EBK THE BASIC PRACTICE OF STATISTICS
EBK THE BASIC PRACTICE OF STATISTICS
8th Edition
ISBN: 8220106747841
Author: Moore
Publisher: YUZU
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Chapter 3, Problem 3.51E

(a)

To determine

To obtain: The proportion of females aged 20-29 weighted under 100 pounds.

To find: The percent of the N(161.58, 48.96) distribution is below 100.

(a)

Expert Solution
Check Mark

Answer to Problem 3.51E

The proportion of females aged 20-29 weighted under 100 pounds is 0.0255.

The percent of the N(161.58, 48.96) distribution is below 100 is 10.38%.

Explanation of Solution

Given info:

The NHANES survey of 2009-2010 includes the weights of 548 females in the United States aged 20-29. The weights of females follow normal distribution with mean 161.58 pounds and standard deviation 48.96 pounds. From the data the number of females aged 20-29 weighted under 100 pounds is 14.

Calculation:

For proportion of females aged 20-29 weighted under 100 pounds:

The formula to find the proportion of females aged 20-29 weighted under 100 pounds is,

Proportion of females = Number of femals aged 20-29 weighted under 100 poundsTotal number of females

Substitute 14 for ‘number of females aged 20-29 weighted under 100 pounds’ and 548 for ‘Total number of females’.

Proportion of females  = 14548=0.0255

Thus, the proportion of females aged 20-29 weighted under 100 pounds is 0.0255.

For percent of the N(161.58, 48.96) distribution is below 100:

Define the random variable x as weights of the females.

The formula for the standardized score is,

z=xμσ

The females aged 20-29 weighted below 100 pounds is denoted as x<100 .

Subtract the mean and then divide by the standard deviation to transform the value of x into standard normal z.

x161.5848.96<100161.5848.96<61.5848.96z<1.26

Where, standardized score z=x161.5848.96

The percent of the N(161.58, 48.96) distribution is below 100 is obtained by finding the area to left of –1.26.

Use Table A: Standard normal cumulative proportions to find the area.

Procedure:

  • Locate –1.2 in the left column of the A-2 Table.
  • Obtain the value in the corresponding row below 0.06.

That is, P(z<1.26)=0.1038

Thus, the percent of the N(161.58, 48.96) distribution is below 100 is 10.38%.

(b)

To determine

To obtain: The proportion of females aged 20-29 weighted over 250 pounds.

To find: The percent of the N(161.58, 48.96) distribution is above 250.

(b)

Expert Solution
Check Mark

Answer to Problem 3.51E

The proportion of females aged 20-29 weighted over 250 pounds is 0.0602.

The percent of the N(161.58, 48.96) distribution is above 250 is 3.51%.

Explanation of Solution

Given info:

The NHANES survey of 2009-2010 includes the weights of 548 females in the United States aged 20-29. From the data the number of females aged 20-29 weighted over 250 pounds is 33. The weights of females follow normal distribution with mean 161.58 pounds and standard deviation 48.96 pounds.

Calculation:

For proportion of females aged 20-29 weighted over 250 pounds:

The formula to find the females aged 20-29 weighted over 250 pounds is,

Proportion of females = Number of femals aged 20-29 weighted over 250 poundsTotal number of females

Substitute 33 for ‘number of females aged 20-29 weighted over 250 pounds’ and 548 for ‘Total number of females’.

Proportion of females  = 33548=0.0602

Thus, the proportion of females aged 20-29 weighted over 250 pounds is 0.0602.

For percent of the N(161.58, 48.96) distribution is above 250:

The females aged 20-29 weighted above 250 pounds is denoted as x>250 .

Subtract the mean and then divide by the standard deviation to transform the value of x into standard normal z.

x161.5848.96>250161.5848.96>88.4248.96z>1.81

Where, standardized score z=x161.5848.96

The percent of the N(161.58, 48.96) distribution is above 250 is obtained by finding the area to right of 1.81. But, the Table A: Standard normal cumulative proportions apply only for cumulative areas from the left.

Use Table A: Standard normal cumulative proportions to find the area to the left of 1.81.

Procedure:

  • Locate 1.8 in the left column of the A-2 Table.
  • Obtain the value in the corresponding row below 0.01.

That is, P(z<1.81)=0.9649

The area to the right of 1.81 is,

P(z>1.81)=1P(z<1.81)=10.9649=0.0351

Thus, the percent of the N(161.58, 48.96) distribution is above 250 is 3.51%.

(c)

To determine

To check: Whether it is a good idea to summarize the distribution of weights by an N(161.58, 48.96) distribution based on answers in part (a) and part (b) or not.

(c)

Expert Solution
Check Mark

Answer to Problem 3.51E

The idea is not good to summarize the distribution of weights by a N(161.58, 48.96) distribution based on answers in part (a) and part (b).

Explanation of Solution

Justification:

The percentage of females aged 20-29 weighted under 100 pounds is 2.55% and the percentage of females aged 20-29 weighted over 250 pounds is 6.02%.

Using normal distribution, the percentage of N(161.58, 48.96) distribution is below 100 is 10.38% and percentage of the N(161.58, 48.96) distribution is above 250 is 3.51%.

Thus, the results suggest that normal distribution not provides a good approximation to the distribution of weights.

Hence, the idea is not good to summarize the distribution of weights by a N(161.58, 48.96) distribution based on answers in part (a) and part (b).

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