91影视

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

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

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

$$\begin{gathered} 渭=鈭�\left[x脳P\left(x\right)\right] \\ =\left(0脳0.659\right)+\left(1脳0.287\right)+\left(3脳0.05\right)+...+\left(5脳0.000\right) \\ =0.403 \end{gathered}$$

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

The standard deviation of the random variable x is computed as

$蟽=\sqrt{鈭�\left[x^2脳P\left(x\right)\right]-{渭}^2}$

The calculations are as follows.

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

The standard deviation is given as

$$\begin{gathered} 蟽=\sqrt{鈭�\left[x^2路P\left(x\right)\right]-{渭}^2} \\ =\sqrt{0.539-{0.403}^2} \\ =0.6136 \\ 鈮�0.6 \end{gathered}$$

Thus, the standard deviation of x is 0.6.