Score: 1 of 3 pts 4 of 14 (13 complete) HW Score: 54.29%, 19 of 35 pts Question

Question

Score: 1 of 3 pts 4 of 14 (13 complete) HW Score: 54.29%, 19 of 35 pts Question Help xP(x) 0 0.034 the X-linked genetic disorder. Determine whether a probability distribution 0.153 20.313 deviation. If a probability distribution is not given, identify the requirements3 0.313 4 0.153 5 0.034 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 is given. If a probability distribution is given, find its mean and standard that are not satisfied YA. Yes, tne tapie snows a proDapility aistribution B. No, the random variable x's number values are not associated with probabilities C. No, the sum of all the probabilities is not equal to 1 No, not every probability is between 0 and 1 inclusive No, the random variable x is categorical instead of numerical D. E. 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 OA. H child(ren) (Round to one decimal place as needed.) O B. The table does not show a probability distribution Click to select and enter your answer(s) and then click Check Answer Clear All Check Answer remaining

Explanation / Answer

First we have that 0 P(x) 1 for 0 x 5 and also P(x) = 1. Therefore, we have a probability distribution.

Mean of the Probability Distribution can be calculated as:

µ = x*P(x) = 0*0.034 + 1*0.153 + 2*0.313 + 3*0.313 + 4*0.153 + 5*.034

µ = 2.5

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