91影视

It is given that 16 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 16.

Thus, the mean value is given as follows:

$$\begin{gathered} 渭=np \\ =\left(16\right)\left(0.5\right) \\ =8.0 \end{gathered}$$

Therefore, the mean number of girls is equal to 8.0.

The standard deviation is computed below:

$$\begin{gathered} 蟽=\sqrt{npq} \\ =\sqrt{\left(16\right)\left(0.5\right)\left(0.5\right)} \\ =2.0 \end{gathered}$$

Therefore, the standard deviation of the number of girls in 16 births is equal to 2.0.

b.

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

$$\begin{gathered} 渭-2蟽=8.0-\left(2\right)\left(2.0\right) \\ =4.0 \end{gathered}$$

Thus, the significantly low number of girls is4.0 or less.

The significantly high number of girls are computed below:

$$\begin{gathered} 渭+2蟽=8.0+\left(2\right)\left(2.0\right) \\ =12.0 \end{gathered}$$

Thus, the significantly high number of girlsis12.0 or more.

Moreover, the values that are not significant will lie between 4.0 and 12.0.

c.

Here, the value of 11lies between 4.0 and 12.0.

Therefore, the value of 11 girls cannot be considered significantly high as it is less than 12.0.

The value of 11 girls suggests that the XSORT method is neither very effective nor ineffective.