Chapter 3.3, Problem 30E

Nuclear power: The following table presents the number of nuclear reactors in a recent year in each country that had one or more reactors.

a. Find the first and third quartiles of these data.

b. Find the median of these data.

c. Find the upper and lower outlier boundaries.

d. Which countries are outliers?

e. Construct a boxplot for these data.

f. Describe the shape of this distribution.

g. What is the 45th percentile?

h. What is the 88th percentile?

i. India has 20 nuclear reactors. What percentile is this?

To determine

Find the first and the third quartiles of the data.

Answer to Problem 30E

The first and the third quartiles are 2 and 18 respectively.

Explanation of Solution

Calculation:

• In recent years the number of nuclear reactors which countries had one or more reactors are given.

Three quartiles:

• The first quartile separates the lowest 25% of the observations from the other 75% of the observations. The first quartile is denoted by $Q_1$.
• The second quartile separates the lower 50% of the observations from the other 50% of the observations. The second quartile is denoted by $Q_2$. The second quartile is same as median.
• The third quartile separates the lowest 75% of the observations from the other 25% of the observations. The third quartile is denoted by $Q_3$.
• Procedure for finding the first and the third quartile:
• Step 1: The observations should be arranged in increasing order.
• Step 2: The size of the data is n.
•                                     For finding first quartile, $L=0.25n$
•                                     For finding third quartile, $L=0.75n$
• Step 3: The quartile will be the average of the observation of the position L and the observation in position $L+1$, if L is a whole number. If the L value is not a whole number, the next higher whole number will be considered. The quartile is the observation in the position of rounded-up value.
• The observations are arranged in increasing order:

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

• The size of the data is $n=30$.
• For finding first quartile,
• $\begin{array}{c}L=0.25\times 30\\ =7.5\end{array}$
• Here, 7.5 is not a whole number, hence the 1^st quartile will be the observation in the 8^th position.
• From the arranged observations the first quartile is 2.
• For finding third quartile,
• $\begin{array}{c}L=0.75\times 30\\ =22.5\end{array}$
• Here, 22.5 is not a whole number, hence the 3^rd quartile will be the observation in the 23^rd position.
• From the arranged observations the third quartile is 18.

Hence, the first and the third quartiles are 2 and 18 respectively.

To determine

Find the median of the data.

Answer to Problem 30E

The median of the data is 5.5.

Explanation of Solution

Calculation:

Median:

Let $x_1,x_2\mathrm{...}{x}_n$ be n values.

The steps for finding the median:

• The all data values should be arranged in ascending order.
• If the total number of data values, n is odd, then the median will be the middle value or if n is even, then the median will be the average of middle two values.
• The observations are arranged in increasing order:
•   

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

• The size of the data is $n=30$.
• Hence, the sample size is even. Therefore, the median is the average of 15^th and 16^th observation.
• From the arranged observations the median is $\frac{5+6}{2}=5.5$

Thus, the median of the data is 5.5.

To determine

Find the upper and lower outlier boundaries.

Answer to Problem 30E

The upper and lower outlier boundaries are –22 and 42 respectively.

Explanation of Solution

Calculation:

Interquartile range:

• The interquartile range is the difference between the third quartile and first quartile. For detecting outlier this measure can be used.
• Interquartile range can be found as, $IQR=Q_3-Q_1$. Where, the first quartile is denoted by $Q_1$ and third quartile is denoted by $Q_3$.

From part (a), the first and the third quartiles are 2 and 18 respectively.

• Substitute these values in the interquartile range formula,
• $\begin{array}{c}IQR=18-2\\ =16\end{array}$
• Outlier boundaries:
• Lower outlier boundary is $Q_1-1.5\times IQR$.
• Upper outlier boundary is $Q_3+1.5\times IQR$.
• Where, the first quartile is denoted by $Q_1$ and third quartile is denoted by $Q_3$.
• Substitute these values in the formulae,
• $\begin{array}{c}\text{Lower outlier boundary}=2-1.5\times 16\\ =2-24\\ =-22\end{array}$.
• $\begin{array}{c}\text{Upper outlier boundary}=18+1.5\times 16\\ =18+24\\ =42\end{array}$

Thus, the upper and lower outlier boundaries are –22 and 42 respectively.

To determine

Find the countries which are outlier.

Answer to Problem 30E

Japan, France and United States are the outlier countries.

Explanation of Solution

• Condition for outlier:
• If any observation is less than the lower outlier boundary, the observation will be outlier.
• If any observation is greater than the upper outlier boundary, the observation will be outlier.

