For a Binomial distribution with parameters n and p = 0.5, what are the mean and variance?

For a Binomial distribution with parameters n and p = 0.5, what are the mean and variance?

• A. Mean = n; Variance = n
• B. Mean = n/2; Variance = n/4
• C. Mean = 0; Variance = n/8
• D. Mean = n/4; Variance = n/2

Answer: (B)

In a Binomial(n, p) distribution, the mean (expected value) is np and the variance is np(1 − p). If p = 0.5, the mean becomes n × 0.5 = n/2, and the variance becomes n × 0.5 × 0.5 = n/4. So the distribution has mean n/2 and variance n/4. Why the other options don’t fit: a mean of n would require p = 1, which isn’t the given p. A zero mean would require p = 0 (or n = 0). A variance of n/2 would imply p(1 − p) = 1/2, which has no real solution for p. A variance of n/8 would imply p(1 − p) = 1/8, which also isn’t achieved by p = 0.5.

In a Binomial(n, p) distribution, the mean (expected value) is np and the variance is np(1 − p). If p = 0.5, the mean becomes n × 0.5 = n/2, and the variance becomes n × 0.5 × 0.5 = n/4. So the distribution has mean n/2 and variance n/4.

Why the other options don’t fit: a mean of n would require p = 1, which isn’t the given p. A zero mean would require p = 0 (or n = 0). A variance of n/2 would imply p(1 − p) = 1/2, which has no real solution for p. A variance of n/8 would imply p(1 − p) = 1/8, which also isn’t achieved by p = 0.5.