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(
X^{2} * P(X) - Mean^{2} )

= Sqrt( 0^{2}* 0.0053 + 1^{2}* 0.0488 +
2^{2}* 0.1811 + 3^{2}* 0.3364 + 4^{2}*
0.3124 + 5^{2}* 0.1160 - 3.2498^{2} )

= 1.0668

**Standard deviation = 1.0668**

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