Chapter PIV, Problem 11RE

(a)

To determine

To find: the mean and standard deviation could be predicted for the number of twin births.

Answer to Problem 11RE

Mean =15.2

Standard deviation = 3.7

Explanation of Solution

Given:

Pregnant women total = 152

X = number of pregnant women that have took clomid and having twins

Formula used:

  $\begin{array}{c}\mu =E\left(X\right)\\ \sigma =\sqrt{npq}\end{array}$

Calculation:

The probability that pregnant women have twins is $\frac{1}{10}\text{=0}\text{.1}$

Number of twin births it could expect is,

  $\begin{array}{c}\mu =E\left(X\right)\\ =np\\ =\left(152\right)\left(0.1\right)\\ =15.2\end{array}$

It could expect that 15.2 twin births.

Standard deviation

  $\begin{array}{c}\sqrt{npq}=\sqrt{\left(152\right)\left(0.1\right)\left(0.9\right)}\\ =3.7\end{array}$

Mean =15.2

Standard deviation = 3.7

(b)

To determine

To explain: In this case, use a normal model.

Explanation of Solution

Given:

  $\begin{array}{c}n=152\\ q=0.9\end{array}$

Calculation:

Yes, we can use normal distribution

1. Trails are independent
2. $\begin{array}{c}np=15.2\\ nq=\left(152\right)\left(0.9\right)\\ =136.8\end{array}$

There is $np\ge 10\text{ and }nq\ge 10$

Therefore, X is norm (15.2, 3.7)

(c)

To determine

To Find: the chance that not greater than 10 women will have twins.

Explanation of Solution

Given:

  $\begin{array}{c}x=10\\ \mu =15.2\\ \sigma =3.7\end{array}$

Calculation:

The chance of no more than 10 women getting twins is

  $\begin{array}{c}P\left[P<10\right]=P\left[Z<\frac{10-15.2}{3.7}\right]\\ =P\left[Z<-1.41\right]\\ =0.079\end{array}$

Graph

  

Z= -1.41