Variance and Standard Deviation Calculator For Ungrouped Data

Variance and Standard deviation for ungrouped data

Variance and standard deviation are the measures of dispersion.

Sample Variance

Let $x_i, i=1,2, \cdots , n$ be $n$ observations.

Sample variance of $X$ is denoted by $s_{x}^2$ and is given by

$s_x^2 =\dfrac{1}{n-1}\sum_{i=1}^{n}(x_i -\overline{x})^2=\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}x_i^2-\frac{\big(\sum_{i=1}^n x_i\big)^2}{n}\bigg)$

where,

  • $\overline{x}=\frac{1}{n}\sum_{i=1}^{n}x_i$ is the sample mean

Sample standard deviation

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

$s_x =\sqrt{s_x^2}$

Variance and Standard deviation calculator for ungrouped data

Use this calculator to find the variance and standard deviation for ungrouped data.

Calculator

Variance and standard deviation Calculator
Enter the X Values (Separated by comma,)
Results
Number of Observations (n):
Sample Mean : ($\overline{x}$)
Sample variance : ($s^2_x$)
Sample std. Deviation : ($s_x$)

How to find variance and standard deviation for ungrouped data?

Step 1 - Enter the (X) values seperated by comma (,)

Step 2 - Click on "Calculate" button to get variance and standard deviation for ungrouped data

Step 3 - Gives the output as number of observations $n$

Step 4 - Calculate sample mean ($\overline{x}$) for ungrouped data

Step 5 - Calculate sample variance ($s^2_x$) for ungrouped data

Step 6 - Calculate sample standard deviation ($s_x$) for ungrouped data

Here are the few solved numerical examples with step by step guide on standard deviation for ungrouped data.

Variance and Standard deviation of ungrouped data Example 1

The age (in years) of 6 randomly selected students from a class are

22,25,24,23,24,20.

Find the variance and standard deviation.

Solution

$x_i$ $x_i^2$
22 484
25 625
24 576
23 529
24 576
20 400
Total 138 3190

Sample mean for ungrouped data

The sample mean of $X$ is

$$ \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{138}{6}\\ &=23\text{ years} \end{aligned} $$

The average of age of students is $23$ years.

Sample variance for ungrouped data

Sample variance of $X$ is

$$ \begin{aligned} s_x^2 &=\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}x_i^2-\frac{\big(\sum_{i=1}^n x_i\big)^2}{n}\bigg)\\ &=\dfrac{1}{5}\bigg(3190-\frac{(138)^2}{6}\bigg)\\ &=\dfrac{1}{5}\big(3190-\frac{19044}{6}\big)\\ &=\dfrac{1}{5}\big(3190-3174\big)\\ &= \frac{16}{5}\\ &=3.2 \end{aligned} $$

Sample standard deviation for ungrouped data

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

The sample standard deviation is

$$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{3.2}\\ &=1.7889 \text{ years} \end{aligned} $$

Thus the standard deviation of age of students is $1.7889$ years.

Variance and Standard deviation for ungrouped data Example 2

The following data gives the hourly wage rates (in dollars) of 10 employees of a company.

20,21,24,25,18,22,24,22,20,22.

Find the variance and standard deviation of hourly wage rates.

Solution

$x_i$ $x_i^2$
20 400
21 441
24 576
25 625
18 324
22 484
24 576
22 484
20 400
22 484
Total 218 4794

Sample mean of ungrouped data

The sample mean of $X$ is

$$ \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{218}{10}\\ &=21.8\text{ dollars} \end{aligned} $$

The average of hourly wage rates is $21.8$ dollars.

Sample variance of ungrouped data

Sample variance of $X$ is

$$ \begin{aligned} s_x^2 &=\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}x_i^2-\frac{\big(\sum_{i=1}^n x_i\big)^2}{n}\bigg)\\ &=\dfrac{1}{9}\bigg(4794-\frac{(218)^2}{10}\bigg)\\ &=\dfrac{1}{9}\big(4794-\frac{47524}{10}\big)\\ &=\dfrac{1}{9}\big(4794-4752.4\big)\\ &= \frac{41.6}{9}\\ &=4.6222 \end{aligned} $$

Sample standard deviation of ungrouped data

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

The sample standard deviation is

$$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{4.6222}\\ &=2.1499 \text{ dollars} \end{aligned} $$

Thus the standard deviation of hourly wage rates is $2.1499$ dollars.

Variance and Standard deviation for ungrouped data Example 3

A random sample of 15 patients yielded the following data on the length of stay (in days) in the hospital.

5, 6, 9, 10, 15, 10, 14, 12, 10, 13, 13, 9, 8, 10, 12.

Compute variance and standard deviation.

Solution

$x_i$ $x_i^2$
5 25
6 36
9 81
10 100
15 225
10 100
14 196
12 144
10 100
13 169
13 169
9 81
8 64
10 100
12 144
Total 156 1734

Sample mean for ungrouped data

The sample mean of $X$ is

$$ \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{156}{15}\\ &=10.4\text{ days} \end{aligned} $$

The average of length of stay in the hospital is $10.4$ days.

