The following table gives the total area in square miles (land and water) of sev
ID: 3350824 • Letter: T
Question
The following table gives the total area in square miles (land and water) of seven states. Complete parts (a) through (c).
State
Area
1
52,300
2
615,400
3
115,000
4
53,200
5
159,300
6
104,500
7
6,400
b. Which state is an outlier on the high end? If you eliminate this state, what are the new mean and median areas for this data set?
State
2
is an outlier on the high end.
The new mean is
81783
square miles.
(Round to the nearest integer as needed.)
The new median is
78850
square miles.
(Round to the nearest integer as needed.)
c. Which state is an outlier on the low end? If you eliminate this state, what are the new mean and median areas for this data set?
State
7
is an outlier on the low end.
The new mean is
square miles.
(Round to the nearest integer as needed.)
State
Area
1
52,300
2
615,400
3
115,000
4
53,200
5
159,300
6
104,500
7
6,400
b. Which state is an outlier on the high end? If you eliminate this state, what are the new mean and median areas for this data set?
State
2
is an outlier on the high end.
The new mean is
81783
square miles.
(Round to the nearest integer as needed.)
The new median is
78850
square miles.
(Round to the nearest integer as needed.)
c. Which state is an outlier on the low end? If you eliminate this state, what are the new mean and median areas for this data set?
State
7
is an outlier on the low end.
The new mean is
square miles.
(Round to the nearest integer as needed.)
Explanation / Answer
Given values:
52300, 615400, 115000, 53200, 159300, 104500, 6400
(b)
To find outlier on high side:
Arranging data in ascending order:
6400,52300,53200,104500,115000,159300,615400
Q1 = First Quartile = (n+1)/4 th item = 8/4 = 2nd item = 52300
Q3 = Third Quartile = (n+1)3/4 th item = 8 X 3/4 = 6th item = 159300
IQR = Q3 - Q1 = 107000
Q3 + 1.5 IQR = 159300 + (1.5 X 107000) = 319800
So, 615400 is greater than 319800
Outlier on the high side:
State 2 is an outlier on the high side.
So,
Eliminate State 2 from the list:
So:
n = 6
New mean = sum of values/6 = 490700/6 = 81783
New Median:
Q2 = (n+1)/2th item = (6+1)/2 = 3.5 = average of 3rd & 4th item = (53200 + 104500)/2 = 78850
(c)
Q1 = (n+1)/4th item = (6+1)/4 = 1.75 = interpolation of 1st & 2nd item = 40825
Q3 = (n+1)3/4th item = (6+1)3/4 = 5.25 = interpolation of 5th & th item = 126075
IQR = Q3 - Q1 = 85250
Q1 - (1.5 X IQR) = 40825 - (1.5 X 85250) = - 87050
No value is less than - 87050
So,
No outlier on the low end.
So, new mean = old mean = 490700/6 = 81783
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