Please help with both problems thank you!
2 answers:
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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.
We are given the following expression:
We can factor 875 as:
Now we will use the following property:
Applying the property:
Solving the cubic root on the left:
Since we get simplify any further this is the answer.