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The lengths of pregnancies are normally distributed.

The mean number of days is$渭=268$.

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

Let x represent the lengths of pregnancies.

a.

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

$$\begin{gathered} P\left(x猢�308\right)=P\left(\frac{x-渭}{蟽}猢�\frac{308-268}{15}\right) \\ =P\left(Z猢�2.67\right) \\ =1-P\left(Z<2.67\right) \\ =1-0.9962 \\ =0.0038 \end{gathered}$$

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.

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,

Mathematically,

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

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

$$\begin{gathered} \frac{x-渭}{蟽}=-1.9 \\ \frac{x-268}{15}=-1.9 \\ x=\left(-1.9脳15\right)+268 \\ =239.5 \\ 鈮�240 \end{gathered}$$

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