PRACTICE OF STATISTICS F/AP EXAM
PRACTICE OF STATISTICS F/AP EXAM
6th Edition
ISBN: 9781319113339
Author: Starnes
Publisher: MAC HIGHER
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Question
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Chapter 2.2, Problem 52E

(a)

To determine

Percentage of the players with 100 plate appearances and batting averages of 0.363 and more.

(a)

Expert Solution
Check Mark

Answer to Problem 52E

0.15% of the Major League Basketball players with 100 plate appearances had batting averages of 0.363 and higher.

Explanation of Solution

Given information:

Mean, μ=0.261

Standard deviation, σ=0.034

According to 68 − 95 − 99.7 rule:

68% of the data of a normal distribution lies with 1 standard deviation from the mean.

95% of the data of a normal distribution lies with 2 standard deviation from the mean.

99.7% of the data of a normal distribution lies with 1 standard deviation from the mean.

Then

The general Normal density graph is represented as:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 2.2, Problem 52E , additional homework tip  1

Note that

0.363 lies 3σ above the mean.

  μ+3σ=0.261+3(0.034)=0.363

According to 68 − 95 − 99.7 rule:

99.7% of the data values lie within 3σ of the mean.

Although,

Data values in total are 100%.

Then

  100%99.7%=0.30%

0.30% of the data values lie more than 3σ from the mean.

We also know that

Normal distribution is symmetric about the mean.

That implies

0.15% of the data values are more than 3σ below the mean.

And

0.15% of the data values are more than 3σ above the mean.

Therefore,

0.15% of the Major League Basketball players with 100 plate appearances had batting averages of 0.363 and higher.

(b)

To determine

Percentile of the player in the distribution with batting average of 0.227.

(b)

Expert Solution
Check Mark

Answer to Problem 52E

The player with batting average of 0.227 is at 16th percentile.

Explanation of Solution

Given information:

Mean, μ=0.261

Standard deviation, σ=0.034

According to 68 − 95 − 99.7 rule:

68% of the data of a normal distribution lies with 1 standard deviation from the mean.

95% of the data of a normal distribution lies with 2 standard deviation from the mean.

99.7% of the data of a normal distribution lies with 1 standard deviation from the mean.

Then

The general Normal density graph is represented as:

  PRACTICE OF STATISTICS F/AP EXAM, Chapter 2.2, Problem 52E , additional homework tip  2

Note that

0.227 lies σ (1 standard deviation) below the mean.

  μσ=0.2610.034=0.227

According to 68 − 95 − 99.7 rule:

68% of the data values lie within σ (1 standard deviation) of the mean.

Although,

Data values in total are 100%.

Then

  100%68%=32%

32% of the data values lie more than σ (1 standard deviation) from the mean.

We also know that

Normal distribution is symmetric about the mean.

That implies

16% of the data values are more than σ (1 standard deviation) above the mean.

And

16% of the data values are more than σ (1 standard deviation) below the mean.

The data value represented by the xth percentile includes x% of the data values below it.

That implies

16% of the players have batting average of 0.227 or less.

Thus,

The player with batting average of 0.227 is at 16th percentile.

Chapter 2 Solutions

PRACTICE OF STATISTICS F/AP EXAM

Ch. 2.1 - Prob. 11ECh. 2.1 - Prob. 12ECh. 2.1 - Prob. 13ECh. 2.1 - Prob. 14ECh. 2.1 - Prob. 15ECh. 2.1 - Prob. 16ECh. 2.1 - Prob. 17ECh. 2.1 - Prob. 18ECh. 2.1 - Prob. 19ECh. 2.1 - Prob. 20ECh. 2.1 - Prob. 21ECh. 2.1 - Prob. 22ECh. 2.1 - Prob. 23ECh. 2.1 - Prob. 24ECh. 2.1 - Prob. 25ECh. 2.1 - Prob. 26ECh. 2.1 - Prob. 27ECh. 2.1 - Prob. 28ECh. 2.1 - Prob. 29ECh. 2.1 - Prob. 30ECh. 2.1 - Prob. 31ECh. 2.1 - Prob. 32ECh. 2.1 - Prob. 33ECh. 2.1 - Prob. 34ECh. 2.1 - Prob. 35ECh. 2.1 - Prob. 36ECh. 2.1 - Prob. 37ECh. 2.1 - Prob. 38ECh. 2.1 - Prob. 39ECh. 2.1 - Prob. 40ECh. 2.2 - Prob. 41ECh. 2.2 - Prob. 42ECh. 2.2 - Prob. 43ECh. 2.2 - Prob. 44ECh. 2.2 - Prob. 45ECh. 2.2 - Prob. 46ECh. 2.2 - Prob. 47ECh. 2.2 - Prob. 48ECh. 2.2 - Prob. 49ECh. 2.2 - Prob. 50ECh. 2.2 - Prob. 51ECh. 2.2 - Prob. 52ECh. 2.2 - Prob. 53ECh. 2.2 - Prob. 54ECh. 2.2 - Prob. 55ECh. 2.2 - Prob. 56ECh. 2.2 - Prob. 57ECh. 2.2 - Prob. 58ECh. 2.2 - Prob. 59ECh. 2.2 - Prob. 60ECh. 2.2 - Prob. 61ECh. 2.2 - Prob. 62ECh. 2.2 - Prob. 63ECh. 2.2 - Prob. 64ECh. 2.2 - Prob. 65ECh. 2.2 - Prob. 66ECh. 2.2 - Prob. 67ECh. 2.2 - Prob. 68ECh. 2.2 - Prob. 69ECh. 2.2 - Prob. 70ECh. 2.2 - Prob. 71ECh. 2.2 - Prob. 72ECh. 2.2 - Prob. 73ECh. 2.2 - Prob. 74ECh. 2.2 - Prob. 75ECh. 2.2 - Prob. 76ECh. 2.2 - Prob. 77ECh. 2.2 - Prob. 78ECh. 2.2 - Prob. 79ECh. 2.2 - Prob. 80ECh. 2.2 - Prob. 81ECh. 2.2 - Prob. 82ECh. 2.2 - Prob. 83ECh. 2.2 - Prob. 84ECh. 2.2 - Prob. 85ECh. 2.2 - Prob. 86ECh. 2.2 - Prob. 87ECh. 2.2 - Prob. 88ECh. 2.2 - Prob. 89ECh. 2.2 - Prob. 90ECh. 2.2 - Prob. 91ECh. 2.2 - Prob. 92ECh. 2 - Prob. R2.1RECh. 2 - Prob. R2.2RECh. 2 - Prob. R2.3RECh. 2 - Prob. R2.4RECh. 2 - Prob. R2.5RECh. 2 - Prob. R2.6RECh. 2 - Prob. R2.7RECh. 2 - Prob. R2.8RECh. 2 - Prob. R2.9RECh. 2 - Prob. T2.1SPTCh. 2 - Prob. T2.2SPTCh. 2 - Prob. T2.3SPTCh. 2 - Prob. T2.4SPTCh. 2 - Prob. T2.5SPTCh. 2 - Prob. T2.6SPTCh. 2 - Prob. T2.7SPTCh. 2 - Prob. T2.8SPTCh. 2 - Prob. T2.9SPTCh. 2 - Prob. T2.10SPTCh. 2 - Prob. T2.11SPTCh. 2 - Prob. T2.12SPTCh. 2 - Prob. T2.13SPT
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