91Ó°ÊÓ

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:$$\begin{gathered} Mean\left(new\right)=\frac{{\displaystyle \frac{heightoffirststudent}{12}}+.......+{\displaystyle \frac{heightoffirststudent}{12}}}{numberofstudents} \\ =\frac{1}{12}(\frac{heightoffirststudent+......+heightoflaststudent}{numberofstudents}) \\ =\frac{1}{12}Mean\left(old\right) \\ =\frac{1}{12}×69.188 \\ =5.77feet \\ Median\left(new\right)=\frac{Median\left(old\right)}{12} \\ =\frac{69.5}{12} \\ =5.79feet \end{gathered}$$

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.