The mean and variance of a random variable having a binomial probability distribution are 4 and 2 respectively. Then P(X=1) is:Question 15Select one:a.1/32b.1/8c.1/16d.1/4

The problem is about a binomial distribution. In a binomial distribution, the mean (μ) is np and the variance (σ^2) is npq, where n is the number of trials, p is the probability of success, and q is the probability of failure (1-p).

Given that the mean (μ) is 4 and the variance (σ^2) is 2, we can set up the following equations:

np = 4 (equation 1) npq = 2 (equation 2)

We can divide equation 2 by equation 1 to get q = 2/4 = 0.5. Since q = 1 - p, we can solve for p to get p = 0.5 as well.

Substituting p = 0.5 into equation 1 gives n = 4 / 0.5 = 8.

Now that we have n = 8 and p = 0.5, we can find P(X=1) using the formula for the binomial probability:

P(X=k) = C(n, k) * (p^k) * (q^(n-k))

where C(n, k) is the combination of n items taken k at a time.

Substituting n = 8, k = 1, p = 0.5, and q = 0.5 gives:

P(X=1) = C(8, 1) * (0.5^1) * (0.5^(8-1)) = 8 * 0.5 * 0.5^7 = 8 * 0.5 * 0.0078125 = 0.03125

So, the closest answer is b. 1/32.