91Ó°ÊÓ

The probability distribution for color blindness in males is provided.

The variable x is the number of males in the group who have a form of color blindness.

The requirements are as follows.

1)The variable x isa numerical random variable.

2)The sum of the probabilities is computed as

$$\begin{gathered} ∑P\left(x\right)=0.659+0.287+0.05+0.004+0.001+0.000 \\ =1.001 \end{gathered}$$

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

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

Thus, the probability distribution is valid.

The mean for the random variable is computed as

$$\begin{gathered} \mathrm{μ}=∑\left[x×P\left(x\right)\right] \\ =\left(0×0.659\right)+\left(1×0.287\right)+\left(3×0.05\right)+...+\left(5×0.000\right) \\ =0.403 \end{gathered}$$

Thus, the mean value of the random variable x is 0.4.

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.

$$\begin{gathered} ∑\left[x^2·P\left(x\right)\right]=\left(0^2×0.659\right)+\left(1^2×0.287\right)+\left(2^2×0.05\right)+...+\left(5^2×0.000\right) \\ =0.539 \end{gathered}$$

The standard deviation is given as

$$\begin{gathered} \mathrm{σ}=\sqrt{∑\left[x^2·P\left(x\right)\right]-{\mathrm{μ}}^2} \\ =\sqrt{0.539-{0.403}^2} \\ =0.6136 \\ ≈0.6 \end{gathered}$$

Thus, the standard deviation of x is 0.6.