Chapter 3.3, Problem 36E

a)

To determine

To find the first and third quartiles of the given data.

Answer to Problem 36E

First quartile = 2 and third quartile = 18.25

Explanation of Solution

Formula:

First quartile Q1:

  $Q_1=\frac{1}{4}{(n+1)}^{th}value$

Third quartile Q3:

  $Q_3=\frac{3}{4}{(n+1)}^{th}value$

Calculation:

Data sorted in ascending order:

$x$
1	3	15
1	4	16
1	4	18
1	4	19
2	5	20
2	6	23
2	7	33
2	8	50
2	9	58
2	10	100

Here, n = 24

First need to find First quartile and third quartile

First Quartile:

  $\begin{array}{l}$ Q_1=\frac{1}{4}{(30+1)}^{th}value$$ Q_1=\frac{1}{4}{(31)}^{th}value$$ Q_1={(7.75)}^{th}value$Q_1=2+0.75\times (2-2)\\ Q_1=2\end{array}$

First quartile is 2

Third quartile:

  $\begin{array}{l}$ Q_3=\frac{3}{4}{(n+1)}^{th}value$$ Q_3=\frac{3}{4}{(30+1)}^{th}value$$ Q_3=\frac{3}{4}{(31)}^{th}value$$ Q_3={(23.25)}^{th}value$Q_3=18+0.25\times (19-18)\\ Q_3=18.25\end{array}$

Third quartile is 18.25

b)

To determine

To find median of the data.

Answer to Problem 36E

Median is 5.5

Explanation of Solution

Formula:

  $median={\left(\frac{n+1}{2}\right)}^{th}value$

Calculation:

  $\begin{array}{l}median={\left(\frac{30+1}{2}\right)}^{th}value\\ median={(}^{15.5}value\\ median=\frac{5+6}{2}\\ median=5.5\end{array}$

c)

To determine

To find upper and lower outlier boundaries

Answer to Problem 36E

Lower outlier boundary is -22.375

Upper outlier boundary is 42.625

Explanation of Solution

  $\begin{array}{l}{Q}_1=2\\ Q_3=18.25\end{array}$

Formula:

IQR:

  $IQR=Q_3-Q_1$

Calculation:

  $\begin{array}{l}IQR=Q_3-Q_1\hfill \\ IQR=\text{ 7}\text{.5-3}\hfill \\ IQR=\text{ 4}\text{.25}\hfill \end{array}$

Therefore,

  $\begin{array}{l}Loweroutlierlimit=Q_1-1.5\times IQR\hfill \\ Loweroutlierlimit=\text{ 2}-1.5\times 16.25\hfill \\ Loweroutlierlimit=\text{ -22}\text{.375}\hfill \\ And\hfill \\ Upperoutlierlimit=Q3+1.5\times IQR\hfill \\ Upperoutlierlimit=\text{ 18}\text{.25}+1.5\times 16.25\hfill \\ Upperoutlierlimit=\text{ 42}\text{.625}\hfill \end{array}$

d)

To determine

To find the list of outlier countries.

Answer to Problem 36E

The outlier countries are Japan, France and United states.

Explanation of Solution

Outliers are those values which are less than Q1-1.5 x IQR and greater than Q3+1.5 x IQR.

Here

Lower outlier boundary is -22.375

Upper outlier boundary is 42.625

There is no any value less than -22.375. But there are three values larger than 42.625 which are 50, 58 and 100. Hence outlier points are 50, 58 and 100. Corresponding countries to the outlier points are Japan, France and United states.

e)

To determine

To construct a boxplot from given data

Explanation of Solution

Boxplot:

  

f)

To determine

To find shape of the distribution

Answer to Problem 36E

The shape of the distribution is positively-skewed.

Explanation of Solution

Here, median is close to first quartile. Hence the shape of the distribution is positively-skewed.

g)

To determine

To find 45^th percentile

Answer to Problem 36E

45^th percentile is 2.54

Explanation of Solution

  $\begin{array}{l}PercentilePosition=\frac{(n+1)P}{100}\\ PercentilePosition=\frac{(30+1)\times 45}{100}\\ PercentilePosition=10.54\end{array}$

Therefore,

  $\begin{array}{l}{P}_{45}=2+0.54\times (3-2)\\ P_{45}=2.54\end{array}$

h)

To determine

To find 88^th percentile

Answer to Problem 36E

88^th percentile is 39.44

Explanation of Solution

  $\begin{array}{l}PercentilePosition=\frac{(n+1)P}{100}\\ PercentilePosition=\frac{(30+1)\times 88}{100}\\ PercentilePosition=27.28\end{array}$

Therefore,

  $\begin{array}{l}{P}_{88}=33+0.28\times (50-27)\\ P_{88}=39.44\end{array}$

i)

To determine

To find percentile rank of 20 nuclear reactors

Answer to Problem 36E

Percentile rank of 20 nuclear reactor is 80%.

Explanation of Solution

Formula:

  $Percentilerankofx=\frac{no.ofvaluesbelowx}{n}\times 100$

Calculation:

  $\begin{array}{l}Percentilerankof20=\frac{24}{30}\times 100\\ Percentilerankof20=80\%\end{array}$