For the data set [tex]6, 5, 10, 11, 13[/tex], the mean, [tex]\bar{x}[/tex], is 9. What is the standard deviation?

The formula for the sample standard deviation is [tex]s=\sqrt{\frac{1}{n-1} \Sigma (x-\bar{x})^2}[/tex].

Use the table to help you.

[tex]\[
\begin{tabular}{|c|c|c|}
\hline
$x$ & $x-\bar{x}$ & $(x-\bar{x})^2$ \\
\hline
6 & -3 & 9 \\
\hline
5 & -4 & 16 \\
\hline
10 & 1 & 1 \\
\hline
11 & 2 & 4 \\
\hline
13 & 4 & 16 \\
\hline
\multicolumn{3}{|c|}{Sum $=46$} \\
\hline
\end{tabular}
\][/tex]

Round your answer to the nearest tenth.

A. 9.2



Answer :

To determine the sample standard deviation for the given data set [tex]\(6, 5, 10, 11, 13\)[/tex], we follow these steps.

Given:
- Data points: [tex]\(6, 5, 10, 11, 13\)[/tex]
- Mean [tex]\(\bar{x} = 9\)[/tex]

First, we calculate the sum of squared differences from the mean using the formula [tex]\(\Sigma (x - \bar{x})^2\)[/tex].

Let's break it down using the table provided:

[tex]\[ \begin{array}{|c|c|c|} \hline x & x - \bar{x} & (x - \bar{x})^2 \\ \hline 6 & 6 - 9 = -3 & (-3)^2 = 9 \\ \hline 5 & 5 - 9 = -4 & (-4)^2 = 16 \\ \hline 10 & 10 - 9 = 1 & 1^2 = 1 \\ \hline 11 & 11 - 9 = 2 & 2^2 = 4 \\ \hline 13 & 13 - 9 = 4 & 4^2 = 16 \\ \hline \end{array} \][/tex]

Summing up the squares of the differences:

[tex]\[ \Sigma (x - \bar{x})^2 = 9 + 16 + 1 + 4 + 16 = 46 \][/tex]

Next, we use the formula for the sample standard deviation, [tex]\(s = \sqrt{\frac{1}{n-1} \Sigma (x - \bar{x})^2}\)[/tex], where [tex]\(n\)[/tex] is the number of data points.

Here [tex]\(n = 5\)[/tex], so [tex]\(n - 1 = 4\)[/tex].

Plugging in the values:

[tex]\[ s = \sqrt{\frac{46}{4}} \][/tex]

Calculating the expression inside the square root:

[tex]\[ \frac{46}{4} = 11.5 \][/tex]

Now, taking the square root:

[tex]\[ s = \sqrt{11.5} \approx 3.391164991562634 \][/tex]

Finally, we round this result to the nearest tenth:

[tex]\[ s \approx 3.4 \][/tex]

Thus, the standard deviation for the given data set is [tex]\(\boxed{3.4}\)[/tex].