When conducting research on color blindness in males, a researcher forms random groups with five males in each group. The random variable x is the number of males in the group who have a form of color blindness. 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.
x P(x)
0 0.669
1 0.278
2 0.048
3 0.004
4 0.001
5 0.000
1. Does the table show a probability distribution?
A Yes, the table shows a probability distribution.
B. No, the random variable x's number values are not associated with probabilities.
C. No, not every probability is between O and 1 inclusive.
D. No, the sum of all the probabilities is not equal to 1.
E. No, the random variable x is categorical instead of numerical.
2. Find the mean of the random variable x.
a. mu = male(s).
b. the table does not show a probability distribution.
3. Find the standard deviation of the random variable x.
a. infinity = male(s).
b. the table does not show a probability distribution.

The random variable x is the number of males in the group who have a form of color blindness. Determine whether a probability distribution is given:

A probability distribution is given if the sum of all probabilities is 1.

We have that:

0.669 + 0.278 + 0.048 + 0.004 + 0.001 + 0 = 1

So yes, the table shows a probability distribution.

2. Find the mean of the random variable x.

We multiply each value by its probability. So

[tex]\mu = 0*0.669 + 1*0.278 + 2*0.048 + 3*0.004 + 4*0.001 + 5*0.000 = 0.39[/tex]

The mean is 0.39 males.

3. Find the standard deviation of the random variable x.

Square root of the subtraction squared of each value and the mean, multiplied by its respective probability. So

[tex]\sigma = \sqrt{0.669*(0-0.39)^2+0.278*(1-0.39)^2+0.048*(2-0.39)^2+0.004*(3-0.39)^2+0.001*(4-0.39)^2+0.000*(5-0.39)^2} = \sqrt{0.3699} = 0.6082[/tex]

The standard deviation is of 0.6082 males.