a The five number summary includes the minimum value, the 1^(st) quartile, the median, the 3^(rd) quartile, and the maximum value.
B
b Start the interval at 1.5
C
c Is it uniform and symmetric? Single or double peaked? Are there any outliers?
A
a Minimum value=1.58
1^(st) Quartile = 2.50
Median = 2.91
3^(rd) Quartile =3.49
Maximum value=4.29
B
bExample Diagram:
C
c See solution.
Practice makes perfect
a The five number summary we want to find are the following.
&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 1.58 and 4.29. Also, the number of values in the data set is 26, an even number, which means the median is the average of the 13^(th) and 14^(th) observation.
|l|
1^(st) 1.58 2^(nd) 1.87 3^(rd) 2.32 4^(th) 2.44 5^(th) 2.46 6^(th) 2.46 7^(th) 2.50 8^(th) 2.63 9^(th) 2.69 [1.25em]
10^(th) 2.72 11^(th) 2.78 12^(th) 2.81 13^(th) 2.83 14^(th) 2.99 15^(th) 3.10 16^(th) 3.16 17^(th) 3.30 18^(th) 3.42 [1.25em]
19^(th) 3.46 20^(th) 3.49 21^(th) 3.54 22^(th) 3.67 23^(th) 3.75 24^(th) 3.87 25^(th) 3.89 26^(th) 4.29
The median is 2.83+2.992=2.91. 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 7^(th) and 20^(th) value respectively.
|l|
1^(st) 1.58 2^(nd) 1.87 3^(rd) 2.32 4^(th) 2.44 5^(th) 2.46 6^(th) 2.46 7^(th) 2.50 8^(th) 2.63 9^(th) 2.69 [1.25em]
10^(th) 2.72 11^(th) 2.78 12^(th) 2.81 13^(th) 2.83 14^(th) 2.99 15^(th) 3.10 16^(th) 3.16 17^(th) 3.30 18^(th) 3.42 [1.25em]
19^(th) 3.46 20^(th) 3.49 21^(th) 3.54 22^(th) 3.67 23^(th) 3.75 24^(th) 3.87 25^(th) 3.89 26^(th) 4.29
Let's summarize what we have found.
&Minimum value=1.58
&1^(st) Quartile = 2.50
&Median = 2.91
&3^(rd) Quartile =3.49
&Maximum value=4.29
b Using an interval of 0.5, we can identify the number of observations.
r|l
Interval & Observations
1.5-2 & 1.58 2.97
2-2.5 & 2.32 2.44 2.46 2.46
2.5-3 & 2.50 2.63 2.69 2.72 2.78 2.81
& 2.83 2.99
3-3.5 & 3.10 3.16 3.30 3.42 3.46 3.49
3.5-4 & 3.54 3.67 3.75 3.87 3.89
4-4.5 & 4.29
Now we have all the information we need to draw the combination histogram and boxplot.
c Examining the diagram from Part B, we notice that the center is around a GPA of 2.9. Also, it is symmetric with a single peak. There are no values that are very far away from the bulk of the data distribution, which means the data set does not contain any outliers.
