The problem is to find the standard deviation and variance of a given set of data manually. The data set, the deviations from the mean, and the squared deviations are already provided in a table. We are also given the mean $\bar{x} = 17.08$.

1. Problem Description

The problem is to find the standard deviation and variance of a given set of data manually. The data set, the deviations from the mean, and the squared deviations are already provided in a table. We are also given the mean $\bar{x} = 17.08$.

2. Solution Steps

The formula for the sample standard deviation $s$ is:

$s = \sqrt{\frac{\sum (x - \bar{x})^2}{n-1}}$

where $x$ represents each data point, $\bar{x}$ is the sample mean, and $n$ is the number of data points.

From the given table, we have the following values for $x$: 25, 21, 20, 19, 19, 18, 17, 16, 15, 14, 11,

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0. The sample mean is given as $\bar{x} = \frac{205}{12} = 17.08$.

The values $(x - \bar{x})$ are also provided in the table.

The values $(x - \bar{x})^2$ are given as 62.73, 15.37, 8.53, 3.69, 3.69, 0.85, 0.01, 1.17, 4.33, 9.49, 36.97, 50.

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3.

We need to find the sum of the squared deviations, $\sum (x - \bar{x})^2$.

$\sum (x - \bar{x})^2 = 62.73 + 15.37 + 8.53 + 3.69 + 3.69 + 0.85 + 0.01 + 1.17 + 4.33 + 9.49 + 36.97 + 50.13 = 196.96$

The number of data points is $n = 12$.

Therefore, the sample standard deviation $s$ is:

$s = \sqrt{\frac{196.96}{12-1}} = \sqrt{\frac{196.96}{11}} = \sqrt{17.90545} \approx 4.23$

The variance $s^2$ is the square of the standard deviation:

$s^2 = (\sqrt{\frac{\sum (x - \bar{x})^2}{n-1}})^2 = \frac{\sum (x - \bar{x})^2}{n-1} = \frac{196.96}{11} = 17.90545 \approx 17.91$

3. Final Answer

Standard deviation: $s = 4.23$

Variance: $s^2 = 17.91$