Question

Durations of PregnanciesThe lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days.

a. In a letter to “Dear Abby,” a wife claimed to have given birth 308 days after a brief visit fromher husband, who was working in another country. Find the probability of a pregnancy lasting308 days or longer. What does the result suggest?

b. If we stipulate that a baby is prematureif the duration of pregnancy is in the lowest 3%,find the duration that separates premature babies from those who are not premature. Premature babies often require special care, and this result could be helpful to hospital administrators in planning for that care.

Short answer

a. The probabilityof a pregnancy lasting 308 days or longer is 0.0038. This result suggests thatit is unusual for a pregnancy to last 308 days or longer.

b. The duration that separates premature babies from those who are not premature is 240 days.

Step-by-step solution

Given information

The lengths of pregnancies are normally distributed.

The mean number of days is$\mu =268$.

The standard deviation of the number of days is$\sigma =15$.

Compute the probability 

Let x represent the lengths of pregnancies.

a.

The probability of a pregnancy lasting 308 days or longer is computed as follows.

$$\begin{gathered} P\left(x⩾308\right)=P\left(\frac{x-\mu }{\sigma }⩾\frac{308-268}{15}\right) \\ =P\left(Z⩾2.67\right) \\ =1-P\left(Z<2.67\right) \\ =1-0.9962 \\ =0.0038 \end{gathered}$$

Therefore, the probability of a pregnancy lasting 308 days or longer is 0.0038.

The results suggest that it is unusual for a pregnancy to last 308 days or longer as the probability is less than 0.05.

Compute the probability 

b.

Given: A baby is premature if the duration of pregnancy is in the lowest 3%.

The probability for the length separating the bottom 3% from the top 97% is given as

$P\left(Z<z\right)=0.03$

Using the standard normal table,

• thearea of 0.03 is observed corresponding to the row value 1 and the column value 0.9, whichimplies that the z-score is 1.9.

Mathematically,

$P\left(Z<-1.9\right)=0.03$

The duration that separates premature babies from those who are not prematureis computed as follows.

$$\begin{gathered} \frac{x-\mu }{\sigma }=-1.9 \\ \frac{x-268}{15}=-1.9 \\ x=\left(-1.9\times 15\right)+268 \\ =239.5 \\ \approx 240 \end{gathered}$$

Thus, the duration that separates premature babies from those who are not premature is 240 days.