91Ó°ÊÓ

It is given that 36 couples have tried the XSORT method to produce a baby. The probability of a girl is equal to 0.5.

a.

Here, the probability of a girl is given to be p=0.5.

The probability of a boy is computed below:

$$\begin{gathered} q=1-p \\ =1-0.5 \\ =0.5 \end{gathered}$$

The number of trials (n) is equal to 36.

Thus, the mean value is given as follows:

$$\begin{gathered} \mathrm{μ}=np \\ =\left(36\right)\left(0.5\right) \\ =18.0 \end{gathered}$$

Therefore, the mean number of girls is equal to 18.0.

The standard deviation is computed below:

$$\begin{gathered} \mathrm{σ}=\sqrt{npq} \\ =\sqrt{\left(36\right)\left(0.5\right)\left(0.5\right)} \\ =3.0 \end{gathered}$$

Therefore, the standard deviation of the number of girls in 36 births is equal to 3.0.

b.

By using the range rule of thumb, the significantly low number of girls is computed below:

$$\begin{gathered} \mathrm{μ}-2\mathrm{σ}=18.0-\left(2\right)\left(3.0\right) \\ =12.0 \end{gathered}$$

Thus, the significantly low number of girls is12.0 or less.

The significantly high number of girls are computed below:

$$\begin{gathered} \mathrm{μ}+2\mathrm{σ}=18.0+\left(2\right)\left(3.0\right) \\ =24.0 \end{gathered}$$

Thus, the significantly high number of girls is24.0 or more.

Moreover, the values that are not significant will lie between 12.0 and 24.0.

Here, the value of 26 can be considered significantly high as it is greater than 24.0.

The value of 26 girls suggests that the XSORT method is effective.