

On pitching injuries, I'll say this: start with rigorous adherence to basic protective guidelines before tackling advanced physics/sports medicine challenges. >100 IP/yr is associated with a 350% increased risk of injury in youth arms. If you can count, you can prevent injuries. — Eric Cressey (@EricCressey) April 8, 2024

Throwing 100 innings as your body is still developing looks like it increases injury risk. But that might also be because of the shape of those innings, the effort the young person is putting into those pitches, the amount of rest they get and how their workload was monitored. If those things are inconsistent in MLB , they’re probably near nonexistent on a concise level across youth leagues. Throwing less has been put out there as a solution … except that it doesn’t prepare them all that well for throwing more in the future. Pitchers are throwing less, everywhere, and they’re injured more.

Are there better mechanics out there that could solve the problem?

There have been findings that have come out of the emergent study of biomechanics . Certain relationships between your landing foot, your trunk rotation and your shoulder movement have been deemed better than others. Some think they’ve got the perfect mechanics that will ensure a way out of this problem. But Casey Mulholland, who runs Kinetic Pro, a private player development lab, outlined a problem with blaming it all on mechanics.

“Let’s say you’ve got a pitcher with a three-quarter arm slot — that means more stress, more valgus torque,” Mulholland said. “He comes to Tampa and I magically change his arm action to produce the same velo more over the top, and now he throws with less torque. Well, with the cleaned-up arm action, he can now throw harder. And the one thing we know that increases stress is velo, sooooo.

“Our brain passes messages to our muscles, forearm flexors in this case, via the central nervous system to contract at just the right moment to offload the stress applied to the UCL. When we become fatigued our brain doesn’t pass this message as well, the muscles don’t contract at the ‘optimal time’ or the ‘optimal amount’ and we end up not being able to offload this stress. The UCL then wears more of a direct stress. Over time, under fatigue, the load of throwing eventually overcomes the tissue tolerance and boom, UCL tear.

“This is why workload management is the only logical answer to slow the injury rate,” thinks Mulholland. “Workload management predicts the possible time at which an athlete might experience too much load.”

What rule changes could incentivize different injury outcomes?

Some proposed rule changes are just not going to happen. Every pitch over 94 is a ball? Just can’t see it. Similarly, the idea that we will limit the number of pitchers on the roster might work to prod teams into getting more innings from each pitching slot. But the players’ union would assuredly be against any reduction in major-league jobs.

Jayson Stark’s proposed “Double Hook” — in which the pitching team loses their designated hitter once the starting pitcher leaves the game — would incentivize teams to acquire players who can go deeper into games. But there’s a little bit of a gap there that will be funky for the pitchers themselves. Until the market shows that it will value lesser quality with more quantity (it hasn’t), the pitcher’s incentives may be misaligned with those of the team as a whole. They’d rather be the guy with better stuff and cleaner numbers if they know that’s what free agency has rewarded in the past.

Limiting the number of active pitchers for a single game seems like a bland rule change that might not mean much. But, in light of Mulholland’s and Sonne’s feelings about the importance of workload monitoring, maybe asking teams to fully rule out some number of pitchers every game could nudge teams into the era of more precise workload management.

In the end, there’s no telling pitchers to throw softer if it isn’t going to help their bottom line — and that’s sort of the simplest way to state the problem. The best we can do as a sport is to provide players with the best possible mechanics that analysis can provide, and be as precise as possible with monitoring their fatigue. At the very top of any game, injuries will happen.

From revision surgery to internal brace procedures, understanding Tommy John surgery today

(Photo of Justin Verlander: Carmen Mandato / Getty Images)

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Eno Sarris is a senior writer covering baseball analytics at The Athletic. Eno has written for FanGraphs, ESPN, Fox, MLB.com, SB Nation and others. Submit mailbag questions to [email protected] . Follow Eno on Twitter @ enosarris

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Switching From iPhone to Android Is Easy. It’s the Aftermath That Stings.

Even if you manage to ditch your iPhone, Apple’s hooks are still there.

By Brian X. Chen

Brian X. Chen is The Times’s lead consumer technology writer and the author of Tech Fix , a column about the social implications of the tech we use.

When I temporarily switched from an iPhone to an Android phone last week, I was bracing for a world of hurt. I’ve owned only Apple phones ever since buying the first-generation iPhone in 2007. And, like many, I’ve bought other Apple products that pair nicely, including AirPods, an Apple Watch and an iPad.

That type of loyalty is the basis of an antitrust case against Apple brought by the Justice Department, which has accused the company of using monopolistic control over the iPhone to harm competitors and deter customers from switching to other phones. To test that theory, I decided to briefly break up with my iPhone.

I was initially surprised by how simple it was to shift my iPhone data to an Android smartphone made by Google. Just by installing an app on my iPhone that Google made to help people switch, I was able to copy my contacts list, photo album and calendar into my Google account. Then, presto — all that data appeared on the Android.

I was almost done. After I called my carrier, Verizon, to transfer my phone number to the Android device, my mission was accomplished: I had become an Android convert.

