Q.The arithmetic mean and standard deviation of a series of 20 items were calculated by a student as 20 cms and 5 cms respectively. But while calculating them, an item 13 was misread as 30. Find the corrected mean and standard deviation.

Standard deviation is defined as the positive square root of variance.

By definition, standard deviation $\sigma = \sqrt{\text{variance}}$. Given variance $= 144$:

$\sigma = \sqrt{144} = 12$ …

The square of the standard deviation is called the variance, option (c).

Standard deviation ($\sigma$) measures the spread of data about the mean. By definition, variance ($\sigma^2$) is exactly the square of the standard deviation: $\text{Variance} = \sigma^2 = (\text{Standard Deviation})^2$.

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Standard deviation $=\sqrt{\text{Variance}}$; here $\sqrt{225}=15$.

By definition, variance is the square of the standard deviation, so standard deviation $=\sqrt{\text{variance}}$.

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Variance is defined as the square of the standard deviation.

By definition, variance $= (\text{standard deviation})^2$.

Given standard deviation $= 0.012$: …

Given standard deviation $=1.6$, the variance is $2.56$.

By definition, variance $=(\text{standard deviation})^2$. Here standard deviation $=1.6$ …

This difference is called the range of the series, the simplest measure of dispersion.

Step 1. Range is defined as: Range $=$ Maximum value $-$ Minimum value.

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