91Ó°ÊÓ

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

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

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

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

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

The standard deviation of the random variable x is computed as

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

The calculations are as follows.

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

The standard deviation is given as

$$\begin{gathered} \mathrm{σ}=\sqrt{∑\left[x^2·P\left(x\right)\right]-{\mathrm{μ}}^2} \\ =\sqrt{7.496-{2.5}^2} \\ =1.1162 \\ ≈1.1 \end{gathered}$$

Thus, the standard deviation of x is 1.1.