Chapter 13.2, Problem 20HP

To determine

To Find:The new data set by adding two data value set that do not affect the median but add $3$ to the mean.

Answer to Problem 20HP

  $6,8,11,24,28,36,38,40,49,60$

Explanation of Solution

Given:Use the stem-and-plot of plant heights. Add two data value set that do not affect the median but add $3$ to the mean.

  

Fromstem-and-leaf plot of the data

Mean:

  $\begin{array}{l}\text{Mean=}\frac{\text{sum of observation}}{\text{total number of observation}}\\ Mean=\frac{6+8+11+28+36+38+40+49}{8}\\ Mean=\frac{216}{8}=27\end{array}$

Mean= $27$

The median is

The number of observations are even

Therefore

  $\begin{array}{l}6,8,11,28,36,38,40,49\\ median=\frac{28+36}{2}=32cm\end{array}$

Median= $32cm$

We have to add two data set values which makes median the same but also add $3$ to the mean.

Therefore

  $\begin{array}{l}\text{New mean= old mean+3}\\ \text{=27+3}\\ \text{=30}\end{array}$

By adding two more numbers

The number of observation becomes $10$

Then

New total sum of data = new mean (number of data)

  $\begin{array}{l}=30\times 10\\ =300\end{array}$

Now

The difference is

Difference = new total sum − old total sum

  $\begin{array}{l}300-216\\ =84\end{array}$

Now

Let the number be $x$

Second number = $84-x$

Then

It make sure the number have to be lesser than $28$ and greater than $36$

So

Taking

  $\begin{array}{l}x=24\\ 84-x=84-24=60\end{array}$

Hence the new data set with mean = $30$ and median = $32$ is

  $6,8,11,24,28,36,38,40,49,60$