Chapter B.2, Problem 25E

To determine

To calculate:

Mean $\left(\overline{x}\right)$ , variance $\left(\nu \right)$ and standard deviation $\left(\sigma \right)$ for the given data.

Answer to Problem 25E

Mean $\left(\overline{x}\right)$ , variance $\left(\nu \right)$ and standard deviation $\left(\sigma \right)$ for the given data is $47,226,15.03$ respectively.

Explanation of Solution

Given information:

Given data: $49,62,40,29,32,70$

Calculation:

For a data set having $n$ , the mean $\overline{x}$ is the sum of the numbers which are divided by $n$ .

  $\begin{array}{l}\overline{x}=\frac{49+62+40+29+32+70}{6}\\ =\frac{282}{6}\\ =47\end{array}$

So mean is $47$ .

The variance $\left(\nu \right)$ of a set of number with a mean of $\left(\overline{x}\right)$ is given by,

  $\nu =\frac{{\left(x_1-\overline{x}\right)}^2+{\left(x_2-\overline{x}\right)}^2+\mathrm{...}+{\left(x_n-\overline{x}\right)}^2}{n}$

Substitute the corresponding values,

  $\begin{array}{l}\nu =\frac{{\left(49-47\right)}^2+{\left(62-47\right)}^2+{\left(40-47\right)}^2+{\left(29-47\right)}^2+{\left(32-47\right)}^2+{\left(70-47\right)}^2}{6}\\ =\frac{{\left(2\right)}^2+{\left(15\right)}^2+{\left(-7\right)}^2+{\left(-18\right)}^2+{\left(-15\right)}^2+{\left(23\right)}^2}{6}\\ =\frac{1356}{6}\\ =226\end{array}$

Variance is $226$ .

The standard deviation of the data set is $\sigma =\sqrt{\nu }$

Now substitute the $226$ for $\nu$

  $\begin{array}{l}\sigma =\sqrt{226}\\ =15.03\end{array}$

The standard deviation is $15.03$ .