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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.

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Answers: 3

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Answers

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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

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Answer from: Quest

a and c are the answeres

ansver

Answer from: Quest

measuring light intensity

light meters can be used to measure light intensity. the meter is held at the soil surface and pointed in the direction of the maximum light intensity, and then the meter is read.

errors can be made when measuring light intensity by accidentally shading the light meter. the reliability of the results can be checked by taking many samples.

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