Chapter 3.3, Problem 26E

Cholesterol levels: The National Health and Nutrition Examination Survey (NHANES) measured the serum HDL cholesterol levels in a large number of women. Following is a sample of 40 HDL levels (in milligrams per deciliter) that are based on the results of that survey. They have 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 20th percentile?
8. h. What is the 67th percentile?
9. i. One woman had a cholesterol level of 58. What percentile is this?

To determine

Find the first and the third quartiles.

Answer to Problem 26E

The first and the third quartiles are 37 and 65.5 respectively.

Explanation of Solution

Calculation:

• The given information is a sample of 40 HDL cholesterol level obtained from the survey.

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:

27	28	30	32	34	36	37	37	37	37
37	40	45	47	48	49	53	53	54	56
57	58	61	62	63	63	64	64	64	65
66	70	72	73	73	74	80	80	81	84

• The size of the data is $n=40$.
• For finding first quartile,
• $\begin{array}{c}L=0.25\times 40\\ =10\end{array}$
• Here, 10 is a whole number, hence the 1^st quartile will be the observation of the average of 10^th and the 11^th observations.
• From the arranged observations the first quartile is $\frac{37+37}{2}=37$ .
• For finding third quartile,
• $\begin{array}{c}L=0.75\times 40\\ =30\end{array}$
• Here, 30 is a whole number, hence the 3^rd quartile will be the observation of the average of 30^th and the 31^st observations.
• From the arranged observations the third quartile is $\frac{65+66}{2}=65.5$.

Hence, the first and the third quartiles are 37 and 65.5 respectively.

To determine

Find the median of the data.

Answer to Problem 26E

The median of the data is 56.5.

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:
•   

27	28	30	32	34	36	37	37	37	37
37	40	45	47	48	49	53	53	54	56
57	58	61	62	63	63	64	64	64	65
66	70	72	73	73	74	80	80	81	84

• The size of the data is 40. That is, $n=40$.
• Hence, the sample size is even. Therefore, the median is the average of 20^th and 21^st observation.
• From the arranged observations the median is $\frac{56+57}{2}=56.5$

Thus, the median of the data is 56.5.

To determine

Find the lower and upper outlier boundaries.

Answer to Problem 26E

The lower and upper outlier boundaries are –5.75 and 108.25 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 37 and 65.5 respectively.
• Substitute these values in the interquartile range formula,
• $\begin{array}{c}IQR=65.5-37\\ =28.5\end{array}$
• 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,
• $\begin{array}{c}\text{Lower outlier boundary}=37-1.5\times 28.5\\ =37-42.75\\ =-5.75\end{array}$.
•  $\begin{array}{c}\text{Upper outlier boundary}=65.5+1.5\times 28.5\\ =65.5+42.75\\ =108.25\end{array}$

Thus, the lower and upper outlier boundaries are –5.75 and 108.25 respectively.

To determine

Find the outliers.

Answer to Problem 26E

There is no outlier in the data.

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 –5.75 and 108.25 respectively.

In the given data, all observations are greater than the lower outlier boundary and less than the upper outlier boundary.

Hence, there is no outlier in the data.

To determine

Draw a boxplot of the data.

Answer to Problem 26E

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 Cholesterol level.
• Click OK.

Output using the MINITAB software is given below:

• From the MINITAB output, it is clear that there is no outlier in the data.

To determine

Explain the shape of the distribution.

Answer to Problem 26E

The data is slightly left skewed or it can be said that approximately symmetric.

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 slightly closer to the 3^rd quartile than the 1^st quartile, in other words, the upper whisker is slightly longer than the lower whisker.

By using the above rule, it can be said that the data is slightly left skewed or it can be said that approximately symmetric.

To determine

Find the 20^th percentile of the data.

Answer to Problem 26E

The 20^th percentile is 37.

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:

27	28	30	32	34	36	37	37	37	37
37	40	45	47	48	49	53	53	54	56
57	58	61	62	63	63	64	64	64	65
66	70	72	73	73	74	80	80	81	84

• The size of the data is $n=40$.
• For finding 20^th percentile,
• $\begin{array}{c}L=\frac{20}{100}\times 40\\ =8\end{array}$
• Here, 8 is a whole number, hence the 20^th percentile will be the average of the observations in the position of 8^th and 9^th.
• From the arranged observations the 20^th percentile is $\frac{37+37}{2}=37$.

Hence, the 20^th percentile is 37.

To determine

Find the 67^th percentile of the data.

Answer to Problem 26E

The 67^th percentile is 64.

Explanation of Solution

Calculation:

• The observations are arranged in increasing order:

27	28	30	32	34	36	37	37	37	37
37	40	45	47	48	49	53	53	54	56
57	58	61	62	63	63	64	64	64	65
66	70	72	73	73	74	80	80	81	84

• The size of the data is $n=40$.
• For finding 67^th percentile,
• $\begin{array}{c}L=\frac{67}{100}\times 40\\ =26.80\end{array}$
• Here, 26.8 is not a whole number, hence the 67^th percentile will be the observation in the 27^th position.
• From the arranged observations the 67^th percentile is 64.

Hence, the 67^th percentile is 64.

To determine

Find the percentile for the given cholesterol level.

Answer to Problem 26E

The value is of 54^th percentile.

Explanation of Solution

Calculation:

It is given that a woman had a cholesterol level of 58.

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:

27	28	30	32	34	36	37	37	37	37
37	40	45	47	48	49	53	53	54	56
57	58	61	62	63	63	64	64	64	65
66	70	72	73	73	74	80	80	81	84

Here, $x=58$.

From the observations, it is clear that there are 21 observations which are less than 58. Total number of observation is 40.

Hence, the percentile is,

$\begin{array}{c}\text{Perentile}=100\times \frac{21+0.5}{\text{40}}\\ =\frac{21.5}{40}\times 100\\ =53.75\end{array}$

• Here, 53.75 is not a whole number, hence value is in 54^th percentile.

Hence, the value is of 54^th percentile.