Chapter 2.2, Problem 2.42E

(a)

To determine

To Calculate: the mean and median.

Answer to Problem 2.42E

  $\begin{array}{c}\text{Mean}=7.27\\ \text{Median}=7.29\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}\mu =69.188\\ \sigma =3.20\\ \text{Min}\text{. = 61}\text{.5}\\ {\text{Q}}_1=67.75\\ \text{Median = 69}\text{.5}\\ {\text{Q}}_3=71\\ \text{Max}\text{.=74}\text{.5}\end{array}$

Calculation:

1 feet = 12 inches, divided mean and median by 12

Mean:

  $\begin{array}{c}=\frac{87.188}{12}\\ =7.27\end{array}$

Median:

  $\begin{array}{c}=\frac{87.5}{12}\\ =7.29\end{array}$

(b)

To determine

To Calculate: the standard deviation and IQR.

Answer to Problem 2.42E

  $\begin{array}{c}\text{Standard deviation}=0.27\\ \text{IQR}=0.27\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}\mu =69.188\\ \sigma =3.20\\ \text{Min}\text{. = 61}\text{.5}\\ {\text{Q}}_1=67.75\\ \text{Median = 69}\text{.5}\\ {\text{Q}}_3=71\\ \text{Max}\text{.=74}\text{.5}\end{array}$

Calculation:

1 feet = 12 inches, divided standard deviation and IQR by 12

Standard deviation:

  $\begin{array}{c}=\frac{3.20}{12}\\ =0.27\end{array}$

IQR:

  $\begin{array}{c}=\frac{3.25}{12}\\ =0.27\end{array}$