Chapter 13.3, Problem 10PPS

To determine

To Find:Measure of variation of his golf scores

Answer to Problem 10PPS

Range = $10$

Upper quartile = $3$

Lower quartile = $-3$

Interquartile range= $6$

No outlier

Explanation of Solution

Given:Rondell’s golf score at the end of the high school golf tournament is $-1,-2,-4,-6,3,-1,-3$

Find the measure of variation of his golf scores. Explain what measures of variation tell you about the data,

Finding range Highest age = $4$

Lowest age = $-6$

Then

Range = highest age − lowest age

  $\begin{array}{l}=4-(-6)\\ =10\end{array}$

The range is $10$ .

Median = $-1$

Finding Upper quartile and lower quartile of rose bowl

The median of the data is $-1$

Arranging the data in the ascending order

  $-6,-3,-2,-1,-1,3,4$

Upper quartile

  $\begin{array}{l}-6,-3,-2,-1,-1,\underset{\_}{3},4\\ =3\end{array}$

Lower quartile

  $\begin{array}{l}-6,\underset{\_}{-3},-2,-1,-1,3,4\\ =-3\end{array}$

For finding interquartile range Interquartile range = Upper quartile − lower quartile

  $\begin{array}{l}=3-(-3)\\ =6\end{array}$

Now,

Multiply the interquartile range by $1.5$

We get

  $\begin{array}{l}=6\times 1.5\\ =9\end{array}$

Now subtract $9$ from the lower quartile and add $9$ to the upper quartile

I.e.

  $\begin{array}{l}-3-9=-12\\ 3+9=12\end{array}$

Thus

There is no outlier in the data