At first, I was happy with my choice — I had upgraded to a fancy Google Pixel phone. But by Day 6, I was ready to switch back.

A bunch of annoyances added up. Even though I could still use most of my Apple products, I started missing my Apple Watch, which requires an iPhone to fully work. For software, I was able to find Android alternatives for all my favorite apps — except for Notes. While switching phones wasn’t technically hard, Apple’s hooks were still in me.

How Apple keeps customers loyal to the iPhone — and whether its practices harm competition — is at the heart of the government’s antitrust suit against the Cupertino giant.

Apple and the Justice Department declined to comment.

In its 88-page complaint, the department said a number of Apple products protected the company’s competitive advantage with the iPhone, including iMessage, Apple’s Wallet app and the Apple Watch. How hard do those perks really make it to ditch your iPhone? Here’s what I found.

Losing iMessage

For the most part, iPhone users and Android users can communicate with each other easily through email, phone calls and apps like Slack, but when it comes to text messaging, there is still an obvious split known as the “green bubble versus blue bubble” disparity .

When iPhone users send texts to other iPhones, the messages appear blue and can tap into exclusive perks like an animation of birthday confetti. But if an iPhone user texts an Android user, the bubble turns green, many features break, and photos and videos deteriorate in quality.

Before transferring my phone number to the Pixel phone, I used my iPhone to send iMessages to my blue bubble comrades warning them that our conversations would soon turn green. “Ew!” a friend replied. But after many remarks made in jest, no one protested, and I soldiered on.

Next, I had to detach my phone number from iMessage on Apple’s website to ensure that my text messages would stop going through Apple’s servers and arrive on my phone. Unless I did this, I would not receive texts from other iPhones. Eventually, the conversations turned green. I prepared myself for humiliation.

But no one gave me a hard time or excluded me. I did notice, however, that many friends had suddenly stopped texting me photos, perhaps because they knew the images would no longer look as good.

For years, some of my closest friends have texted me only through Signal, the third-party messaging app with strong privacy protections and many of the same features as iMessage. Signal is also available on Android, preserving that tradition.

Apple has announced that later this year, it will improve texts between iPhone and Android users by adopting rich communication services, a standard that Google and others integrated into their messaging apps years ago. Texts sent between iPhones and Androids will remain green, but images and videos will be higher quality.

Losing Apple Wallet

For iPhones, the go-to app for making mobile payments in stores is Apple Wallet, and for Android users, the equivalent app is Google Wallet. The experience of using each wallet app was identical: I loaded up my credit cards and Clipper card for the Bay Area’s rapid transit service.

The Justice Department’s criticism of Apple Wallet centers on how Apple gives only its app access to the iPhone’s payment chip, preventing competing wallet services from using that chip to make payments. But the way Apple designed its Wallet app had no impact on my ability to switch to an Android.

Losing the Apple Watch and other products

For an iPhone owner, a main incentive to buy more Apple products is that they work seamlessly together. A Mac laptop, for instance, uses many of the same apps for messaging, note taking and reminders as the iPhone, and the data is synchronized among the devices with Apple’s iCloud. In theory, the more invested you are in Apple’s ecosystem — and the more that Apple restricts its products from working with competing devices, the Justice Department says — the tougher it is to switch from an iPhone.

After I switched to an Android phone, my feelings about using other Apple products ranged from moderate annoyance to deep frustration:

The iPad worked independently from the iPhone, but I could no longer see my text messages on the tablet anymore. This was minor because I don’t do much texting on my iPad.

My AirPods Pro were OK — they connected quickly with the Pixel for playing music. But the downside is that the AirPods use Adaptive EQ, a technology that tunes sound quality to the shape of your ear, and it works only with software on the iPhone. So audio doesn’t sound as good.

I could not use my Android phone to locate my AirTags, the tiny Apple trackers I use to find my wallet and keys, on a map. But when my AirTags were in my pocket, the Android phone showed an alert that an “unknown tracker” was moving around with me, a safety feature for combating stalkers.

The Apple Watch requires an iPhone to set up, but its fitness tracking can work independently. Because I had already set up my watch, I could continue to use it at the gym alongside my Android phone. But I could no longer see my detailed workout data.

I ran into other annoyances not specifically called out in the lawsuit and finally reached peak frustration when I tried to find a replacement for Apple’s Notes, which I use regularly on my Mac, iPad and phone for work and personal errands. I used alternatives but didn’t like them, and combined with the aforementioned issues, it was all too much.

The upshot: Switching is easy, until it’s not

My experience isn’t universal. Some people would care more than others about how certain Apple products would change if they switched phones. Younger people would probably care a lot about lacking iMessage in schools, where a green bubble has been known to be an invitation for mockery and exclusion, according to education experts . Parents who use AirTags to track their children would view losing access to those as a deal breaker.

The upshot from this experiment is that while it’s not technically hard to switch to a different phone, there are plenty of things that could make you regret it.

Brian X. Chen is the lead consumer technology writer for The Times. He reviews products and writes Tech Fix , a column about the social implications of the tech we use. More about Brian X. Chen

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