Five males with an x-linked genetic disorder have one child each. the random variable x is the number of children among the five who inherit the x-linked genetic disorder. 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 0.029
1 0.147
2 0.324
3 0.324
4 0.147
5 0.029
a. does the table show a probability distribution?
b. find the mean of the random variable x.

Answer and Explanation:

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

To find :

a. Does the table show a probability distribution?

b. Find the mean and standard deviation of the random variable x.

Solution :

a) To determine that table shows a probability distribution we add up all six probabilities if the sum is 1 then it is a valid distribution.

Yes it is a probability distribution.

b) First we create the table as per requirements,

x      P(x)         xP(x)           x²            x²P(x)

0    0.029         0              0                0

1     0.147        0.147           1              0.147

2    0.324       0.648         4              1.296

3    0.324       0.972         9              2.916

4    0.147        0.588        16              2.352

5    0.029       0.145         25            0.725

   ∑P(x)=1      ∑xP(x)=2.5               ∑x²P(x)=7.436

The mean of the random variable is

The standard deviation of the random sample is

Therefore, The mean is 2.5 and the standard deviation is 1.08.

answer: i'm not sure

step-by-step explanation: