In Exercises 15–20, refer to the accompanying table,which describes results from groups of 8 births from 8 differentsets of parents. The random variable x represents the number ofgirls among 8 children.

Find the mean and standarddeviation for the numbers of girls in 8 births.

Number of girls x	P(x)
0	0.004
1	0.031
2	0.109
3	0.219
4	0.273
5	0.219
6	0.109
7	0.031
8	0.004

The probability distribution for the number of girls among 8 children is provided.

The variable x is the number of girls among 8 children.

The requirements are as follows:

1)The variable x is anumerical random variable.

2)The sum of the probabilities is computed as:

$$\begin{gathered} \sum P\left(x\right)=0.004+0.031+0.109+...+0.004 \\ =0.999 \end{gathered}$$

Therefore,the sum of the probabilities is approximately equal to 1 with a round of error as 0.001.

3) Each value of P(x) is between 0 and 1.

Thus, all the requirements are satisfied.

The mean for the random variable x is computed as:

$$\begin{gathered} \mu =\sum \left[x\times P\left(x\right)\right] \\ =\left(0\times 0.004\right)+\left(1\times 0.031\right)+\left(2\times 0.109\right)+...+\left(8\times 0.004\right) \\ =3.996 \\ \approx 4.0 \end{gathered}$$

Thus, the mean number of girls in 8 births is 4.0.

The standard deviation of the random variable x is computed as:

$\sigma =\sqrt{\sum \left[x^2\times P\left(x\right)\right]-{\mu }^2}$

The calculations are as follows:

$$\begin{gathered} \sum \left[x^2·P\left(x\right)\right]=\left(0^2\times 0.004\right)+\left(1^2\times 0.031\right)+\left(2^2\times 0.109\right)+...+\left(8^2\times 0.004\right) \\ =17.98 \end{gathered}$$

The standard deviation is given as:

$$\begin{gathered} \sigma =\sqrt{\sum \left[x^2·P\left(x\right)\right]-{\mu }^2} \\ =\sqrt{17.98-{3.996}^2} \\ =1.41 \\ \approx 1.4 \end{gathered}$$

Thus, the standard deviation of girls in 8 births is 1.4.