Question

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.
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.
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.

Explanation:

Step1: Calculate the variance

The formula for variance \(\sigma^{2}=\sum(x^{2}\cdot P(x))-\mu^{2}\).
First, calculate \(\sum(x^{2}\cdot P(x))\):

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We know \(\mu = 2.5\), then \(\mu^{2}=2.5^{2}=6.25\)
So \(\sigma^{2}=7.524 - 6.25=1.274\)

Step2: Calculate the standard deviation

The formula for standard deviation \(\sigma=\sqrt{\sigma^{2}}\)
Since \(\sigma^{2}=1.274\), then \(\sigma=\sqrt{1.274}\approx1.1\)

Answer:

\(\sigma = 1.1\) child(ren)