Variance and Standard Deviation for Grouped Data
Variance and standard deviation are the measures of dispersion.
Let $(x_i,f_i), i=1,2, \cdots , n$
be the observed frequency distribution.
Sample variance for Grouped Data
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
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
$s_x =\sqrt{s_x^2}$
Variance and Standard Deviation Calculator for Grouped Data
Use this calculator to find the variance and standard deviation for grouped data.
Calculator
Variance and standard deviation Calculator | |
---|---|
Type of Frequency Distribution | DiscreteContinuous |
Enter the Classes for X (Separated by comma,) | |
Enter the frequencies (f) (Separated by comma,) | |
Results | |
Number of Observation (n): | |
Sample Mean : ($\overline{x}$) | |
Sample variance : ($s^2_x$) | |
Sample std. Deviation : ($s_x$) | |
How to find variance and standard deviation for grouped data?
Step 1 - Select type of frequency distribution (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" for variance and standard deviation calculation
Step 5 - 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 ($s_x$) for grouped data
Below are the numerical examples with step by step guide solution on variance and standard deviation for grouped data.
Variance and standard deviation for grouped data Example 1
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 and standard deviation 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 |
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 .
Sample variance for grouped data
Sample variance of $X$ is
$$ \begin{aligned} s_x^2 &=\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}f_ix_i^2-\frac{\big(\sum_{i=1}^n f_ix_i\big)^2}{n}\bigg)\\ &=\dfrac{1}{29}\bigg(342-\frac{(96)^2}{30}\bigg)\\ &=\dfrac{1}{29}\big(342-\frac{9216}{30}\big)\\ &=\dfrac{1}{29}\big(342-307.2\big)\\ &= \frac{34.8}{29}\\ &=1.2 \end{aligned} $$
Sample standard deviation for grouped data
The standard deviation is the positive square root of the variance.
The sample standard deviation is
$$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{2.5}\\ &=1.0954 \text{ accidents } \end{aligned} $$
Thus the standard deviation of no.of car accidents is $1.0954$ accidents .
Variance and Standard Deviation for Grouped Data Example 2
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 and standard deviation 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 |
Sample mean for geouped 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 .
Sample variance for grouped data
Sample variance of $X$ is
$$ \begin{aligned} s_x^2 &=\dfrac{1}{N-1}\bigg(\sum_{i=1}^{n}f_ix_i^2-\frac{\big(\sum_{i=1}^n f_ix_i\big)^2}{N}\bigg)\\ &=\dfrac{1}{39}\bigg(3112-\frac{(314)^2}{40}\bigg)\\ &=\dfrac{1}{39}\big(3112-\frac{98596}{40}\big)\\ &=\dfrac{1}{39}\big(3112-2464.9\big)\\ &= \frac{647.1}{39}\\ &=16.5923 \end{aligned} $$
Sample standard deviation for grouped data
The standard deviation is the positive square root of the variance.
The sample standard deviation is
$$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{22.5}\\ &=4.0734 \text{ days } \end{aligned} $$
Thus the standard deviation of total number of man days lost is $4.0734$ days .
Variance and Standard Deviation for Grouped Data Example 3
The following data shows the distribution of maximum loads in short tons supported by certain cables produced by a company:
Maximum load | No. of Cables |
---|---|
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.
Variance and Standard Deviation for Grouped Data Example 4
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 data.
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).
Variance and Standard Deviation for Grouped Data Example 5
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
In this tutorial, you learnt about formula for variance and standard deviation for grouped data and how to calculate variance and standard deviation for grouped data.
You also learnt about how to solve numerical problems on variance and standard deviation for grouped data.
To learn more about other descriptive statistics measures, please refer to the following tutorials:
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