five males with an x - linked genetic disorder have one child each. the random variable x is the number of…

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.\n|x|p(x)|\n|0|0.028|\n|1|0.156|\n|2|0.316|\n|3|0.316|\n|4|0.156|\n|5|0.028|\ndoes the table show a probability distribution? select all that apply.\n□a. yes, the table shows a probability distribution.\n□b. no, the random variable xs number values are not associated with probabilities.\n□c. no, the random variable x is categorical instead of numerical.\n□d. no, the sum of all the probabilities is not equal to 1\n□e. no, not every probability is between 0 and 1 inclusive\nfind the mean of the random variable x. select the correct choice below and, if necessary, fill in the answer box to complete your choice.\n○a. μ = □ child(ren) (round to one decimal place as needed.)\n○b. the table does not show a probability distribution.\nfind 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.

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.\n|x|p(x)|\n|0|0.028|\n|1|0.156|\n|2|0.316|\n|3|0.316|\n|4|0.156|\n|5|0.028|\ndoes the table show a probability distribution? select all that apply.\n□a. yes, the table shows a probability distribution.\n□b. no, the random variable xs number values are not associated with probabilities.\n□c. no, the random variable x is categorical instead of numerical.\n□d. no, the sum of all the probabilities is not equal to 1\n□e. no, not every probability is between 0 and 1 inclusive\nfind the mean of the random variable x. select the correct choice below and, if necessary, fill in the answer box to complete your choice.\n○a. μ = □ child(ren) (round to one decimal place as needed.)\n○b. the table does not show a probability distribution.\nfind 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.

Answer

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. $\sigma = 1.1$