In Exercises 7–14, determine whether a probability

distribution is given. If a probability distribution is given, find its mean and standard deviation. If a probability distribution is not given, identify the requirements that are not satisfied.

Five males with an X-linked genetic disorder have one child each. The random variable xis the number of children among the five who inherit the X-linked genetic disorder.

x	P(x)
0	0.031
1	0.156
2	0.313
3	0.313
4	0.156
5	0.031

The probability distribution for the five males with an X-linked genetic disorder, having one child each, is provided.

The variable x is the number of children among the five who inherit the X-linked genetic disorder.

The requirements are as follows.

1) Variable x is anumerical random variable.

2)The sum of the probabilities is computed as

$$\begin{gathered} \sum P\left(x\right)=0.031+0.156+0.313+0.313+0.156+0.031 \\ =1 \end{gathered}$$

Therefore,the sum of the probabilities is equal to 1.

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

Thus, the stated distribution is a valid probability distribution.

The mean for the random variable is computed as

$$\begin{gathered} \mu =\sum \left[x·P\left(x\right)\right] \\ =\left(0\times 0.031\right)+\left(1\times 0.156\right)+\left(2\times 0.313\right)+...+\left(5\times 0.155\right) \\ =2.5 \end{gathered}$$

Thus, the mean value of the random variable x is 2.5.

The standard deviation of the random variable x is computed as

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

The calculations are as follows.

$$\begin{gathered} \sum \left[x^2·P\left(x\right)\right]=\left(0^2\times 0.031\right)+\left(1^2\times 0.156\right)+\left(2^2\times 0.313\right)+...+\left(5^2\times 0.155\right) \\ =7.496 \end{gathered}$$

The standard deviation is given as

$$\begin{gathered} \sigma =\sqrt{\sum \left[x^2·P\left(x\right)\right]-{\mu }^2} \\ =\sqrt{7.496-{2.5}^2} \\ =1.1162 \\ \approx 1.1 \end{gathered}$$

Thus, the standard deviation of x is 1.1.