Chapter 10.3, Problem 24PFA

To determine

To find mean and standard deviation

Answer to Problem 24PFA

Mean is 16.9 and Standard deviation is 0.9361

Explanation of Solution

Given information:

Temperature is {-16.0, -18.3, -15.8, -16.6, -18.2, -17.4, -16.5}

Formula used:

Formula of mean,

Mean= $\frac{{\displaystyle \sum _{i=1}^n{X}_i}}{n}$

Where, n=total number of terms,

  ${\displaystyle \sum _{i=1}^n{X}_i}$ = sum of all terms Formula of standard deviation,

Standard deviation, ${\sigma }_{}=\sqrt{\frac{{\displaystyle \sum _{}^{}{(x_i^{}-\mu )}^2}}{N}}$

Where $x_i^{}$ = each value from data

  $\mu$ =mean of data

  $N$ = size of data

Calculation:

For mean,

Substituting values,

= $\frac{-16.0+\text{(}-18.3)+\text{(}-15.8)+\text{(}-16.6)+\text{(}-18.2)+\text{(}-17.4)+(-16.5)}{7}$

On solving,

= $\frac{-118.8}{7}$

On solving,

=-16.9

values	Difference= $x_{}^{}-\mu$	${(x_{}^{}-\mu )}^2$
-16.0	0.97	0.94
-18.3	-1.32	1.76
-15.8	1.17	1.37
-16.6	0.37	0.13
-18.2	-1.22	1.50
-17.4	-0.42	0.18
-16.5	0.47	0.22
	${\displaystyle \sum _{}^{}{(x_{}^{}-\mu )}^2}$ =6.134

For Standard deviation substituting value,

  $\sigma$ = $\sqrt{\frac{6.134}{7}}$

On solving,

  $\sigma$ =0.9361