Question

Five males with a particular genetic disorder have one child each. The random variable x is...

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)

0.00020.0002

0.00530.0053

0.04500.0450

0.19190.1919

0.40890.4089

0.34870.3487

Find the mean of the random variable x. Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

A.

muμequals=nothing

​(Round to one decimal place as​ needed.)

B.

The table is not a probability distribution.

Find the standard deviation of the random variable x. Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

A.

sigmaσequals=nothing

​(Round to one decimal place as​ needed.)

B.

The table is not a probability distribution.

Click to select and enter your answer(s).

Homework Answers

Answer #1

To be a valid probability distribution, the sum of all probability must be equal to one.

Sum = 0.0002 + 0.0053 + 0.045 + 0.1919 + 0.4089 + 0.3487 = 1

µ = E(X) = 0 * 0.0002 + 1 * 0.0053 + 2 * 0.045 + 3 * 0.1919 + 4 * 0.4089 + 5 * 0.3487 = 4.0501 = 4.1 (ans)

Option-A) µ = 4.1

E(X2) = 02 * 0.0002 + 12 * 0.0053 + 22 * 0.045 + 32 * 0.1919 + 42 * 0.4089 + 52 * 0.3487 = 17.1723

Variance = E(X2) - (E(X))2 = 17.1723 - 4.12 = 0.3623

Standard deviation = sqrt(0.3623) = 0.6

Option-A) σ = 0.6

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