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9. The probability of a child being born with light-colored eyes to two particular parents is 0.25.
The parents intend to have 5 children. Find the variance of this binomial distribution.
0.9
1.0
1.3
3.8


Sagot :

If X is a random variable representing the number of children that end up having light-colored eyes, then X follows a binomial distribution with success probability p = 0.25 and number of trials n = 5.

The variance of a binomial distribution with these parameters is n p (1 - p), so the variance in this case would be

5 • 0.25 • (1 - 0.25) = 0.9375 ≈ 0.9