91Ó°ÊÓ

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:

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

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:

$$\begin{gathered} \mathrm{μ}=∑\left[x×P\left(x\right)\right] \\ =\left(0×0.004\right)+\left(1×0.031\right)+\left(2×0.109\right)+...+\left(8×0.004\right) \\ =3.996 \\ ≈4.0 \end{gathered}$$

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

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

$\mathrm{σ}=\sqrt{∑\left[x^2×P\left(x\right)\right]-{\mathrm{μ}}^2}$

The calculations are as follows:

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

The standard deviation is given as:

$$\begin{gathered} \mathrm{σ}=\sqrt{∑\left[x^2·P\left(x\right)\right]-{\mathrm{μ}}^2} \\ =\sqrt{17.98-{3.996}^2} \\ =1.41 \\ ≈1.4 \end{gathered}$$

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