Sample variance for ungrouped data

Sample variance of $X$ is

$$ \begin{aligned} s_x^2 &=\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}x_i^2-\frac{\big(\sum_{i=1}^n x_i\big)^2}{n}\bigg)\\ &=\dfrac{1}{14}\bigg(1734-\frac{(156)^2}{15}\bigg)\\ &=\dfrac{1}{14}\big(1734-\frac{24336}{15}\big)\\ &=\dfrac{1}{14}\big(1734-1622.4\big)\\ &= \frac{111.6}{14}\\ &=7.9714 \end{aligned} $$

Sample standard deviation for ungrouped data

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

The sample standard deviation is

$$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{7.9714}\\ &=2.8234 \text{ days} \end{aligned} $$

Thus the standard deviation of length of stay in the hospital is $2.8234$ days.

Variance and Standard deviation of ungrouped data Example 4

Blood sugar level (in mg/dl) of a sample of 20 patients admitted to the hospitals are as follows:

75,89,72,78,87, 85, 73, 75, 97, 87, 84, 76,73,79,99,86,83,76,78,73.

Calculate variance and standard deviation for the given data.

Solution

$x_i$ $x_i^2$
75 5625
80 6400
72 5184
78 6084
82 6724
85 7225
73 5329
75 5625
97 9409
87 7569
84 7056
76 5776
73 5329
79 6241
99 9801
86 7396
83 6889
76 5776
78 6084
73 5329
Total 1611 130851

Sample mean for ungrouped data

The sample mean of $X$ is

$$ \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{1611}{20}\\ &=80.55\text{ mg/dl} \end{aligned} $$

The average of blood sugar level is $80.55$ mg/dl.

Sample variance for ungrouped data

Sample variance of $X$ is

$$ \begin{aligned} s_x^2 &=\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}x_i^2-\frac{\big(\sum_{i=1}^n x_i\big)^2}{n}\bigg)\\ &=\dfrac{1}{19}\bigg(130851-\frac{(1611)^2}{20}\bigg)\\ &=\dfrac{1}{19}\big(130851-\frac{2595321}{20}\big)\\ &=\dfrac{1}{19}\big(130851-129766.05\big)\\ &= \frac{1084.95}{19}\\ &=57.1026 \end{aligned} $$

Sample standard deviation for ungrouped data

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

The sample standard deviation is

$$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{57.1026}\\ &=7.5566 \text{ mg/dl} \end{aligned} $$

Thus the standard deviation of blood sugar level is $7.5566$ mg/dl.

Variance and Standard deviation for ungrouped data Example 5

The rice production (in Kg) of 10 acres is given as: 1120, 1240, 1320, 1040, 1080, 1720, 1600, 1470, 1750, and 1885. Find variance and standard deviation for the given data.

Solution

$x_i$ $x_i^2$
1120 1254400
1240 1537600
1320 1742400
1040 1081600
1080 1166400
1720 2958400
1600 2560000
1470 2160900
1750 3062500
1885 3553225
Total 14225 21077425

Sample mean for ungrouped data

The sample mean of $X$ is

$$ \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{14225}{10}\\ &=1422.5\text{ Kg} \end{aligned} $$

The average of rice production is $1422.5$ Kg.

Sample variance for ungrouped data

Sample variance of $X$ is

$$ \begin{aligned} s_x^2 &=\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}x_i^2-\frac{\big(\sum_{i=1}^n x_i\big)^2}{n}\bigg)\\ &=\dfrac{1}{9}\bigg(21077425-\frac{(14225)^2}{10}\bigg)\\ &=\dfrac{1}{9}\big(21077425-\frac{202350625}{10}\big)\\ &=\dfrac{1}{9}\big(21077425-20235062.5\big)\\ &= \frac{842362.5}{9}\\ &=93595.8333 \end{aligned} $$

Sample standard deviation for ungrouped data

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

The sample standard deviation is

$$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{93595.8333}\\ &=305.9344 \text{ Kg} \end{aligned} $$

Thus the standard deviation of rice production is $305.9344$ Kg.

Conclusion

In this tutorial, you learnt about formula for variance and standard deviation for ungrouped data and how to calculate variance and standard deviation for ungrouped data.

You also learnt about how to solve numerical problems on variance and standard deviation for ungrouped data.

To learn more about other descriptive statistics measures, please refer to the following tutorials:

Descriptive Statistics

Let me know in the comments if you have any questions on Variance and standard deviation calculator for ungrouped data with examples and your thought on this article.

VRCBuzz co-founder and passionate about making every day the greatest day of life. Raju is nerd at heart with a background in Statistics. Raju looks after overseeing day to day operations as well as focusing on strategic planning and growth of VRCBuzz products and services. Raju has more than 25 years of experience in Teaching fields. He gain energy by helping people to reach their goal and motivate to align to their passion. Raju holds a Ph.D. degree in Statistics. Raju loves to spend his leisure time on reading and implementing AI and machine learning concepts using statistical models.

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