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3 years ago

I need help please and thanks

Mathematics

1 answer:

NikAS [45]3 years ago

8 0

Answer:

SAS

Step-by-step explanation:

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Five males with an X-linked genetic disorder have one child each. The random variable x is the number of children among the fiv

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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

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