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 (a) mean and (b) standard deviation. If a probability distribution is not​ given, identify the requirements that are not satisfied.

X | P(x)

0 | 0.033

1|0.156

2 | 0.311

3 | 0.311

4 | 0.156

5 | 0.033

Find the mean of the random variable x.

Find the Standard deviation of the random variable x.

Answer 1

Solution :

yes table show the probability distribution

total to probabilities are 1

Mean = = X * P(X)

= 0 *0.033 + 1 * 0.156 + 2 *0.311 + 3 *0.311 + 4 * 0.156+5 *0.033

= (0 +0.156 +0.622 +0.933+0..624+0.165 )

= 2.5

Standard deviation =

=X ² * P(X) - ²

=  [ 0²  0.033 + 1² * 0.156 + 2² *0.311 + 3² *0.311 + 4² * 0.156+5² *0.033] -2.5 ²

= [( 0+ 0.156 + 1.244+2.799+2.496+0.825 ) )] -6.25

=7.52 -6.25

=1.27

=1.1269