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:
| 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.
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.
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