1st Edition

ISBN: 9780618800766

Author: Ron Larson, Lee Stiff, Laurie Boswell, Timothy Kanold

Publisher: Houghton Mifflin-High School

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The method used for finding the corresponding z-critical value in a normal distribution using the known probability is said to be an inverse normal distribution. The inverse normal distribution is a continuous probability distribution with a family of tw…

Mean, Median, Mode

It is a descriptive summary of a data set. It can be defined by using some of the measures. The central tendencies do not provide information regarding individual data from the dataset. However, they give a summary of the data set. The central tendency o…

Z-Scores

A z-score is a unit of measurement used in statistics to describe the position of a raw score in terms of its distance from the mean, measured with reference to standard deviation from the mean. Z-scores are useful in statistics because they allow compar…

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Question

Chapter 5.5, Problem 50E

To determine

To find the deviation of the shoe sizes sold from the average shoe size.

Expert Solution

Answer to Problem 50E

The deviations for the shoe sizes are $−2,1,−3,112,212,0,12 and −12$

A positive deviation from the average size indicates that the recorded shoe size is greater than the average shoe size.

A negative deviation from the average size indicates that the recorded shoe size is smaller than the average shoe size.

Explanation of Solution

Given information:

The average women’s shoe size is $812$ .

The record of the shoe sizes sold is $612,912,512,10,11,812,9 and 8$

Calculation :

The deviation of the recorded shoe size from the average shoe size can be calculated by:

$612−812=132−172=−42=−2912−812=192−172=22=1512−812=112−172=−62=−310−812=202−172=32=112 [Using Least Common Divisor]11−812=222−172=52=212 [Using Least Common Divisor]812−812=172−172=09−812=182−172=12 [Using Least Common Divisor]8−812=162−172=−12 [Using Least Common Divisor]$

The deviations for the shoe sizes are $−2,1,−3,112,212,0,12 and −12$

A positive deviation from the average size indicates that the recorded shoe size is greater than the average shoe size.

A negative deviation from the average size indicates that the recorded shoe size is smaller than the average shoe size.

To determine

To find the mean of the deviation.

Expert Solution

Answer to Problem 50E

The mean of the deviation is $0$

The store owner can conclude that the average women’s shoe size i.e. $812$ , was correct because the mean of the deviation is $0$

Explanation of Solution

Given information:

The average women’s shoe size is $812$ .

The record of the shoe sizes sold is $612,912,512,10,11,812,9 and 8$

The deviations for the shoe sizes are $−2,1,−3,112,212,0,12 and −12$

Formula used:

The following formula is used:

$Mean=Sum of observationsNumber of observation$

Calculation :

The mean of the deviations can be calculated by:

$Mean=Sum of observationsNumber of observationMean=(−2+1−3+112+212+0+12−12)8Mean=(−42+22−62+32+52+0+12−12)8 [write the fraction using Least Common Divisor]Mean=(−4+2−6+3+5+0+1−12)8Mean=(11−112)8Mean=08Mean=0$

Hence, the mean of the deviation is $0$

From the above mean, the store owner can conclude that the average women’s shoe size i.e. $812$ , was correct because the mean of the deviation is $0$

Chapter 5.5, Problem 49E

Chapter 5.6, Problem 1C

Chapter 5 Solutions

Pre-Algebra

Ch. 5.1 - Prob. 1C Ch. 5.1 - Prob. 2C Ch. 5.1 - Prob. 3C Ch. 5.1 - Prob. 4C Ch. 5.1 - Prob. 5C Ch. 5.1 - Prob. 1E Ch. 5.1 - Prob. 2E Ch. 5.1 - Prob. 3E Ch. 5.1 - Prob. 4E Ch. 5.1 - Prob. 5E

Ch. 5.1 - Prob. 6E Ch. 5.1 - Prob. 7E Ch. 5.1 - Prob. 8E Ch. 5.1 - Prob. 9E Ch. 5.1 - Prob. 10E Ch. 5.1 - Prob. 11E Ch. 5.1 - Prob. 12E Ch. 5.1 - Prob. 13E Ch. 5.1 - Prob. 14E Ch. 5.1 - Prob. 15E Ch. 5.1 - Prob. 16E Ch. 5.1 - Prob. 17E Ch. 5.1 - Prob. 18E Ch. 5.1 - Prob. 19E Ch. 5.1 - Prob. 20E Ch. 5.1 - Prob. 21E Ch. 5.1 - Prob. 22E Ch. 5.1 - Prob. 23E Ch. 5.1 - Prob. 24E Ch. 5.1 - Prob. 25E Ch. 5.1 - Prob. 26E Ch. 5.1 - Prob. 27E Ch. 5.1 - Prob. 28E Ch. 5.1 - Prob. 29E Ch. 5.1 - Prob. 30E Ch. 5.1 - Prob. 31E Ch. 5.1 - Prob. 32E Ch. 5.1 - Prob. 33E Ch. 5.1 - Prob. 34E Ch. 5.1 - Prob. 35E Ch. 5.1 - Prob. 36E Ch. 5.1 - Prob. 37E Ch. 5.1 - Prob. 38E Ch. 5.1 - Prob. 39E Ch. 5.1 - Prob. 40E Ch. 5.1 - Prob. 41E Ch. 5.1 - Prob. 42E Ch. 5.1 - Prob. 43E Ch. 5.1 - Prob. 44E Ch. 5.1 - Prob. 45E Ch. 5.1 - Prob. 46E Ch. 5.1 - Prob. 47E Ch. 5.1 - Prob. 48E Ch. 5.1 - Prob. 49E Ch. 5.1 - Prob. 50E Ch. 5.1 - Prob. 51E Ch. 5.1 - Prob. 52E Ch. 5.1 - Prob. 53E Ch. 5.1 - Prob. 54E Ch. 5.1 - Prob. 55E Ch. 5.1 - 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