Question

Mean and variance pair is given below. A pair representing the data of a binomial distribution is

• A. \((7,4)\)
• B. \((12,5)\)
• C. \((15,6)\)
• D. \((18,10)\)

Hint:

For a binomial distribution, \[ \boxed{ \mu=np,\qquad \sigma^2=npq. } \] To verify a given mean-variance pair, \[ \boxed{ p=1-\frac{\sigma^2}{\mu}, \qquad n=\frac{\mu}{p}. } \] A valid binomial distribution requires \[ \boxed{n} \] to be a positive integer.

Answer 1

The Correct Option is C

Step 1: Recall the formulas for a binomial distribution. For a binomial distribution, \[ \boxed{ \mu=np, } \] and \[ \boxed{ \sigma^2=npq, } \] where \[ q=1-p. \] Hence, \[ \sigma^2=\mu q. \] Since \[ 0<q<1, \] it follows that \[ \boxed{ \sigma^2<\mu. } \]

Step 2: Check the given pairs. For option (A), \[ 4<7, \] but \[ p=\frac{3}{7}, \] giving \[ n=\frac{7}{3/7}=\frac{49}{3}, \] which is not an integer. Hence, not possible. For option (B), \[ 5<12, \] \[ p=\frac{7}{12}, \] so \[ n=\frac{12}{7/12}=\frac{144}{7}, \] not an integer. Hence, not possible. For option (C), \[ 6<15, \] \[ q=\frac{6}{15}=\frac25, \qquad p=\frac35. \] Thus, \[ n=\frac{15}{3/5}=25, \] which is an integer. Hence, this pair is valid. For option (D), \[ 10<18, \] \[ p=\frac49, \] giving \[ n=\frac{18}{4/9}=40.5, \] not an integer. Hence, not possible. Therefore, \[ \boxed{(15,6)} \] is the correct pair. Thus, \[ \boxed{(C)} \] is the correct answer.