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 $N\left(161.58,48.96\right)$ distribution is below 100.

Answer to Problem 3.51E

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

The percent of the $N\left(161.58,48.96\right)$ 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,

$\text{Proportion of females = }\frac{\text{Number of femals aged 20-29 weighted under 100 pounds}}{\text{Total number of females}}$

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

$\begin{array}{c}\text{Proportion of females = }\frac{14}{548}\\ =0.0255\end{array}$

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

For percent of the $N\left(161.58,48.96\right)$ distribution is below 100:

Define the random variable x as weights of the females.

The formula for the standardized score is,

$z=\frac{x-\mu }{\sigma }$

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.

$\begin{array}{c}\frac{x-161.58}{48.96}<\frac{100-161.58}{48.96}\\ <-\frac{61.58}{48.96}\\ z<-1.26\end{array}$

Where, standardized score $z=\frac{x-161.58}{48.96}$

The percent of the $N\left(161.58,48.96\right)$ 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, $P\left(z<-1.26\right)=0.1038$

Thus, the percent of the $N\left(161.58,48.96\right)$ 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 $N\left(161.58,48.96\right)$ distribution is above 250.

Answer to Problem 3.51E

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

The percent of the $N\left(161.58,48.96\right)$ 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,

$\text{Proportion of females = }\frac{\text{Number of femals aged 20-29 weighted over 250 pounds}}{\text{Total number of females}}$

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

$\begin{array}{c}\text{Proportion of females = }\frac{33}{548}\\ =0.0602\end{array}$

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

For percent of the $N\left(161.58,48.96\right)$ 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.

$\begin{array}{c}\frac{x-161.58}{48.96}>\frac{250-161.58}{48.96}\\ >\frac{88.42}{48.96}\\ z>1.81\end{array}$

Where, standardized score $z=\frac{x-161.58}{48.96}$

The percent of the $N\left(161.58,48.96\right)$ 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, $P\left(z<1.81\right)=0.9649$

The area to the right of 1.81 is,

$\begin{array}{c}P\left(z>1.81\right)=1-P\left(z<1.81\right)\\ =1-0.9649\\ =0.0351\end{array}$

Thus, the percent of the $N\left(161.58,48.96\right)$ 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 $N\left(161.58,48.96\right)$ distribution based on answers in part (a) and part (b) or not.

Answer to Problem 3.51E

The idea is not good to summarize the distribution of weights by a $N\left(161.58,48.96\right)$ 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 $N\left(161.58,48.96\right)$ distribution is below 100 is 10.38% and percentage of the $N\left(161.58,48.96\right)$ 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 $N\left(161.58,48.96\right)$ distribution based on answers in part (a) and part (b).