Question

Which of the following is not the correct data of a binomial distribution?

• A. Mean \(6\), variance \(2\)
• B. Mean \(16\), variance \(12\)
• C. Mean \(20\), variance \(16\)
• D. Mean \(15\), variance \(5\)

Hint:

For a binomial distribution, \[ \boxed{p=1-\frac{\sigma^2}{\mu},\qquad n=\frac{\mu}{p}.} \] A pair of mean and variance is valid only if the resulting value of \[ \boxed{n} \] is a positive integer.

Answer 1

The Correct Option is D

Step 1: Recall the formulas for a binomial distribution. For a binomial distribution, \[ \mu=np, \] and \[ \sigma^2=npq, \] where \[ q=1-p. \] Hence, \[ p=1-\frac{\sigma^2}{\mu}. \] Also, \[ n=\frac{\mu}{p}. \] For a valid binomial distribution, \[ n \] must be a positive integer.

Step 2: Check each option. \[ \boxed{(A) \] \[ p=1-\frac26=\frac23,\qquad n=\frac6{2/3}=9. \] Valid. \[ \boxed{(B) \] \[ p=1-\frac{12}{16}=\frac14,\qquad n=\frac{16}{1/4}=64. \] Valid. \[ \boxed{(C) \] \[ p=1-\frac{16}{20}=\frac15,\qquad n=\frac{20}{1/5}=100. \] Valid. \[ \boxed{(D) \] \[ p=1-\frac5{15}=\frac23,\qquad n=\frac{15}{2/3} =\frac{45}{2} =22.5. \] Since \[ n \] is not an integer, this is not possible.

Step 3: Write the final answer. Hence, \[ \boxed{\text{Mean }15,\ \text{variance }5} \] is not the correct data of a binomial distribution. Thus, \[ \boxed{(D)} \] is the correct answer.