Tall or short? Refer to Exercise 19. Mr. Walker converts his students’ original heights from inches to feet.

(a) Find the mean and median of the students’ heights in feet. Show your work.

(b) Find the standard deviation and IQR of the students’ heights in feet. Show your work.

The formula used: $z=\frac{x-mean}{s\tan darddeviation}$

$1$foot equals $12$inches, as we all know. We divide each measurement by $12$converting inches to feet. As a result, the new mean and median are as follows:$$\begin{gathered} Mean\left(new\right)=\frac{{\displaystyle \frac{heightoffirststudent}{12}}+.......+{\displaystyle \frac{heightoffirststudent}{12}}}{numberofstudents} \\ =\frac{1}{12}(\frac{heightoffirststudent+......+heightoflaststudent}{numberofstudents}) \\ =\frac{1}{12}Mean\left(old\right) \\ =\frac{1}{12}\times 69.188 \\ =5.77feet \\ Median\left(new\right)=\frac{Median\left(old\right)}{12} \\ =\frac{69.5}{12} \\ =5.79feet \end{gathered}$$

As a result, the new mean and median are 5.77 and 5.79 feet, respectively.

Subtract 12 from the original deviance.

As a result, the revised standard deviation is as follows: $$\begin{gathered} std.dev.\left(new\right)=\sqrt{\frac{{\left(firstdeviation\left(ft\right)\right)}^2+....+{\left(lastdeviation(ft\right))}^2}{n-1}} \\ =\sqrt{\frac{{\left(\frac{firstdeviation\left(in\right)}{12}\right)}^2+.....+{\left(\frac{lastdeviation\left(in\right)}{12}\right)}^2}{n-1}} \\ =\frac{1}{12}std.dev.\left(old\right) \end{gathered}$$

$$\begin{gathered} =\frac{3.2}{12} \\ ==0.27feet \end{gathered}$$

The medians of the first and second half of the data are still the first and third quartiles; the devalues must only be changed to feet. Divide the first and third quartiles of the original data set by $12$to achieve this:

$$\begin{gathered} Q_3\left(new\right)=\frac{Q_1\left(old\right)}{12} \\ =\frac{67.75}{12} \\ =5.65feet \\ {Q}_3\left(new\right)=\frac{Q_3\left(old\right)}{12} \\ =\frac{71}{12} \\ =5.92feet \end{gathered}$$

Thus the new inter quartile range is obtained as follows:

$IQR\left(new\right)=Q_3\left(new\right)-Q_1\left(new\right)$

$$\begin{gathered} =5.92-5.65 \\ =0.27feet \end{gathered}$$

The revised standard deviation and interquartile range of the height of the students are$0.27$feet.