Chapter 7, Problem 121CR

Let x denote the duration of a randomly selected pregnancy (the time elapsed between conception and birth). Accepted values for the mean value and standard deviation of x are 266 days and 16 days, respectively. Suppose that the probability distribution of x is (approximately) normal.

1. a. What is the probability that the duration of pregnancy is between 250 and 300 days?
2. b. What is the probability that the duration of pregnancy is at most 240 days?
3. c. What is the probability that the duration of pregnancy is within 16 days of the mean duration?
4. d. A “Dear Abby” column dated January 20, 1973, contained a letter from a woman who stated that the duration of her pregnancy was exactly 310 days. (She wrote that the last visit with her husband, who was in the Navy, occurred 310 days before birth.) What is the probability that duration of pregnancy is 310 days or more? Does this probability make you a bit skeptical of the claim?
5. e. Some insurance companies will pay the medical expenses associated with childbirth only if the insurance has been in effect for more than 9 months (275 days). This restriction is designed to ensure that the insurance company pays benefits for only those pregnancies for which conception occurred during coverage. Suppose that conception occurred 2 weeks after coverage began. What is the probability that the insurance company will refuse to pay benefits because of the 275-day insurance requirement?

To determine

Obtain the probability that the duration of pregnancy is between 250 and 300 days.

Answer to Problem 121CR

The probability that the duration of pregnancy is between 250 and 300 days is 0.8247.

Explanation of Solution

Calculation:

It is given that the duration of pregnancy is normally distributed with mean and standard deviation as 266 days and 16 days, respectively.

That is, $\mu =266\text{ days and }\sigma =16\text{ days}$

Define the random variable x as duration of randomly selected pregnancy.

The general formula to x value to z score is $z*=\frac{x*-\mu }{\sigma }$

Standardize $a=250$ using the above formula.

$\begin{array}{c}z*=\frac{a*-\mu }{\sigma }\\ =\frac{250-266}{16}\\ =\frac{-16}{16}\\ =-1\end{array}$

Standardize $b=300$ using the above formula.

$\begin{array}{c}z*=\frac{b*-\mu }{\sigma }\\ =\frac{300-266}{16}\\ =\frac{\text{34}}{16}\\ =2.13\end{array}$

The required probability is obtained as given below:

$P\left(-1<z<2.13\right)=P\left(z<2.13\right)-P\left(z<-1\right)$

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z<-1\right)$.

Procedure:

• In the $z*$ row locate −1.0.
• In column locate .00.
• The intersection of the row −1.0 and column .00 gives 0.1587.

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z<2.13\right)$.

Procedure:

• In the $z*$ row locate 2.1.
• In column locate .03.
• The intersection of the row 2.1 and column .03 gives 0.9834.

The required probability is obtained as given below:

$\begin{array}{c}P\left(-1<z<2.13\right)=P\left(z<2.13\right)-P\left(z<-1\right)\\ =0.9834-0.1587\\ =\text{0}\text{.8247}\end{array}$

Thus, the probability that the duration of pregnancy is between 250 and 300 days is 0.8247.

To determine

Find the probability that the duration of pregnancy is at most 240 days.

Answer to Problem 121CR

The probability that the duration of pregnancy is at most 240 days is 0.0516.

Explanation of Solution

Calculation:

Standardize $c=240$ using the z* formula.

$\begin{array}{c}z*=\frac{c*-\mu }{\sigma }\\ =\frac{240-266}{16}\\ =\frac{-26}{16}\\ =-1.63\end{array}$

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z<-1.63\right)$.

Procedure:

• In the $z*$ row locate −1.6.
• In column locate .03.
• The intersection of the row −1.6 and column .03 gives 0.0516.

The required probability is obtained as given below:

$P\left(z<-1.63\right)=\text{0}\text{.0516}$

Thus, the probability that the duration of pregnancy is at most 240 days is 0.0516.

To determine

Find the probability that the duration of pregnancy is within 16 days of the mean duration.

Answer to Problem 121CR

The probability that the duration of pregnancy is within 16 days of the mean duration is 0.6826.

Explanation of Solution

Calculation:

The duration of 16 days of pregnancy represents one standard deviation from the mean.

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z<-1\right)$.

Procedure:

• In the $z*$ row locate −1.0.
• In column locate .00.
• The intersection of the row −1.0 and column .00 gives 0.1587.

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z<1\right)$.

Procedure:

• In the $z*$ row locate 1.0.
• In column locate .00.
• The intersection of the row 1.0 and column .00 gives 0.8413.

The required probability is obtained as given below:

$\begin{array}{c}P\left(-1\le z\le 1\right)=\text{P}\left(z\le 1\right)-P\left(z\le 1\right)\\ =0.8413-0.1587\\ =\text{0}\text{.6826}\end{array}$

Thus, the probability that the duration of pregnancy is within 16 days of the mean duration is 0.6826.

To determine

Find the probability that duration of pregnancy is 310 days or more.

Check whether the probability is a bit skeptical of the claim.

Answer to Problem 121CR

The probability that duration of pregnancy is 310 days or more is 0.003.

The claim is skeptical as the probability is very low.

Explanation of Solution

Calculation:

It is given that a letter from a woman stated that the duration of her pregnancy was exactly 310 days.

Standardize $d=310$ using the z* formula.

$\begin{array}{c}z*=\frac{d*-\mu }{\sigma }\\ =\frac{310-266}{16}\\ =\frac{44}{16}\\ =\text{2}\text{.75}\end{array}$

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z<2.75\right)$.

Procedure:

• In the $z*$ row locate 2.7.
• In column locate .05.
• The intersection of the row 2.7 and column .05 gives 0.9970.

The required probability is obtained as given below:

$\begin{array}{c}P\left(z>2.75\right)=1-P\left(z\le 2.75\right)\\ \text{=1}-0.9970\\ =\text{0}\text{.003}\end{array}$

Thus, the probability that duration of pregnancy is 310 days or more is 0.003.

Here, the probability is very low and hence duration of pregnancy is 310 days or more is unlikely to happen.

Thus, the claim is skeptical.

To determine

Find the probability that the insurance company will refuse to pay benefits.

Answer to Problem 121CR

The probability that the insurance company will refuse to pay benefits is 0.3783.

Explanation of Solution

Calculation:

The medical expenses associated with childbirth are paid by some insurance companies only when the insurance has been for more than 9 months (275 days). This restriction is made because the insurance company pays benefits for the pregnancies for which the conception occurred during the coverage.

Here, the conception occurs two weeks after the coverage began. Hence, the insurance company would refuse to pay if the pregnancy is less than or equal to 261 $\left(=275-14\right)$ days.

Standardize $e=261$ using the z* formula.

$\begin{array}{c}z*=\frac{e*-\mu }{\sigma }\\ =\frac{261-266}{16}\\ =\frac{-5}{16}\\ =-0.31\end{array}$

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z<-0.31\right)$.

Procedure:

• In the $z*$ row locate −0.3.
• In column locate .01.
• The intersection of the row −0.3 and column .01 gives 0.3783.

The required probability is obtained as given below:

$P\left(z<-0.31\right)=\text{0}\text{.3783}$.

Thus, the probability that the insurance company will refuse to pay benefits is 0.3783.