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:
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.
