Question

Five males with a particular genetic disorder have one child each. The random variable x is the number of children among the five who inherit the genetic disorder. Determine whether the table describes a probability distribution. If it? does, find the mean and standard deviation.

X 0 1 2 3 4 5

?P(x) 0.0053 0.0488 0.1811 0.3364 0.3124 0.1160

Answer 1

The table describes valid probability distribution if sum of probabilities equal to 1

That is

P(X) = 1

From above table,

P(X) = 0.0053 + 0.0488 + 0.1811 + 0.3364 + 0.3124 + 0.1160

= 1

Table describes valid probability distribution.

Mean = X * P(X)

= 0 * 0.0053 + 1 * 0.0488 + 2 * 0.1811 + 3 * 0.3364 + 4 * 0.3124 + 5 * 0.1160

= 3.2498

Mean = 3.2498

Standard deviation = Sqrt( X² * P(X) - Mean² )

= Sqrt( 0²* 0.0053 + 1²* 0.0488 + 2²* 0.1811 + 3²* 0.3364 + 4²* 0.3124 + 5²* 0.1160 - 3.2498² )

= 1.0668

Standard deviation = 1.0668