Chapter B.2, Problem 23E

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 23E

Mean $\left(\overline{x}\right)$ , variance $\left(\nu \right)$ and standard deviation $\left(\sigma \right)$ for the given data is $24,10,3.16$ respectively.

Explanation of Solution

Given information:

Given data: $4,10,8,2$

Calculation:

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

  $\begin{array}{l}\overline{x}=\frac{4+10+8+2}{4}\\ =24\end{array}$

So mean is $24$ .

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,

  $\begin{array}{l}\nu =\frac{{\left(4-6\right)}^2+{\left(10-6\right)}^2+{\left(8-6\right)}^2+{\left(2-6\right)}^2}{4}\\ =\frac{{\left(-2\right)}^2+{\left(4\right)}^2+{\left(2\right)}^2+{\left(-4\right)}^2}{4}\\ =10\end{array}$

Variance is $10$ .

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

Now substitute the $10$ for $\nu$

  $\begin{array}{l}\sigma =\sqrt{10}\\ =3.16\end{array}$

The standard deviation is $3.16$ .