Variance and Standard Deviation of Ungrouped Data Calculator
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 Observation (n): | |
| Sample Mean : ($\overline{x}$) | |
| Sample variance : ($s^2_x$) | |
| Sample Standard Deviation : ($s_x$) | |
How to use Variance and Standard Deviation of Ungrouped Data?
Step 1 - Enter the set of numerical values (X) seperated by ,
Step 2 - Click on Calculate button to calculate sample mean, sample variance and sample standard deviation
Step 3 - Calculate number of observation (n)
Step 4 - Calculate sample mean for ungrouped data
Step 5 - Calculate sample variance for ungrouped data
Step 6 - Calculate sample standard deviation for ungrouped data
Variance and Standard Deviation for Ungrouped Data
Let $x_i, i=1,2, \cdots , n$ be $n$ observations
Variance and Standard Deviation formula for ungrouped data
mean of $X$ is denoted by $\overline{X}$ and is given by
$$ \begin{eqnarray*} \overline{X}& =\frac{1}{n}\sum_{i=1}^{n}x_i \end{eqnarray*} $$
Variance of $X$ is denoted by $s_{x}^2$ and is given by
$$ \begin{aligned} s_x^2 & =\frac{1}{n-1}\sum_{i=1}^{n}(x_i -\overline{x})^2\\ &=\frac{1}{n-1}\bigg(n\sum_{i=1}^{n} x_i^2-\big(\sum_{i=1}^nx_i\big)^2\bigg) \end{aligned} $$
The standard deviation of $X$ is defined as the positive square root of variance. The standard deviation of $X$ is given by
$$ \begin{eqnarray*} s_x & =\sqrt{s_x^2} \end{eqnarray*} $$
Below are the numerical examples with step by step guide solution on variance and 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 ungroped 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 of ungrouped data
The standard deviation is the positive square root of the variance.
The sample standard deviation is
$$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{3.2}\\ &=1.7889 \text{ years} \end{aligned} $$
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 of 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 for given data
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 for 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 of 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
Hope you like above article about formula for variance and standard deviation for ungrouped data and how to calculate variance and standard deviation for ungrouped data.
