In: Statistics and Probability
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 0.164
2 0.307
3 0.307
4 0.164
5 0.029
Does the table show a probability distribution? Select all that apply.
A
Yes table shows a probability distribution.
B.
No, the random variable x's number values are not associated with probabilities.
C.
No, the random variable x is categorical instead of numerical.
D.
No, not every probability is between 0 and 1 inclusive.
E.
No, the sum of all the probabilities is not equal to 1.
Find the mean of the random variable x. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
nothing
A. u =. child(ren) (Round to one decimal place as needed
B. The table does not show a probability distribution.
Find the standard deviation of the random variable x. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. O= child(ren) (Round to one decimal place as
needed.)
B. The table does not show a probability distribution.
Solution :
x | P(x) | x * P(x) | x2 * P(x) |
0 | 0.029 | 0 | 0 |
1 | 0.164 | 0.164 | 0.164 |
2 | 0.307 | 0.614 | 1.228 |
3 | 0.307 | 0.921 | 2.763 |
4 | 0.164 | 0.656 | 2.624 |
5 | 0.029 | 0.145 | 0.725 |
Sum | 1 | 2.5 | 7.504 |
A) Yes table shows a probability distribution.
Mean = = X * P(X) = 2.5
Standard deviation =
=X 2 * P(X) - 2
= 7.504 - 2.52
= 1.1