Chapter 7.6, Problem 72E

Consider babies born in the “normal” range of 37-43 weeks gestational age. The paper referenced in Example 7.27 ("Fetal Growth Parameters and Birth Weight: Their Relationship to Neonatal Body Composition," Ultrasound in Obstetrics and Gynecology [2009]: 441-446) suggests that a normal distribution with mean /µ = 3500 grams and standard deviation σ = 600 grams is a reasonable model for the probability distribution of the continuous numerical variable x = birth weight of a randomly selected full- term baby.

1. a. What is the probability that the birth weight of a randomly selected full-term baby exceeds 4000 g?
2. b. What is the probability that the birth weight of a randomly selected full-term baby is between 3000 and 4000 g?
3. c. What is the probability that the birth weight of a randomly selected full-term baby is either less than 2000 g or greater than 5000 g?
4. d. What is the probability that the birth weight of a randomly selected full-term baby exceeds 7 pounds? (Hint: 1 lb = 453.59 g.)
5. e. How would you characterize the most extreme 0.1% of all full-term baby birth weights?
6. f. If x is a random variable with a normal distribution and a is a numerical constant (a ≠ 0), then y = ax also has a normal distribution. Use this formula to determine the distribution of full-term baby birth weight expressed in pounds (shape, mean, and standard deviation), and then recalculate the probability from Part (d). How does this compare to your previous answer?

To determine

Find the probability that the birth weight of a randomly selected full-term baby exceeds 4,000 g.

Answer to Problem 72E

The probability that the birth weight of a randomly selected full-term baby exceeds 4,000 g is 0.2033.

Explanation of Solution

Calculation:

It is given that the birth weight is normally distribution with mean $\mu =3,500\text{ grams}$ and standard deviation $\sigma =600\text{ grams}$.

The z value for random variable x with mean $\mu$ and standard deviation $\sigma$ can be obtained as follows:

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

Let x denote the birth weight of a randomly selected full-term baby.

First calculate the value of z score as given below:

$\begin{array}{c}a*=\frac{a-\mu }{\sigma }\\ =\frac{4,000-3,500}{600}\\ =\frac{500}{600}\\ =0.83\end{array}$

The required probability is obtained as given below.

$\begin{array}{c}P\left(x>4,000\right)=P\left(z>0.83\right)\\ =z\text{ curve area to the right of 0}\text{.83}\\ =1-P\left(z<0.83\right)\end{array}$

Use the Appendix- TABLE 2, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the required probability.

• Locate the value 0.8 in row.
• Locate the value 0.03 in column.
• The intersecting value of row and column is 0.7967.

$\begin{array}{c}P\left(x>4,000\right)=1-0.7967\\ =0.2033\end{array}$

Thus, the probability that the birth weight of a randomly selected full-term baby exceeds 4,000 g is 0.2033.

To determine

Find the probability that the birth weight of a randomly selected full-term baby is between 3,000 and 4,000 g.

Answer to Problem 72E

The probability that the birth weight of a randomly selected full-term baby is between 3,000 g and 4,000 g is 0.5934.

Explanation of Solution

Calculation:

First calculate the value of z score as given below:

$\begin{array}{c}a*=\frac{a-\mu }{\sigma }\\ =\frac{3,000-3,500}{600}\\ =\frac{-500}{600}\\ =-0.83\end{array}$

$\begin{array}{c}b*=\frac{b-\mu }{\sigma }\\ =\frac{4,000-3,500}{600}\\ =\frac{500}{600}\\ =0.83\end{array}$

The required probability is obtained as given below:

$\begin{array}{c}P\left(3,000<x<4,000\right)=P\left(-0.83<z<0.83\right)\\ =z\text{ curve area between }-\text{0}\text{.83 and 0}\text{.83}\\ =P\left(z<0.83\right)-P\left(z<-0.83\right)\end{array}$

Use the Appendix- TABLE 2, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the probability.

