Chapter 14.5, Problem 35E

To determine

Find the mean and the standard deviation.

Answer to Problem 35E

  $8.25,19.377$

Explanation of Solution

Given information:

Given: $\left\{44,72,58,61,71,49,55,68\right\}$

Calculation:

Consider the data,

  $\left\{44,72,58,61,71,49,55,68\right\}$

Now find mean,

  $\begin{array}{l}mean=\frac{44+72+58+61+71+49+55+68}{8}=\frac{478}{8}\\ =59.75\end{array}$

Now,

  $\begin{array}{cc}value& dis\tan ceoffrommean\left(59.75\right)\\ 44& 15.75\\ 72& 12.25\\ 58& 1.75\\ 61& 1.25\\ 71& 11.25\\ 49& 10.75\\ 55& 4.75\\ 68& 8.25\end{array}$

  $\begin{array}{l}sumofdis\tan ce=66\\ meandeviation=66/n\\ =66/8=8.25\end{array}$

  $meandeviation=8.25$

Now find standard deviation,

Add and square the given data and divide the sum by number of items,

  $\begin{array}{l}44+72+58+61+71+49+55+68\\ ={\left(478\right)}^2\\ =228484/8\\ =28560.5\end{array}$

Now square and sum the original data numbers,

  $\begin{array}{l}\left(44\times 44\right)+\left(72\times 72\right)+\left(58\times 58\right)+\left(61\times 61\right)+\left(71\times 71\right)+\left(49\times 49\right)+\left(55\times 55\right)+\left(68\times 68\right)=25932\\ now,\\ 28560.5-25932=2628.5\\ S\tan darddeviation=\frac{2628.5}{n-1}=\frac{2628.5}{7}=375.5\\ S\tan darddeviation\left(\sigma \right)=\sqrt{375.5}=19.377\end{array}$

Hence, mean and the standard deviation is $8.25,19.377$