Chapter 3.3, Problem 25E

Hazardous waste: Following is a list of the number of hazardous waste sites in each of the 50 states of the United States in a recent year. The list has been sorted into numerical order.

1. a. Find the first and third quartiles of these data.
2. b. Find the median of these data.
3. c. Find the upper and lower outlier boundaries.
4. d. Are there any outliers? If so, list them.
5. e. Construct a boxplot for these data.
6. f. Describe the shape of this distribution.
7. g. What is the 30th percentile?
8. h. What is the 85th percentile?
9. i. The state of Georgia has 16 hazardous waste sites. What percentile is this?

To determine

Find the first and the third quartiles.

Answer to Problem 25E

The first and the third quartiles are 11 and 32 respectively.

Explanation of Solution

Calculation:

• The number of hazardous waste sites of 50 states of United States is 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:

0	1	2	2	3	5	6	9	9	9
9	9	11	12	12	12	12	12	13	13
14	14	14	14	15	15	15	16	19	19
20	21	25	26	29	30	32	32	32	38
40	48	49	49	52	67	86	97	97	116

• The size of the data is $n=50$.
• For finding first quartile,
• $\begin{array}{c}L=0.25\times 50\\ =12.5\end{array}$
• Here, 12.5 is not a whole number, hence the 1^st quartile will be the observation in the 13^th position.
• From the arranged observations the first quartile is 11.
• For finding third quartile,
• $\begin{array}{c}L=0.75\times 50\\ =37.5\end{array}$
• Here, 37.5 is not a whole number, hence the 3^rd quartile will be the observation in the 38^th position.
• From the arranged observations the third quartile is 32.

Hence, the first and the third quartiles are 11 and 32 respectively.

To determine

Find the median of the data.

Answer to Problem 25E

The median of the data is 15.

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:
•   

0	1	2	2	3	5	6	9	9	9
9	9	11	12	12	12	12	12	13	13
14	14	14	14	15	15	15	16	19	19
20	21	25	26	29	30	32	32	32	38
40	48	49	49	52	67	86	97	97	116

• The size of the data is $n=50$.
• Hence, the sample size is even. Therefore, the median is the average of 25^th and 26^th observation.
• From the arranged observations the median is $\frac{15+15}{2}=15$

Thus, the median of the data is 15.

To determine

Find the lower and upper outlier boundaries.

Answer to Problem 25E

The lower and upper outlier boundaries are –20.5 and 63.5 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 11 and 32 respectively.
• Substitute these values in the interquartile range formula,
• $\begin{array}{c}IQR=32-11\\ =21\end{array}$
• 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,
• $\begin{array}{c}\text{Lower outlier boundary}=11-1.5\times 21\\ =11-31.5\\ =-20.5\end{array}$.
• $\begin{array}{c}\text{Upper outlier boundary}=32+1.5\times 21\\ =32+31.5\\ =63.5\end{array}$

Thus, the lower and upper outlier boundaries are –20.5 and 63.5 respectively.

To determine

Find the outliers.

Answer to Problem 25E

There are 5 outliers and are  $67,86,97,97,116$.

Explanation of Solution

Calculation:

• 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 –20.5 and 63.5 respectively

In the given data, there are five values which are greater than the upper outlier boundary, that is, $67,86,97,97,116>63.5$ . Other observations are greater than the lower outlier boundary and less than the upper outlier boundary.

Hence, there are 5 outliers and are $67,86,97,97,116>63.5$.

To determine

Draw a boxplot of the data.

Answer to Problem 25E

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 Data.
• Click OK.

Output using the MINITAB software is given below:

• From the MINITAB output, it is clear that there were  five outliers in the data.

To determine

Explain the shape of the distribution.

Answer to Problem 25E

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, in other words, 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 30^th percentile of the data.

Answer to Problem 25E

The 30^th percentile is 12.

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. Where $1<p<100$ .

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:

0	1	2	2	3	5	6	9	9	9
9	9	11	12	12	12	12	12	13	13
14	14	14	14	15	15	15	16	19	19
20	21	25	26	29	30	32	32	32	38
40	48	49	49	52	67	86	97	97	116

• The size of the data is $n=50$.
• For finding 30^th percentile,
• $\begin{array}{c}L=\frac{30}{100}\times 50\\ =15\end{array}$
• Here, 15 is a whole number, hence the 30^th percentile will be the average of the observations in the position of 15^th and 16^th.
• From the arranged observations the 30^th percentile is $\frac{12+12}{2}=12$.

Hence, the 30^th percentile is 12.

To determine

Find the 85^th percentile of the data.

Answer to Problem 25E

The 85^th percentile is 49.

Explanation of Solution

Calculation:

• The observations are arranged in increasing order:

0	1	2	2	3	5	6	9	9	9
9	9	11	12	12	12	12	12	13	13
14	14	14	14	15	15	15	16	19	19
20	21	25	26	29	30	32	32	32	38
40	48	49	49	52	67	86	97	97	116

• The size of the data is $n=50$.
• For finding 85^th percentile,
• $\begin{array}{c}L=\frac{85}{100}\times 50\\ =42.5\end{array}$
• Here, 42.5 is not a whole number, hence the 85^th percentile will be the observation in the 43^rdposition.
• From the arranged observations the 85^th percentile is 49.

Hence, the 85^th percentile is 49.

To determine

Find the percentile for the given hazardous waste sites.

Answer to Problem 25E

The value is of 55^th percentile.

Explanation of Solution

Calculation:

It is given that the state of Georgia has 16 hazardous waste sites

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:

0	1	2	2	3	5	6	9	9	9
9	9	11	12	12	12	12	12	13	13
14	14	14	14	15	15	15	16	19	19
20	21	25	26	29	30	32	32	32	38
40	48	49	49	52	67	86	97	97	116

Here, $x=16$.

From the observations, it is clear that there are 27 observations which are less than 16. Total number of observation is 50.

Hence, the percentile is,

$\begin{array}{c}\text{Perentile}=100\times \frac{27+0.5}{\text{50}}\\ =\frac{27.5}{50}\times 100\\ =55\end{array}$

• Here, 55 is a whole number.

Hence, the value is of 55^th percentile.