• Locate the value 0.8 in row.
• Locate the value 0.03 in column.
• The intersecting value of row and column is 0.7967.
• Locate the value –0.8 in row.
• Locate the value 0.03 in column.
• The intersecting value of row and column is 0.2033.

$\begin{array}{c}P\left(3,000<x<4,000\right)=0.7967-0.2033\\ =0.5934\end{array}$

Thus, the probability that the birth weight of a randomly selected full-term baby is between 3,000 g and 4,000 g is 0.5934.

To determine

Find the probability that the birth weight of a randomly selected full-term baby is either less than 2,000 g or greater than 5,000 g.

Answer to Problem 72E

The probability that the birth weight of a randomly selected full-term baby is either less than 2,000 g or greater than 5,000 g is 0.0124.

Explanation of Solution

Calculation:

First calculate the value of z score as given below:

$\begin{array}{c}b*=\frac{b-\mu }{\sigma }\\ =\frac{2,000-3,500}{600}\\ =\frac{-1500}{600}\\ =-2.5\end{array}$

$\begin{array}{c}a*=\frac{a-\mu }{\sigma }\\ =\frac{5,000-3,500}{600}\\ =\frac{1500}{600}\\ =2.5\end{array}$

The required probability is obtained as given below:

$\begin{array}{c}P\left(x<2000\text{ or }x>5000\right)=P\left(z<-2.5\text{ or }z>2.5\right)\\ =\left[\begin{array}{l}\left(z\text{ curve area to the left of }-2.5\right)+\\ \left(z\text{ curve area to the right of 2}\text{.5}\right)\end{array}\right]\\ =P\left(z<-2.5\right)+1-P\left(z<2.5\right)\end{array}$

Use the Appendix- TABLE 2, Standard Normal Probabilities (Cumulative z Curve Areas) to obtain the required probability.

• Locate the value –2.5 in row.
• Locate the value 0.00 in column.
• The intersecting value of row and column is 0.0062.
• Locate the value 2.5 in row.
• Locate the value 0.00 in column.
• The intersecting value of row and column is 0.9938.

The required probability is obtained as given below:

$\begin{array}{c}P\left(x<2000\text{ or }x>5000\right)=0.0062+\left(1-0.9938\right)\\ =0.0062+0.0062\\ =0.0124\end{array}$

Thus, the probability that the birth weight of a randomly selected full-term baby is either less than 2,000 g or greater than 5,000 g is 0.0124.

To determine

Find the probability that the birth weight of a randomly selected full-term baby born in Canada exceeds 7 pounds.

Answer to Problem 72E

The probability that the birth weight of a randomly selected full-term baby exceeds 7 pounds is 0.7054.

Explanation of Solution

Calculation:

It is given that the birth weights are normally distributed with mean $\mu =3,500\text{ grams}$ and standard deviation $\sigma =600\text{ grams}$.

Standardize the score using the formula given below:

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

Convert 7 pounds into grams, that is, $7\times 453.59=3,175.13$ grams.

First calculate the value of z score as follows:

$\begin{array}{c}a*=\frac{a-\mu }{\sigma }\\ =\frac{3,175.13-3,500}{600}\\ =\frac{-324.87}{600}\\ =-0.54\end{array}$

Then, the required probability is obtained as given below:

$\begin{array}{c}P\left(x>3,175.13\right)=P\left(z>-0.54\right)\\ =z\text{ curve area to the right of }-\text{0}\text{.54}\\ =1-\left(z\text{ curve area to the left of }-0.54\right)\end{array}$

Use the Appendix-TABLE 2, Standard Normal Probabilities (Cumulative z Curve Areas): to obtain the probability.

• Locate the value –0.5 in row.
• Locate the value 0.04 in column.
• The intersecting value of row and column is 0.2946.

$\begin{array}{c}P\left(x>3,175.13\right)=1-0.2946\\ =0.7054\end{array}$

Thus, the probability that the birth weight of a randomly selected full-term baby exceeds 7 pounds is 0.7054.

