## Variance and Standard Deviation for Grouped Data Calculator

Use Variance and Standard Deviation for Grouped Data calculator to calculate sample mean,sample variance and sample standard deviation for grouped data based on data provided in class groups and type of frequency distribution.

Below article on standard deviation for grouped data calculator provides step by step procedure about how to use variance for grouped data calculator with detailed standard deviation for grouped data examples.

### Calculator

Variance and standard deviation for grouped data Calculator
Type of Frequncy Distribution DiscreteContinuous
Enter the Classes for X (Separated by comma,)
Enter the frequencies (f) (Separated by comma,)
Results
Number of Observations (n):
Sample Mean : ($\overline{x}$)
Sample variance : ($s^2_x$)
Sample Standard Deviation : ($s_x$)

### How to use Variance and Standard Deviation for Grouped Data Calculator?

Step 1 - Select type of frequency distribution either Discrete or continuous

Step 2 - Enter the Range or classes (X) seperated by comma (,)

Step 3 - Enter the Frequencies (f) seperated by comma

Step 4 - Click on “Calculate” button to calculate sample standard deviation for grouped data

Step5 - Gives output as number of observation (n)

Step 6 - Calculate Sample mean ($\overline{x}$) for grouped data

Step 7 - Calculate Sample Variance ($s^2_x$) for grouped data

Step 8 - Calculate Sample Standard Deviation for Grouped Data ($s_x$)

## Variance and Standard Deviation for Grouped Data

Let $(x_i,f_i), i=1,2, \cdots , n$ be the observed frequency distribution.

## Sample Variance Calculation

The sample variance of $X$ is denoted by $s_x^2$ and is given by

$s_x^2 =\dfrac{1}{N-1}\sum_{i=1}^{n}f_i(x_i -\overline{x})^2$

OR

$s_x^2 =\dfrac{1}{N-1}\bigg(\sum_{i=1}^{n}f_ix_i^2-\frac{\big(\sum_{i=1}^n f_ix_i\big)^2}{N}\bigg)$

where,

• $N=\sum_{i=1}^n f_i$ is the total number of observations,
• $\overline{x}$ is the sample mean.

## Sample Standard Deviation for Grouped Data Calculation

The sample standard deviation of $X$ is defined as the positive square root of sample variance. The sample standard deviation of $X$ is given by

$s_x =\sqrt{s_x^2}$

Below are solved examples on Variance and Standard Deviation of Grouped Data to understand how to calculate sample mean, sample variance and sample standard deviation.

## Example - 1 Variance and standard deviation for grouped data

Following tables shows a frequency distribution of daily number of car accidents at a particular cross road during a month of April.

No.of car accidents ($x$) 2 3 4 5 6
No. of days ($f$) 9 11 6 3 1

Calculate variance of grouped data and standard deviation for above frequency table of number of car accidents.

### Solution

$x_i$ $f_i$ $f_i*x_i$ $f_ix_i^2$
2 9 18 36
3 11 33 99
4 6 24 96
5 3 15 75
6 1 6 36
Total 30 96 342

Step 1 - Sample mean for grouped data

The sample mean of $X$ is

\begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n f_ix_i\\ &=\frac{96}{30}\\ &=3.2\text{ accidents } \end{aligned}

The average of no.of car accidents is $3.2$ accidents .

Step 2 - Sample variance for grouped data

Sample variance of $X$ is

\begin{aligned} s_x^2 &=\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}f_ix_i^2-\frac{\big(\sum_{i=1}^n f_ix_i\big)^2}{n}\bigg)\\ &=\dfrac{1}{29}\bigg(342-\frac{(96)^2}{30}\bigg)\\ &=\dfrac{1}{29}\big(342-\frac{9216}{30}\big)\\ &=\dfrac{1}{29}\big(342-307.2\big)\\ &= \frac{34.8}{29}\\ &=1.2 \end{aligned}

Step 3 - Sample standard deviation for grouped data

The standard deviation is the positive square root of the variance.

The sample standard deviation of grouped data calculated as

\begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{2.5}\\ &=1.0954 \text{ accidents } \end{aligned}

Thus the standard deviation of no.of car accidents is $1.0954$ accidents.

