Respuesta :
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.
[tex]\sum P(X)=0.029+0.147+0.324+0.324+0.147+0.029[/tex]
[tex]\sum P(X)=1[/tex]
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
[tex]\mu=\sum xP(x)=2.5[/tex]
The standard deviation of the random sample is
[tex]Variance=\sigma^2[/tex]
[tex]\sigma^2=\sum x^2P(x)-\mu^2[/tex]
[tex]\sigma^2=7.436-(2.5)^2[/tex]
[tex]\sigma^2=7.436-6.25[/tex]
[tex]\sigma^2=1.186[/tex]
[tex]\sigma=\sqrt{1.186}[/tex]
[tex]\sigma=1.08[/tex]
Therefore, The mean is 2.5 and the standard deviation is 1.08.