From part (c), the lower and upper outlier boundaries are –22 and 42 respectively.

• The observations are arranged in increasing order:
•   

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

There are 3 values which are greater than the upper outlier boundary that is, $50,58,100>42$. Other all values are less than the upper outlier boundary and greater than the lower outlier boundary. Hence, 50, 58 and 100 are outliers.

The countries corresponding to the three values are,

Value	Country
50	Japan
58	France
100	United States

Hence, Japan, France and United States are the outlier countries.

To determine

Draw a boxplot of the data.

Answer to Problem 30E

The boxplot is given below,

Explanation of Solution

Calculation:

• Boxplot:

Software procedure:

• Step-by-step procedure to draw a boxplot using the MINITAB software:
• Choose Graph > Boxplot.
• Choose Simple. Click OK.
• In Graph variables, enter the data of Number of reactors.
• Click OK.

Output using the MINITAB software is given below:

• From the MINITAB output, it is clear that there are three outliers in the data.

To determine

Explain the shape of the distribution.

Answer to Problem 30E

The data is right skewed.

Explanation of Solution

The rule for determining the skewness from the Boxplot:

• The data is right skewed if the median is closer to the 1^st quartile than the 3^rd quartile or the upper whisker is longer than the lower whisker.
• The data is left skewed if the median is closer to the 3^rd quartile than the 1^st quartile or the lower whisker is longer than the upper whisker.
• The data is approximately symmetric if the median is the middle point of 1^st quartile and the 3^rd quartile or the length of the upper whisker and the lower whisker is approximately same.

From the Boxplot of part (e), is clear that the median is closer to the 1^st quartile than the 3^rd quartile or the upper whisker is longer than the lower whisker.

By using the above rule, it can be said that the data is right skewed.

To determine

Find the 45^th percentile of the data.

Answer to Problem 30E

The 45^th percentile is 4.

Explanation of Solution

Calculation:

p^th percentile:

The p^th percentile separates the lowest p% of the observations from the highest $\left(100-p\right)\%$ of the observations. $1<p<100$ .

Procedure for finding p^th percentile:

• Step 1: The observations should be arranged in increasing order.
• Step 2: The size of the data is n.

                              For finding p^th percentile, $L=\frac{p}{100}n$

• Step 3: The p^th percentile will be the average of the observation of the position L and the observation in position $L+1$, if L is a whole number. If the L value is not a whole number, the next higher whole number will be considered. The p^th percentile is the observation in the position of rounded-up value.
• The observations are arranged in increasing order:

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

• The size of the data is $n=30$.
• For finding 45^th percentile,
• $\begin{array}{c}L=\frac{45}{100}\times 30\\ =13.5\end{array}$
• Here, 13.5 is not a whole number, hence the 45^th percentile will be the value of 14^th position.
• From the arranged observations the 45^th percentile is 4.

Hence, the 45^th percentile is 4.

To determine

Find the 88^th percentile of the data.

Answer to Problem 30E

The 88^th percentile is 33.

Explanation of Solution

Calculation:

• The observations are arranged in increasing order:

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

• The size of the data is $n=30$.
• For finding 88^th percentile,
• $\begin{array}{c}L=\frac{88}{100}\times 30\\ =26.4\end{array}$
• Here, 26.4 is not a whole number, hence the 88^th percentile will be the observation in the 27^th position.
• From the arranged observations the 88^th percentile is 33.

Hence, the 88^th percentile is 33.

To determine

Find the percentile for the given data.

Answer to Problem 30E

The value is of 82^nd percentile.

Explanation of Solution

Calculation:

It is given that India has 20 nuclear reactors.

Procedure for finding the percentile for a given observation:

• The observations should be arranged in ascending order.
• If the observation is x, then the percentile of the observation is,

                                       $\text{Perentile}=100\times \frac{\left(\text{Number of values less than }x\right)+0.5}{\text{Number of values in the data set}}$

If the value is not a whole number the nearest whole number will be the percentile.

• The observations are arranged in increasing order:

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, $x=20$.

From the observations, it is clear that there are 24 observations which are less than 20. Total number of observation is 30.

Hence, the percentile is,

$\begin{array}{c}\text{Perentile}=100\times \frac{24+0.5}{\text{30}}\\ =\frac{24.5}{30}\times 100\\ =81.67\end{array}$

• Here, 81.67 is not a whole number, hence value is of 82^nd percentile.

Hence, the value is of 82^nd percentile.