Question

A random variable \(X\) follows the binomial distribution and \[ X\sim B(n,0.3). \] If the mean of \(X\) is three times as large as the standard deviation of \(X\), then \(n=\)

• A. \(9\)
• B. \(21\)
• C. \(27\)
• D. \(3\)

Hint:

For binomial distribution, always remember: \[ \text{Mean}=np,\qquad \text{Variance}=npq,\qquad \text{S.D.}=\sqrt{npq}. \]

Answer 1

The Correct Option is B

Step 1: State the mean and standard deviation for a binomial distribution.
For X ~ B(n, p), the mean is μ = np and the standard deviation is σ = √(npq), where q = 1 - p. Here p = 0.3 and q = 0.7.

Step 2: Translate the given condition into an equation.
The mean equals three times the standard deviation: np = 3√(npq). Substituting the known probabilities gives 0.3n = 3√(0.21n).

Step 3: Square both sides and solve for n.
Squaring: 0.09n² = 9(0.21n) = 1.89n. Dividing both sides by n (n>0): 0.09n = 1.89. Thus n = 1.89/0.09 = 21.

Step 4: Final conclusion.
The number of trials is 21.