Chapter 2.1, Problem 21E

(a)

To determine

To find: the mean and median of the students’ heights in feet.

Answer to Problem 21E

The new mean is 5.77 feet and the new median is 5.79 feet.

Explanation of Solution

Given:

  

Calculation:

We know that 1 foot = 12 inches. For converting inches into feet, we divide each measure with 12. Thus the new mean and median are obtained as follows:

  $\begin{array}{l}\text{Mean(new)=}\frac{\frac{\text{height of first student}}{\text{12}}\text{+}\mathrm{...}\text{+}\frac{\text{height of last student}}{\text{12}}}{\text{number of students}}\\ =\frac{1}{12}\left(\frac{\text{height of first student}+\mathrm{...}+\text{height of last student}}{\text{number of students}}\right)\\ =\frac{1}{12}\text{Mean}\left(old\right)\\ =\frac{1}{12}\times 69.188\\ =5.77\text{feet}\end{array}$

  $\begin{array}{l}\text{Median(new)=}\frac{\text{Median(old)}}{\text{12}}\\ =\frac{69.5}{12}\\ =5.79\text{feet}\end{array}$

Conclusion:

Therefore the new mean is 5.77 feet and the new median is 5.79 feet.

(b)

To determine

To find: the standard deviation and IQR of the students’ heights in feet.

Answer to Problem 21E

The new standard deviation and new inter quartile range of student's height are equal to 0.27 feet.

Explanation of Solution

Calculation:

Divide the original deviation by 12.

Thus the new standard deviation is obtained as follows:

  $\begin{array}{l}\text{std}\text{.dev}\text{.(new)}=\sqrt{\frac{\left({\text{first deviation(ft)}}^{\text{2}}\right)\text{+}\mathrm{...}\text{+}\left({\text{last deviation(ft)}}^{\text{2}}\right)}{\text{n-1}}}\\ =\sqrt{\frac{{\left(\frac{\text{first deviation(in)}}{\text{12}}\right)}^{\text{2}}\text{+}\mathrm{...}\text{+}{\left(\frac{\text{last deviation(in)}}{\text{12}}\right)}^2}{n-1}}\\ =\frac{1}{12}\sqrt{\frac{{\left(\text{first deviation(in)}\right)}^2+\mathrm{...}+{\left(\text{last deviation(in)}\right)}^2}{n-1}}\\ =\frac{1}{12}\text{std}\text{.dev}\text{.(old)}\\ \text{=}\frac{3.2}{12}\\ =0.27\text{feet}\end{array}$

The first and third quartiles are still the medians of the first and second halves of the data; these values must simply be converted to feet. To do this, divide the first and third quartiles of the original data set by 12:

  $\begin{array}{l}{Q}_1(new)=\frac{Q_1\left(old\right)}{12}\\ =\frac{67.75}{12}\\ =5.65\text{feet}\\ Q_3(new)=\frac{Q_3\left(old\right)}{12}\\ =\frac{71}{12}\\ =5.92\text{feet}\end{array}$

Thus the new inter quartile range is obtained as follows:

  $\begin{array}{l}\text{IQR}\left(new\right)=Q_3\left(new\right)-Q_1\left(new\right)\\ =5.92-5.65\\ =0.27\text{feet}\end{array}$

Conclusion:

The new standard deviation and new inter quartile range of student's height are equal to 0.27 feet.