a To create the boxplot, we need to know the lowest and highest value, the lower and upper quartile, and the median.
B
b Is the distribution skewed? Are there any outliers?
C
c What does the presence of the outlier do to the mean?
D
d What does the presence of the outlier do to the standard deviation?
E
e Remove the outlier and calculate the mean once more.
A
aDiagram:
B
b See solution.
C
c Median, see solution.
D
d IQR, see solution.
E
e Gather data for a few more weeks. Then make your decision.
Practice makes perfect
a To create the boxplot, we need to determine the following for the data set.
&Minimum value
&1^(st) Quartile
&Median
&3^(rd) Quartile
&Maximum value
Examining the observations, we notice that they have been ordered from least to greatest. Therefore, we can immediately identify the minimum and maximum value as 0 and 36.5. Also, the number of values in the data set is 25, an odd number, which means the median must be the 13^(th) observation.
|ccccc|
0, & 0, & 0, & 0, & 1.7,
2.6, & 2.9, & 4.2, & 4.4, & 5.1,
5.6, & 6.4, & 8.0, & 8.9, & 9.7,
10.1, & 11.2, & 13.6, & 15.1, & 16.3,
17.7, & 21.4, & 22.0, & 22.2, & 36.5
To find the 1^(st) and 3^(rd) Quartile, we have to identify the middle value of the lower and upper half, which will be the average of the 6^(th) and 7^(th) value for the lower half and of the 18^(th) and 19^(th) value for the upper half.
|ccccc|
0, & 0, & 0, & 0, & 1.7,
2.6, & 2.9, & 4.2, & 4.4, & 5.1,
5.6, & 6.4, & 8.0, & 8.9, & 9.7,
10.1, & 11.2, & 13.6, & 15.1, & 16.3,
17.7, & 21.4, & 22.0, & 22.2, & 36.5
Having identified the relevant values, we can calculate the quartiles.
Upper Quartile:& 2.6+2.9/2 = 2.75 [1em]
Lower Quartile:& 15.1+16.3/2 = 15.7
Let's summarize what we have found.
&Minimum value=0
&1^(st) Quartile = 2.75
&Median = 8.0
&3^(rd) Quartile = 15.7
&Maximum value=36.5
Now we have everything we need to draw the boxplot. Using a bin width of 6, we can identify the number of observations in each interval.
r|l
Interval& Observations
0-6 & 0, 0, 0, 0, 1.7, 2.6, 2.9, 4.2
& 4.4, 5.1, 5.6
6-12 & 6.4, 8.0, 8.9, 9.7, 10.1, 11.2
12-18 & 13.5, 15.1, 16.3, 17.7
18-24 & 21.4, 22.0, 22.2
24-30 &
30-36 &
36-42 & 36.5
Now we have all the information we need to draw the combination histogram and boxplot.
b Observing the diagram from Part A, we see that it has a left skew and an outlier at 36.5. Therefore, the center is best described by its median at 8 pounds. Also, knowing the upper and lower quartile, we can calculate the interquartile range.
IQR: 15.7-2.75 = 12.95
c The mean is the sum of all observations divided by the number of observations.
Since the data set contains an outlier at 36.5, this will make the mean greater compared to if the outlier was removed. Since the outlier is not representative of the population in general, the median is a better measure, as it is unaffected by outliers.
d Similar to the mean, the standard deviation considers all observations when calculating the spread of the population. The IQR, however, measures the difference between the upper and lower quartile. Since there is an outlier present in the data set, the IQR is better, since it is unaffected by the outlier.
e As explained in Part B, there is an outlier present in the data set which increases the mean. If we remove the outlier, the mean would fall to something more representative for the population as a whole.
When removing the outlier, the mean drops to 8.71. Therefore, based on this data, the manager should not order the dumpsters. Instead, he might want to wait a few more weeks and gather more data. Then he can make the decision.
