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.