## Example - 2 Variance and standard deviation for grouped data

The table below shows the total number of man-days lost to sickness during one week’s operation of a small chemical plant.

Days Lost 1-3 4-6 7-9 10-12 13-15
Frequency 8 7 10 9 6

Calculate the variance of grouped data and standard deviation for above frequency table of the number of lost days.

### Solution

Class Interval Class Boundries mid-value ($x_i$) Freq ($f_i$) $f_i*x_i$ $f_ix_i^2$
1-3 0.5-3.5 2 8 16 32
4-6 3.5-6.5 5 7 35 175
7-9 6.5-9.5 8 10 80 640
10-12 9.5-12.5 11 9 99 1089
13-15 12.5-15.5 14 6 84 1176
Total 40 314 3112

Step 1 - Sample mean for grouped data

The sample mean of $X$ is

\begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{314}{40}\\ &=7.85\text{ days } \end{aligned}

The average of total number of man days lost is $7.85$ days .

Step 2 - Sample Variance for grouped data

Sample variance of $X$ is

\begin{aligned} s_x^2 &=\dfrac{1}{N-1}\bigg(\sum_{i=1}^{n}f_ix_i^2-\frac{\big(\sum_{i=1}^n f_ix_i\big)^2}{N}\bigg)\\ &=\dfrac{1}{39}\bigg(3112-\frac{(314)^2}{40}\bigg)\\ &=\dfrac{1}{39}\big(3112-\frac{98596}{40}\big)\\ &=\dfrac{1}{39}\big(3112-2464.9\big)\\ &= \frac{647.1}{39}\\ &=16.5923 \end{aligned}

Step 3 - Sample standard deviation for grouped data

The standard deviation for grouped data is the positive square root of the variance.

The sample standard deviation for grouped data calculated as

\begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{22.5}\\ &=4.0734 \text{ days } \end{aligned}

Thus the standard deviation of total number of man days lost is $4.0734$ days .

## Example - 3 Variance and Standard Deviation for Grouped Data

The following data shows the distribution of maximum loads in short tons supported by certain cables produced by a company:

9.25-9.75 2
9.75-10.25 5
10.25-10.75 12
10.75-11.25 17
11.25-11.75 14
11.75-12.25 6
12.25-12.75 3
12.75-13.25 1

Compute variance and standard deviation for the above frequency distribution.

### Solution

Class Interval Class Boundries mid-value ($x_i$) Freq ($f_i$) $f_i*x_i$ $f_ix_i^2$
9.25-9.75 9.25-9.75 9.5 2 19 180.5
9.75-10.25 9.75-10.25 10 5 50 500
10.25-10.75 10.25-10.75 10.5 12 126 1323
10.75-11.25 10.75-11.25 11 17 187 2057
11.25-11.75 11.25-11.75 11.5 14 161 1851.5
11.75-12.25 11.75-12.25 12 6 72 864
12.25-12.75 12.25-12.75 12.5 3 37.5 468.75
12.75-13.25 12.75-13.25 13 1 13 169
Total 60 665.5 7413.75

Sample mean for grouped data

The sample mean of $X$ is

\begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{665.5}{60}\\ &=11.0917\text{ tons} \end{aligned}

The average of maximum load is $11.0917$ tons.

Sample variance for grouped data

Sample variance of $X$ is

\begin{aligned} s_x^2 &=\dfrac{1}{N-1}\bigg(\sum_{i=1}^{n}f_ix_i^2-\frac{\big(\sum_{i=1}^n f_ix_i\big)^2}{N}\bigg)\\ &=\dfrac{1}{59}\bigg(7413.75-\frac{(665.5)^2}{60}\bigg)\\ &=\dfrac{1}{59}\big(7413.75-\frac{442890.25}{60}\big)\\ &=\dfrac{1}{59}\big(7413.75-7381.50417\big)\\ &= \frac{32.24583}{59}\\ &=0.5465 \end{aligned}

Sample standard deviation for grouped data

The standard deviation is the positive square root of the variance.

The sample standard deviation is

\begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{1.5}\\ &=0.7393 \text{ tons} \end{aligned}

Thus the standard deviation of maximum load is $0.7393$ tons.

