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
c. no, not every probability is between 0 and 1 inclusive
d. no, the random variable x is categorical instead of numerical
e. no, the random variable xs number values are not associated with probabilities
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
a. μ = 2.5 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. σ = child(ren) (round to one decimal place as needed.)
b. the table does not show a probability distribution

Explanation:

Step1: Check probability - distribution requirements

All probabilities $P(x)$ are between 0 and 1 inclusive, and $\sum_{x = 0}^{5}P(x)=0.033 + 0.163+0.304 + 0.304+0.163 + 0.033=1$. The random - variable $x$ is numerical and its values are associated with probabilities, so it is a probability distribution.

Step2: Calculate the mean $\mu$

The formula for the mean of a discrete probability distribution is $\mu=\sum_{x}x\cdot P(x)$. So, $\mu=0\times0.033 + 1\times0.163+2\times0.304 + 3\times0.304+4\times0.163 + 5\times0.033=0 + 0.163+0.608+0.912 + 0.652+0.165 = 2.5$.

Step3: Calculate the variance $\sigma^{2}$

The formula for the variance is $\sigma^{2}=\sum_{x}(x - \mu)^{2}\cdot P(x)$.
$(0 - 2.5)^{2}\times0.033=( - 2.5)^{2}\times0.033 = 6.25\times0.033 = 0.20625$
$(1 - 2.5)^{2}\times0.163=( - 1.5)^{2}\times0.163 = 2.25\times0.163 = 0.36675$
$(2 - 2.5)^{2}\times0.304=( - 0.5)^{2}\times0.304 = 0.25\times0.304 = 0.076$
$(3 - 2.5)^{2}\times0.304=(0.5)^{2}\times0.304 = 0.25\times0.304 = 0.076$
$(4 - 2.5)^{2}\times0.163=(1.5)^{2}\times0.163 = 2.25\times0.163 = 0.36675$
$(5 - 2.5)^{2}\times0.033=(2.5)^{2}\times0.033 = 6.25\times0.033 = 0.20625$
$\sigma^{2}=0.20625+0.36675 + 0.076+0.076+0.36675+0.20625 = 1.298$

Step4: Calculate the standard deviation $\sigma$

$\sigma=\sqrt{\sigma^{2}}=\sqrt{1.298}\approx1.1$

Answer:

The mean $\mu = 2.5$ child(ren).
The standard deviation $\sigma\approx1.1$ child(ren)