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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Chapter 2.1, Problem 25E

(a)

To determine

Shape of the resulting distribution.

(a)

Expert Solution
Check Mark

Answer to Problem 25E

The distribution is roughly symmetric along with a single peak.

Explanation of Solution

Given information:

  100cm=1m

In order to convert the corrected long − jump distances from centimeters to meters, every distance value in centimeters needs to be divided by 100.

In the previous problem,

We came to know that the shape of the distribution of corrected long − jump distances in centimeters was a roughly symmetric distribution along with a single peak.

When every value in the data is divided by 100, then the shape of the distribution will not be affected because there will be no change in the relationships between every data pair.

Thus,

The shape of the distribution of corrected long − jump distances in meters in this case is also a roughly symmetric distribution along with a single peak.

(b)

To determine

Mean of the distribution of the corrected long − jump distance in meters.

(b)

Expert Solution
Check Mark

Answer to Problem 25E

Mean of the distribution of the corrected long − jump distance is 5.573 m

Explanation of Solution

Given information:

  100cm=1m

  x¯N=557.3cm

In order to convert the corrected long − jump distances from centimeters to meters, every distance value in centimeters needs to be divided by 100.

When we divide every data value by 100, then the center of the distribution also to be divided by 100, because the mean is the measure of the center.

In the previous problem,

We came to know that the mean of the distribution of the corrected long − jump distance was 557.3 cm. For converting the mean of the distribution of corrected long − jump distances from centimeters to meters, we need to divide the mean in centimeters by 100.

  x¯N=557.3100=5.573m

Thus,

The mean of the distribution of corrected long − jump distance in meters is 5.573 m.

(c)

To determine

Standard deviation of the distribution of the corrected long − jump distance in meters.

(c)

Expert Solution
Check Mark

Answer to Problem 25E

Standard deviation of the distribution of the corrected long − jump distance is 0.04713 m.

Explanation of Solution

Given information:

  100cm=1m

  SDN=4.713cm

In order to convert the corrected long − jump distances from centimeters to meters, every distance value in centimeters needs to be divided by 100.

In the previous problem,

We came to know that the standard deviation of the distribution of the corrected long − jump distance was 4.713 m. For converting the standard deviation of the distribution ofcorrected long − jump distances from centimeters to meters, we need to divide the standard deviation in centimeters by 100.

  SDN=4.713100=0.04713m

Thus,

The standard deviation of the distribution of corrected long − jump distance in meters is 0.0473 m.

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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