Question

Select ALL the correct answers.

A pitcher for a professional baseball team allows runs in the first nine games he starts this season. Let A be the set of the number of

runs allowed by the pitcher in his first nine starts.

A = {1, 4, 2, 2, 3, 1, 1, 2, 1}

In the tenth game he starts, he allows 9 runs. Let B represent the set of the number of runs allowed in all ten games he has started.

Select the true statements.

The median of B is 1 run more than the median of A.

The interquartile range of B is greater than the interquartile range of A.

The interquartile range of A is 1 less than the interquartile range of B.

The median of A is the same as the median of B.

Including the runs allowed in the tenth game does not cause the spread of the data to

change.

Answer 1

(b) The interquartile range of B is greater than the interquartile range of A.

(d) The median of A is the same as the median of B.

Step-by-step explanation:

Given

`A = \{1, 4, 2, 2, 3, 1, 1, 2, 1\}`

`10th\ run = 9`

So:

`B = \{1, 4, 2, 2, 3, 1, 1, 2, 1,9\}`

Required

Select all true statements

(a) & (d) Median Comparisons

`A = \{1, 4, 2, 2, 3, 1, 1, 2, 1\}`
`B = \{1, 4, 2, 2, 3, 1, 1, 2, 1,9\}`

`n = 9`
`n = 10`

Arrange the data:

`A = \{1, 1, 1, 1, 2, 2, 2,3, 4\}`
`B = \{1,1,1,1,2,2,2,3,4,9\}`

`Median = (n + 1)/(2)th`

`Median = (9 + 1)/(2)th`
`Median = (10 + 1)/(2)th`

`Median = (10)/(2)th`
`Median = (11)/(2)th`

`Median = 5th`
`Median = 5.5}th` --- average of 5th and 6th

`Median = 2`
`Median = (2+2)/(2) = 2`

Option (d) is correct because both have a median of: 2

(b) & (c) Interquartile Range Comparisons

`A = \{1, 1, 1, 1, 2, 2, 2,3, 4\}`
`B = \{1,1,1,1,2,2,2,3,4,9\}`

`n = 9`
`n = 10`

First, calculate the lower quartile (Q1)

`Q_1 = (n + 1)/(4)th`[Odd n]
`Q_1 = (n)/(4)th` [Even n]

`Q_1 = (9 + 1)/(4)th`
`Q_1 = (10)/(4)th`

`Q_1 = (10)/(4)th`
`Q_1 = 2.5`

`Q_1 = 2.5th`

This means that:

`Q_1 = 2nd + 0.5(3rd - 2nd)`
`Q_1 = 2nd + 0.5(3rd - 2nd)`

`Q_1 = 1 + 0.5(1- 1)`
`Q_1 = 1+ 0.5(1 - 1)`

`Q_1 = 1`
`Q_1 = 1`

Next, calculate the upper quartile (Q3)

`Q_3 = (3)/(4)(n + 1)th` [Odd n]
`Q_3 = (3)/(4)(n)th` [Even n]

`Q_3 = (3)/(4)(9 + 1)th`
`Q_3 = (30)/(4)th`

`Q_3 = (30)/(4)th`
`Q_3 = 7.5th`

`Q_3 = 7.5th`

This means that:

`Q_3 = 7th + 0.5(8th- 7th)`
`Q_3 = 7th + 0.5(8th- 7th)`

`Q_3 = 2 + 0.5(3- 2)`
`Q_3 = 2+ 0.5(4 - 2)`

`Q_3 = 2.5`
`Q_3 = 3`

The interquartile range is
`IQR = Q_3 - Q_1`

So, we have:

`IQR = 2.5 - 1`
`IQR = 3 - 1`

`IQR = 1.5`
`IQR =2`

(b) is true because B has a greater IQR than A

(e) This is false because some spread measures (which include quartiles and the interquartile range) changed when the 10th data is included.

The upper quartile and the interquartile range of A and B are not equal