Chapter 10.2, Problem 18STP

To determine

To find:The incorrect answer for the given information.

Answer to Problem 18STP

The incorrect option is D.

Explanation of Solution

Given information:The table below shows the ascending order of the total number of wins a women’s team had in basketball tournaments for the last 15 years.

Total Wins

29

31

31

32

33

33

34

34

34

34

35

36

37

39

39

Formula used:

For a given set of data

  $range=\text{largest}\text{value}-\text{least}\text{value}$

  $\begin{array}{l}\text{First quartile}=\text{median of the lower half of the data}\text{.}\hfill \\ \hfill \\ \text{Third quartile}=\text{median of the upper half of the data}\text{.}\hfill \\ \hfill \\ \text{Interquartile Range}=\text{Third quartile}-\text{First quartile}\hfill \end{array}$

The Median of a given set of values is the middle value of the ordered set.

  $Median=\left\{\begin{array}{l}{\left(\frac{n+1}{2}\right)}^{th}\text{term,}\text{if}n\text{is}\text{odd}\\ \\ \frac{{\left(\frac{n}{2}\right)}^{th}\text{term}+{\left(\frac{n}{2}+1\right)}^{th}\text{term}}{2},\text{if}n\text{is}\text{even}\end{array}\right\}$

Calculation:

Total Wins

29

31

31

32

33

33

34

34

34

34

35

36

37

39

39

The median of the given table is

  $\begin{array}{l}Median={\left(\frac{n+1}{2}\right)}^{th}term,i\text{f}\text{n}\text{is}\text{odd}\\ \text{Here}n=\text{15}\left(odd\right)\\ Median={\left(\frac{15+1}{2}\right)}^{th}term=8^{th}term\\ Median=34\end{array}$

The range of the given table is

  $\begin{array}{l}range=\text{largest}\text{value}-\text{least}\text{value}=\text{39}-\text{29}\\ range=10\end{array}$

The interquartile range of the table is

  $\begin{array}{c}\text{First quartile}=\text{median of the lower half of the data}\\ \text{=median}\left\{29,31,31,32,33,33,34\right\}\\ ={\left(\frac{7+1}{2}\right)}^{th}term=4^{th}term\\ =32\\ \\ \text{Third quartile}=\text{median of the upper half of the data}\\ \text{=median}\left\{34,34,35,36,37,39,39\right\}\\ ={\left(\frac{7+1}{2}\right)}^{th}term=4^{th}term\\ =36\\ \\ \text{Interquartile Range}=\text{Third quartile}-\text{First quartile}\\ \text{=36}-32=4\end{array}$

  $\begin{array}{l}\text{To}\text{find}\text{the}\text{outlier},\text{Mutltipy}\text{Interquartile}\text{Range}\text{by}\text{1}\text{.5}\\ \text{=1}\text{.5}\times \text{4=6}\\ \\ \text{then}\\ 32-6=26\\ 36+6=42\\ \text{Since}\text{no}\text{value}\text{lies}\text{outside}26\&\text{42}\text{therefore}\text{there}\text{are}\text{no}\text{outlier}\text{.}\end{array}$

The lower half of the given table is $\left\{29,31,31,32,33,33,34\right\}$ and the upper half of the table is $\left\{34,34,35,36,37,39,39\right\}$ , that means one half of the teams won more than 34 games and the other half won less than 34 games, therefore option A is correct.

Since Range of the given table is = 10,therefore option B is correct.

Also the list of one fourth of the given table is $\left\{29,31,31,32\right\}$ that means about one fourth of the teams won 32 or fewer games, therefore option C is correct.

And since there are no outliers for the given table, therefore option D is incorrect.

Hence the incorrect option is D.