Essential Statistics
Essential Statistics
2nd Edition
ISBN: 9781259570643
Author: Navidi
Publisher: MCG
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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.

Chapter 3.3, Problem 25E, Hazardous waste: Following is a list of the number of hazardous waste sites in each of the 50 states

  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?

a.

Expert Solution
Check Mark
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 Q1.
  • The second quartile separates the lower 50% of the observations from the other 50% of the observations. The second quartile is denoted by Q2. 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 Q3.
  • 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:
0122356999
991112121212121313
14141414151515161919
20212526293032323238
404849495267869797116
  • The size of the data is n=50.
  • For finding first quartile,
  • L=0.25×50=12.5
  • Here, 12.5 is not a whole number, hence the 1st quartile will be the observation in the 13th position.
  • From the arranged observations the first quartile is 11.
  • For finding third quartile,
  • L=0.75×50=37.5
  • Here, 37.5 is not a whole number, hence the 3rd quartile will be the observation in the 38th position.
  • From the arranged observations the third quartile is 32.

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

b.

Expert Solution
Check Mark
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 x1,x2...xn 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:
  •   
0122356999
991112121212121313
14141414151515161919
20212526293032323238
404849495267869797116
  • The size of the data is n=50.
  • Hence, the sample size is even. Therefore, the median is the average of 25th and 26th observation.
  • From the arranged observations the median is 15+152=15

Thus, the median of the data is 15.

c.

Expert Solution
Check Mark
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=Q3Q1. Where, the first quartile is denoted by Q1 and third quartile is denoted by Q3.
  • From part (a), the first and the third quartiles are 11 and 32 respectively.
  • Substitute these values in the interquartile range formula,
  • IQR=3211=21
  • Outlier boundaries:
  • Lower outlier boundary is Q11.5×IQR.
  • Upper outlier boundary is Q3+1.5×IQR.
  • Where, the first quartile is denoted by Q1 and third quartile is denoted by Q3.
  • Substitute these values in the formulae,
  • Lower outlier boundary=111.5×21=1131.5=20.5.
  • Upper outlier boundary=32+1.5×21=32+31.5=63.5

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

d.

Expert Solution
Check Mark
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.

e.

Expert Solution
Check Mark
To determine

Draw a boxplot of the data.

Answer to Problem 25E

The boxplot is given below,

Essential Statistics, Chapter 3.3, Problem 25E , additional homework tip  1

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:

Essential Statistics, Chapter 3.3, Problem 25E , additional homework tip  2

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

f.

Expert Solution
Check Mark
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 1st quartile than the 3rd quartile or the upper whisker is longer than the lower whisker.
  • The data is left skewed if the median is closer to the 3rd quartile than the 1st quartile or the lower whisker is longer than the upper whisker.
  • The data is approximately symmetric if the median is the middle point of 1st quartile and the 3rd 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 1st quartile than the 3rd 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.

g.

Expert Solution
Check Mark
To determine

Find the 30th percentile of the data.

Answer to Problem 25E

The 30th percentile is 12.

Explanation of Solution

Calculation:

pth percentile:

The pth percentile separates the lowest p% of the observations from the highest (100p)% of the observations. Where 1<p<100 .

Procedure for finding pth percentile:

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

                              For finding pth percentile, L=p100n

  • Step 3: The pth 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 pth percentile is the observation in the position of rounded-up value.
  • The observations are arranged in increasing order:
0122356999
991112121212121313
14141414151515161919
20212526293032323238
404849495267869797116
  • The size of the data is n=50.
  • For finding 30th percentile,
  • L=30100×50=15
  • Here, 15 is a whole number, hence the 30th percentile will be the average of the observations in the position of 15th and 16th.
  • From the arranged observations the 30th percentile is 12+122=12.

Hence, the 30th percentile is 12.

h.

Expert Solution
Check Mark
To determine

Find the 85th percentile of the data.

Answer to Problem 25E

The 85th percentile is 49.

Explanation of Solution

Calculation:

  • The observations are arranged in increasing order:
0122356999
991112121212121313
14141414151515161919
20212526293032323238
404849495267869797116
  • The size of the data is n=50.
  • For finding 85th percentile,
  • L=85100×50=42.5
  • Here, 42.5 is not a whole number, hence the 85th percentile will be the observation in the 43rdposition.
  • From the arranged observations the 85th percentile is 49.

Hence, the 85th percentile is 49.

i.

Expert Solution
Check Mark
To determine

Find the percentile for the given hazardous waste sites.

Answer to Problem 25E

The value is of 55th 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,

                                                Perentile=100×(Number of values less than x)+0.5Number 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:
0122356999
991112121212121313
14141414151515161919
20212526293032323238
404849495267869797116

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,

Perentile=100×27+0.550=27.550×100=55

  • Here, 55 is a whole number.

Hence, the value is of 55th percentile.

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Chapter 3 Solutions

Essential Statistics

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