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 0.0321 0.1522 0.3163 0.3164 0.1525 0.032
Accepted Solution
A:
Answer:a) It is probability distributionb) mean = 2.5c) Standard Deviation = 1.11Step-by-step explanation:We are given the following in the question:x 0 1 2 3 4 5P(x) 0.033 0.149 0.318 0.318 0.149 0.033Probability distribution:[tex]\displaystyle\sum P_i(x_i) = 0.033+ 0.149 +0.318+ 0.318+ 0.149 +0.033 = 1[/tex]Since, the summation of all the probabilities is 1 and each value of probability is between 0 and 1, the given is a probability distribution.Mean of probability distribution:[tex]E(x) = \displaystyle\sumx_iP(x_i)\\E(x) = 0(0.033)+ 1(0.149)+2(0.318)+3(0.318) +4(0.149) + 5(0.033)\\E(x) = 2.5[/tex]Thus, the mean of distribution is 2.5Standard deviation of probability distribution:[tex]E(x^2) = 0^2(0.033)+ 1^2(0.149)+2^2(0.318)+3^2(0.318) +4^2(0.149) + 5^2(0.033)\\E(x^2) = 7.492\\\sigma^2 = E(x^2) - (E(x))^2\\\sigma^2 = 7.492 - (2.5)^2 = 1.242\\\sigma= \sqrt{1.242} = 1.11[/tex]Thus, the standard deviation of the probability distribution is 1.11