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.)

Answer 1

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