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 mean and standard deviation. If a probability distribution is not​ given, identify the requirements that are not satisfied. x ​P(x) 0 0.034 1 0.153 2 0.313 3 0.313 4 0.153 5 0.034 Does the table show a probability​ distribution? Select all that apply.

Answer 1

X	0	1	2	3	4	5
P(X)	0.034	0.153	0.313	0.313	0.153	0.034

Conditions to probability distribution is valid are

1) All probabilities must be positive and it shoul be between 0 and 1 inclusive.

2) Sum of all probabilities must be equal to 1.

In given example,

All probabilities are non-negative and between 0 and 1

So first condition is satisfied.

For second condition,

P(X) = 0.034 + 0.153 + 0.313 + 0.313 + 0.153 + 0.034

= 1

Therefore second condition is satisfied.

Given probability distribution is valid.

E(X) = X * P(X)  

= 0 * 0.034 + 1 * 0.153 + 2 * 0.313 + 3 * 0.313 + 4 * 0.153 + 5 * 0.034

= 2.5

Mean = 2.5

Standard deviation (X) = Sqrt( X² *P(X) - mean² )

= Sqrt (0² * 0.034 + 1² * 0.153 + 2² * 0.313 + 3² * 0.313 + 4² * 0.153 + 5² * 0.034 - 2.5² )

Standard deviation = 1.1269