Biology, 30.03.2021 01:20 GreenHerbz206

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.

Answers: 3

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Answers

Answer from: loloroyroy264

(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.

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:

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

$Median = \frac{n + 1}{2}th$

$Median = \frac{9 + 1}{2}th$ $Median = \frac{10 + 1}{2}th$

$Median = \frac{10}{2}th$ $Median = \frac{11}{2}th$

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

Median = 2 $Median = \frac{2+2}{2} = 2$

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)

$Q_1 = \frac{n + 1}{4}th$ [Odd n] $Q_1 = \frac{n}{4}th$ [Even n]

$Q_1 = \frac{9 + 1}{4}th$ $Q_1 = \frac{10}{4}th$

$Q_1 = \frac{10}{4}th$ Q_1 = 2.5

Q_1 = 2.5th

This means that:

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)

$Q_3 = \frac{3}{4}(n + 1)th$ [Odd n] $Q_3 = \frac{3}{4}(n)th$ [Even n]

$Q_3 = \frac{3}{4}(9 + 1)th$ $Q_3 = \frac{30}{4}th$

$Q_3 = \frac{30}{4}th$ Q_3 = 7.5th

Q_3 = 7.5th

This means that:

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

Answer from: Quest

To kill bacteria from our body especially from stomach

Answer from: Quest

request for reexamination

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