Question

3. 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. XP(x) 0 0.027 1 0.154 0.319 3 0.319 4 0.154 5 0.027 21 Does the table show a probability distribution? Select all that apply. A. Yes, the table shows a probability distribution. B. No, the random variable x is categorical instead of nu C. No, the sum of all the probabilities is not equal to 1. D. No, not every probability is between 0 and 1 inclusive. E. No, the random variable x's number values are not associated with probabilities. Find the mean of the random variable x. Select the correct choice below and, necessary, fill in the answer box to complete your choice. O A. = child(ren) (Round to one decimal place as needed.) OB. 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. O A. 0 = child(ren) (Round to one decimal place as needed.) B. The table does not show a probability distribution. O

Answer 1

of who p(x) 0.027 0154 0.319 0.319 6.100 0.027 A Mes x shows a probability distributen because the total of all probability are I and all different value. have probability between 0 to 1 and x is the no of chiliden so it should be categorized numbers y so A is connect 4 nozo M = suplul = Ot 10.154 f 2x 0.319 + Se 3x0.319 + 4x0.104 + 5x0.027 [vcx) = Eripen) - 4² & ot IXG-15 4 + 4 X0.319€ 9X0:319 + 16 X0-154 + 25 X 0027 - Men 6:25 = 7.286 - 6:27 1,036

as I = Lucky E = (1.036 = 1.05448