91Ó°ÊÓ

A sample of 20 births is considered. The proportion of boys (p) is equal to 0.512.

It is required that$npâ\copyright ¾5$ and $nqâ\copyright ¾5$.

The values are computed below:

$$\begin{gathered} np=\left(20\right)\left(0.512\right) \\ =10.24 \\ â\copyright ¾5 \end{gathered}$$

$$\begin{gathered} nq=\left(20\right)\left(1-0.512\right) \\ =9.76 \\ â\copyright ¾5 \end{gathered}$$

As the requirement is fulfilled, the normal approximation can be applied to compute the probability value.

The mean value is equal to:

$$\begin{gathered} \mathrm{μ}=np \\ =20×0.512 \\ =10.24 \end{gathered}$$

The standard deviation is equal to

$$\begin{gathered} \mathrm{σ}=\sqrt{npq} \\ =\sqrt{20×0.512×0.488} \\ =2.235 \end{gathered}$$

Let x represent the number of boys in the sample.

Here, x is equal to 8.

The value of x is transformed as follows:

$$\begin{gathered} \left(x-0.5,x+0.5\right)=\left(8-0.5,8+0.5\right) \\ =\left(7.5,8.5\right) \end{gathered}$$

It is required to compute the probability of fewer than 8 boys. Thus, the probability to the left of 7.5 needs to be computed.

The z score value is equal to:

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{7.5-10.24}{2.235} \\ =-1.23 \end{gathered}$$

The z-score is -1.23.

Referring to the standard normal distribution, the area to the left of -1.23 is equal to 0.1093.

Therefore, the probability of fewer than 8 boys is equal to 0.1093.