## 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 frequencies.

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 (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 for Grouped Data** to understand how to calculate sample mean, sample variance and sample standard deviation.

## 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 |

**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 for 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.

## 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 |

**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 .

## 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.

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 to help you.