Question

In a statistic class, 11 scores were randomly selected with the following results were obtained: 68, 74, 66, 37, 52, 71, 90, 65, 76, 73, 22. What are the outer fences?

A)-6.0, 140.0

B)-10.0, 92.0

C)2.0, 128.0

D)2.0, 162.0

E)37.0, 107.0

Answer 1

First, we will arrange the data in ascending order

22, 37, 52, 65, 66, 68, 71, 73, 74, 76, 90

Dividing the data into two equal halves, we get

22, 37, 52, 65, 66 lower half and 71, 73, 74, 76, 90 upper half

now, we need to find the median for both halves

22, 37, 52, 65, 66. odd data number

first quartile = median of lower half = center value = 52

Similarly

71, 73, 74, 76, 90

odd data number

third quartile = median of upper half = center value = 74

we know that IQR = third quartile - first quartile = 74-52 = 22

Formula for upper outer fence is Q3+ 3(IQR)

we have Q3 = 74 and IQR = 22

so, upper outer fence = 74 + 3*22 = 74 + 66 = 140

Formula for lower outer fence is Q1- 3(IQR)

we have Q1 = 52 and IQR = 22

so, lower outer fence = 52 - 3*22 = 52 - 66 = -14

So, outer fence limit is (-14,140)

it is not matching with any of the options.

I think option A is correct answer because it has same upper outer fencing. Check the values correctly for first option, it might be (-14,140)