To determine

Find the most extreme 0.1% of all full-term baby birth weights for babies.

Answer to Problem 72E

The most extreme 0.1% of all full-term baby birth weights for babies is greater than 5,474 grams and less than 1,526 grams.

Explanation of Solution

Calculation:

The formula for converting a z score $z*$ back to an x value $x*$ is as follows:

$x*=\mu +z*\sigma$

The value of $z*$ is obtained as given below:

$\begin{array}{c}P\left(z>z*\text{ or }z<-z*\right)=0.001\\ 2P\left(z>z*\right)=0.001\\ P\left(z>z*\right)=0.0005\\ P\left(z<z*\right)=0.9995\end{array}$

Use the Appendix-TABLE 2, Standard Normal Probabilities (Cumulative z Curve Areas): to obtain the $z*$ value:

• Locate the probability value of 0.9995.
• Move left until the first column is reached. Note the value as 3.2.
• Move upward until the first row is reached. Note the value as 0.09.
• The value of $z*$ is 3.29.

The required value is obtained as given below:

$P\left(z>3.29\text{ or }z<-3.29\right)=0.001$

The most extreme 0.1% values are greater than 3.29 and less than –3.29.

Then, the required weights are obtained as given below:

$\begin{array}{c}x*=3,500+\left(3.29\right)\times 600\\ =3,500+1,974\\ =5,474\end{array}$

$\begin{array}{c}x*=3,500-\left(3.29\right)\times 600\\ =3,500-1,974\\ =1,526\end{array}$

Thus, the most extreme 0.1% of all full-term baby birth weights for babies is greater than 5,474 grams and less than 1,526 grams.

To determine

Find the probability that the birth weight expressed in pounds of a randomly selected full-term baby exceeds 7 pounds.

Compare the result with Part (d).

Answer to Problem 72E

The probability that the birth weight expressed in pounds of a randomly selected full-term baby born in Canada exceeds 7 pounds is 0.7054.

The results are the same.

Explanation of Solution

Calculation:

It is given that a normal distribution is with mean $\mu =3,500\text{ grams}$ and standard deviation $\sigma =600\text{ grams}$.

Let x denote the birth weight of a randomly selected full-term baby.

Consider $y=ax$ also has a normal distribution and denotes the birth weight of a randomly selected full-term baby expressed in pounds.

The mean of y is obtained as given below:

$\begin{array}{c}{\mu }_y=\frac{3,500}{453.59}\\ =7.71622\end{array}$

The standard deviation of y is obtained as given below:

$\begin{array}{c}{\sigma }_y=\frac{600}{453.59}\\ =1.32278\end{array}$

First calculate the value of z score as follows:

$\begin{array}{c}a*=\frac{a-\mu }{\sigma }\\ =\frac{7-7.71622}{1.32278}\\ =\frac{-0.71622}{1.32278}\\ =-0.54\end{array}$

The required probability is obtained as given below:

$\begin{array}{c}P\left(y>7\right)=P\left(z>-0.54\right)\\ =z\text{ curve area to the right of }-\text{0}\text{.54}\\ =1-\left(z\text{ curve area to the left of }-0.54\right)\end{array}$

Use the Appendix- TABLE 2, Standard Normal Probabilities (Cumulative z Curve Areas): to obtain the probability.

• Locate the value –0.5 in row.
• Locate the value 0.04 in column.
• The intersecting value of row and column is 0.2946.

Now,

$\begin{array}{c}P\left(y>7\right)=1-0.2946\\ =0.7054\end{array}$

Thus, the probability that the birth weight expressed in pounds of a randomly selected full-term baby born in Canada exceeds 7 pounds is 0.7054.

From Part (d), the probability that the birth weight of a randomly selected full-term baby exceeds 7 pounds is 0.7054, which indicates that the probability value expressed in grams is the same as expressing in pounds.

Thus, the result is the same by expressing in pounds.