## Example - 4 Variance and Standard Deviation for Grouped Data

Following table shows the weight of 100 pumpkin produced from a farm :

Weight (‘00 grams) Frequency
$4 \leq x < 6$ 4
$6 \leq x < 8$ 14
$8 \leq x < 10$ 34
$10 \leq x < 12$ 28
$12 \leq x < 14$ 20

Calculate variance and standard deviation for the given frequency distribution.

### Solution

Class Interval Class Boundries mid-value ($x_i$) Freq ($f_i$) $f_i*x_i$ $f_ix_i^2$
4-6 4-6 5 4 20 100
6-8 6-8 7 14 98 686
8-10 8-10 9 34 306 2754
10-12 10-12 11 28 308 3388
12-14 12-14 13 20 260 3380
Total 100 992 10308

Sample mean for grouped data

The sample mean of $X$ is

\begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{992}{100}\\ &=9.92\text{ ('00 grams)} \end{aligned}

The average of weight of pumpkin is $9.92$ (‘00 grams).

Sample variance for grouped data

Sample variance of $X$ is

\begin{aligned} s_x^2 &=\dfrac{1}{N-1}\bigg(\sum_{i=1}^{n}f_ix_i^2-\frac{\big(\sum_{i=1}^n f_ix_i\big)^2}{N}\bigg)\\ &=\dfrac{1}{99}\bigg(10308-\frac{(992)^2}{100}\bigg)\\ &=\dfrac{1}{99}\big(10308-\frac{984064}{100}\big)\\ &=\dfrac{1}{99}\big(10308-9840.64\big)\\ &= \frac{467.36}{99}\\ &=4.7208 \end{aligned}

Sample standard deviation for grouped data

The standard deviation is the positive square root of the variance.

The sample standard deviation is

\begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{10}\\ &=2.1727 \text{ ('00 grams)} \end{aligned}

Thus the standard deviation of weight of pumpkin is $2.1727$ (‘00 grams).

## Example - 5 Variance and Standard Deviation for Grouped Data

The following table gives the amount of time (in minutes) spent on the internet each evening by a group of 56 students. Compute variance and standard deviation for the following frequency distribution.

Time spent on Internet ($x$) No. of Students ($f$)
10-12 3
13-15 12
16-18 15
19-21 24
22-24 2

### Solution

Class Interval Class Boundries mid-value ($x_i$) Freq ($f_i$) $f_i*x_i$ $f_ix_i^2$
10-12 9.5-12.5 11 3 33 363
13-15 12.5-15.5 14 12 168 2352
16-18 15.5-18.5 17 15 255 4335
19-21 18.5-21.5 20 24 480 9600
22-24 21.5-24.5 23 2 46 1058
Total 56 982 17708

Sample mean for grouped data

The sample mean of $X$ is

\begin{aligned} \overline{x} &=\frac{1}{N}\sum_{i=1}^n f_ix_i\\ &=\frac{982}{56}\\ &=17.5357\text{ minutes} \end{aligned}

The average of amount of time (in minutes) spent on the internet is $17.5357$ minutes.

Sample variance for grouped data

Sample variance of $X$ is

\begin{aligned} s_x^2 &=\dfrac{1}{N-1}\bigg(\sum_{i=1}^{n}f_ix_i^2-\frac{\big(\sum_{i=1}^n f_ix_i\big)^2}{N}\bigg)\\ &=\dfrac{1}{55}\bigg(17708-\frac{(982)^2}{56}\bigg)\\ &=\dfrac{1}{55}\big(17708-\frac{964324}{56}\big)\\ &=\dfrac{1}{55}\big(17708-17220.07143\big)\\ &= \frac{487.92857}{55}\\ &=8.8714 \end{aligned}

Sample standard deviation for grouped data

The standard deviation is the positive square root of the variance.

The sample standard deviation is

\begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{22.5}\\ &=2.9785 \text{ minutes} \end{aligned}

Thus the standard deviation of amount of time (in minutes) spent on the internet is $2.9785$ minutes.

## Conclusion

I hope, you may like above article on Variance and Standard Deviation for Grouped Data Calculator with step by step guide on how to use variance for grouped data calculator with supportive examples.