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 0 1 2 3 4 5P(x) 0.033 0.149 0.318 0.318 0.149 0.033

Answer 1

Answer:

a) It is probability distribution

b) mean = 2.5

c) Standard Deviation = 1.11

Step-by-step explanation:

We are given the following in the question:

x          0           1           2         3          4          5

P(x)  0.033   0.149   0.318   0.318   0.149   0.033

Probability distribution:

$\displaystyle\sum P_i(x_i) = 0.033+ 0.149 +0.318+ 0.318+ 0.149 +0.033 = 1$

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:

$E(x) = \displaystyle\sumx_iP(x_i)\\E(x) = 0(0.033)+ 1(0.149)+2(0.318)+3(0.318) +4(0.149) + 5(0.033)\\E(x) = 2.5$

Thus, the mean of distribution is 2.5

Standard deviation of probability distribution:

$E(x^2) = 0^2(0.033)+ 1^2(0.149)+2^2(0.318)+3^2(0.318) +4^2(0.149) + 5^2(0.033)\\E(x^2) = 7.492\\\sigma^2 = E(x^2) - (E(x))^2\\\sigma^2 = 7.492 - (2.5)^2 = 1.242\\\sigma= \sqrt{1.242} = 1.11$

Thus, the standard deviation of the probability distribution is 1.11