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(B) For a binomial distribution, mean $\mu = np = 1$ and variance $\sigma^2 = npq = 0.75$.Dividing variance by mean: $\frac{npq}{np} = \frac{0.75}{1}

Vedclass · Accepted Answer

$P(X = 1) = 2P(X = 2)$

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(B) For a binomial distribution, mean $\mu = np = 1$ and variance $\sigma^2 = npq = 0.75$.Dividing variance by mean: $\frac{npq}{np} = \frac{0.75}{1} \implies q = 0.75 = \frac{3}{4}$.Since $p + q = 1$, $p = 1 - 0.75 = 0.25 = \frac{1}{4}$.Using $np = 1$, $n(\frac{1}{4}) = 1 \implies n = 4$.The probability mass function is $P(X = k) = \binom{4}{k} (\frac{1}{4})^k (\frac{3}{4})^{4-k}$.For option $(C)$: $P(X = 3) = \binom{4}{3} (\frac{1}{4})^3 (\frac{3}{4})^1 = 4 \cdot \frac{1}{64} \cdot \frac{3}{4} = \frac{12}{256}$.$P(X = 4) = \binom{4}{4} (\frac{1}{4})^4 (\frac{3}{4})^0 = 1 \cdot \frac{1}{256} \cdot 1 = \frac{1}{256}$.Thus, $P(X = 3) = 12 \cdot P(X = 4)$. Checking option $(B)$: $P(X = 1) = \binom{4}{1} (\frac{1}{4})^1 (\frac{3}{4})^3 = 4 \cdot \frac{1}{4} \cdot \frac{27}{64} = \frac{27}{64}$.$P(X = 2) = \binom{4}{2} (\frac{1}{4})^2 (\frac{3}{4})^2 = 6 \cdot \frac{1}{16} \cdot \frac{9}{16} = \frac{54}{256} = \frac{27}{128}$.Since $P(X = 1) = 2 \cdot P(X = 2)$, option $(B)$ is correct.

If the mean and the variance of a binomial variate $X$ are $1$ and $0.75$ respectively, then which of the following is true?

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