Answer:
[tex]Q_1=56\\ \\Q_2=57\\ \\Q_3=62\\ \\Q_3-Q_1=9\ \text{Interquartile range}[/tex]
There exists at least one outlier in the data set at 76
Step-by-step explanation:
First, arrange the data
58 62 63 55 57 62 54 61 56 57 61 56 57 57 76
in ascending order:
54 55 56 56 57 57 57 57 58 61 61 62 62 63 76
This list of numbers contains 15 numbers, the middle number is 57, then
[tex]Q_2=57[/tex]
This number divides the list of numbers in two parts: left part of 7 numbers and right part of 7 numbers.
The middle number of left part is 56, so
[tex]Q_1=56[/tex]
The middle number of right part is 62, so
[tex]Q_3=62[/tex]
The interquartile range is
[tex]Q_3-Q_1=62-56=6[/tex]
Multiply the interquartile range by 1.5:
[tex]1.5(Q_3-Q_1)=1.5\cdot 6=9[/tex]
Now subtract it from [tex]Q_1[/tex] and add it to [tex]Q_3:[/tex]
[tex]Q_1-9=56-9=47\\ \\Q_3+9=62+9=71[/tex]
The minimum number in the list, 54, is greater than 47, so there is no lower outlier.
The maximum number in the list, 76, is greater than 71, so there is upper outlier.
