Q.Find the mean and standard deviation using the short-cut method for the following data: $x_i$: $60, 61, 62, 63, 64, 65, 66, 67, 68$; $f_i$: $2, 1, 12, 29, 25, 12, 10, 4, 5$ OR Find the mean and variance for the following frequency distribution: Class: $0\text{-}10$, $10\text{-}20$, $20\text{-}30$, $30\text{-}40$, $40\text{-}50$; Frequency: $5, 8, 15, 16, 6$

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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