Chapter 18, Problem 37E

(a)

To determine

To find out what percentage of pregnancies should last between $2\text{70 and 280}$ days.

Answer to Problem 37E

  $21.1\%$ .

Explanation of Solution

It is given in the question that we have assume that the duration of human pregnancies can be described by a normal model with mean $266$ days and standard deviation $16$ days. Thus, let us first find the z -scores as:

  $\begin{array}{l}{z}_{270}=\frac{270-266}{16}\\ =0.25\\ z_{280}=\frac{280-266}{16}\\ =0.875\end{array}$

Thus, the percentage of pregnancies should last between $2\text{70 and 280}$ days is as:

  $\begin{array}{l}P(0.25\le z\le 0.875)=normalcdf(0.25,0.875,0,1)\\ =0.211\end{array}$

Thus, the percentage of pregnancies should last between $2\text{70 and 280}$ days is $21.1\%$ .

(b)

To determine

To find out at least how many days should the longest $25\%$ of all pregnancies last.

Answer to Problem 37E

At least $276.792$ days should the longest $25\%$ of all pregnancies last.

Explanation of Solution

It is given in the question that we have assume that the duration of human pregnancies can be described by a normal model with mean $266$ days and standard deviation $16$ days. Thus, using graphing calculator to find $invNorm(0.75,0,1)$ to find z -score of $0.674$ . Thus, plug in z -score to find x , we have,

  $\begin{array}{l}0.674=\frac{x-266}{16}\\ \Rightarrow 0.674\times 16=x-266\\ \Rightarrow x=266+0.674\times 16\\ \Rightarrow x=276.792\end{array}$

Thus, we conclude that at least $276.792$ days should the longest $25\%$ of all pregnancies last.

(c)

To determine

To find out what is the distribution of this sample mean $\overline{y}$ and specify the model, mean and the standard deviation.

Answer to Problem 37E

The mean is $266$ days and the standard deviation is $2.066$ days.

Explanation of Solution

It is given in the question that we have assume that the duration of human pregnancies can be described by a normal model with mean $266$ days and standard deviation $16$ days. Thus, the conditions are as:

Random condition: It is satisfied as we are assuming that the $60$ women were randomly selected.

  $10\%$ condition: The sample size less than $10\%$ of the population size.

Thus, the conditions are met. And the model of this sampling distribution is the Normal model and the mean and the standard deviation is as follows:

  $\begin{array}{l}\mu ={\mu }_{\overline{y}}=266\\ {\sigma }_{\overline{y}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{16}{\sqrt{60}}\\ =2.066\end{array}$

(d)

To determine

To find out what is the probability that the mean duration of these patients’ pregnancies will be less than $260$ days.

Answer to Problem 37E

  $0.002$ .

Explanation of Solution

It is given in the question that we have assume that the duration of human pregnancies can be described by a normal model with mean $266$ days and standard deviation $16$ days. Now, let us find the z -score as:

  $\begin{array}{l}{z}_{260}=\frac{260-266}{2.066}\\ =-2.904\end{array}$

Thus, using the graphical method we will find the probability that the mean duration of these patients’ pregnancies will be less than $260$ days as:

  $\begin{array}{l}P(z<-2.904)=normalcdf(-E99,-2.904,0,1)\\ =0.002\end{array}$