Question

Tall or short? Mr. Walker measures the heights (in inches) of the students in one of his classes.

He uses a computer to calculate the following numerical summaries: Next, Mr. Walker has his entire class stand on their chairs, which are $18$ inches off the ground. Then he measures the distance from the top of each student’s head to the floor.

(a) Find the mean and median of these measurements. Show your work.

(b) Find the standard deviation and IQR of these measurements. Show your work.

Short answer

Part (a) The new mean is $87.188$ and the new median is $87.5$

Part (b) The deviations did not change and the IQR, $Q_1$ and $Q_2$, both increase by $18$.

Step-by-step solution

Step 1. Given

Step 2. Concept

The formula used: $Mean\left(new\right)=\frac{Sumofstudentheightss\tan dingonchairs}{numberofstudents}$

Step 3. Calculation

The new mean and median both increase by $18$ and can be found in the table below.

$$\begin{gathered} Mean\left(new\right)=Sumofstudentheightss\tan dingonchairs/numberofstudents \\ =(heightoffirststudent+18)+...+(heightoflaststudent+18)/numberofstudents \\ =Sumofstudentheightss\tan dingonfloor+18\times numberofstudents/numberofstudents \end{gathered}$$$$\begin{gathered} =Mean\left(old\right)+18 \\ =69.188+18 \\ =87.188 \end{gathered}$$$$\begin{gathered} Median\left(new\right)=Median\left(old\right)+18 \\ =69.5+18 \\ =87.5 \end{gathered}$$

The new mean and median are $87.188$ and$87.5$, respectively.

Therefore, the new mean is $87.188$ and the new median is $87.5$

Step 1. Explanation

While the mean grew by $18$, the standard deviation increased by $18$ as well, resulting in the deviations remaining unchanged. $Q_1$ and $Q_3$ both rise by $18$ for the IQR, therefore their difference is the same as it was in the original data set.