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.

x	p(x)
0	0.028
1	0.156
2	0.316
3	0.316
4	0.156
5	0.028

does the table show a probability distribution? select all that apply.
□a. yes, the table shows a probability distribution.
□b. no, the random variable xs number values are not associated with probabilities.
□c. no, the random variable x is categorical instead of numerical.
□d. no, the sum of all the probabilities is not equal to 1
□e. no, not every probability is between 0 and 1 inclusive
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 = □ 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.

Explanation:

Step1: Check probability - distribution requirements

We need to check if $\sum_{x}P(x)=1$ and $0\leq P(x)\leq1$ for all $x$.
Calculate $\sum_{x = 0}^{5}P(x)=0.028 + 0.156+0.316 + 0.316+0.156 + 0.028=1$. Also, $0\leq P(x)\leq1$ for $x = 0,1,\cdots,5$. So it is a probability - distribution.

Step2: Calculate the mean $\mu$

The formula for the mean of a discrete probability distribution is $\mu=\sum_{x}x\cdot P(x)$.
$\mu=0\times0.028 + 1\times0.156+2\times0.316 + 3\times0.316+4\times0.156 + 5\times0.028$
$=0 + 0.156+0.632+0.948 + 0.624+0.14$
$=2.5$.

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

The formula for the variance is $\sigma^{2}=\sum_{x}(x - \mu)^{2}\cdot P(x)$.
$(0 - 2.5)^{2}\times0.028+(1 - 2.5)^{2}\times0.156+(2 - 2.5)^{2}\times0.316+(3 - 2.5)^{2}\times0.316+(4 - 2.5)^{2}\times0.156+(5 - 2.5)^{2}\times0.028$
$=( - 2.5)^{2}\times0.028+( - 1.5)^{2}\times0.156+( - 0.5)^{2}\times0.316+(0.5)^{2}\times0.316+(1.5)^{2}\times0.156+(2.5)^{2}\times0.028$
$=6.25\times0.028 + 2.25\times0.156+0.25\times0.316+0.25\times0.316+2.25\times0.156+6.25\times0.028$
$=0.175+0.351 + 0.079+0.079+0.351+0.175$
$=1.21$.

Step4: Calculate the standard - deviation $\sigma$

The standard - deviation $\sigma=\sqrt{\sigma^{2}}$. Since $\sigma^{2}=1.21$, then $\sigma=\sqrt{1.21}=1.1$.

Answer:

A. Yes, the table shows a probability distribution.
A. $\mu = 2.5$ child(ren)
A. $\sigma = 1.1$