Question

The sum and product of the mean and variance of a binomial distribution are 24 and 128 respectively. The distribution is -

• A. $\left( \frac { 1 } { 7 } + \frac { 1 } { 8 } \right) ^ { 12 }$
• B. $\left( \frac { 1 } { 4 } + \frac { 3 } { 4 } \right) ^ { 16 }$
• C. $\left( \frac { 1 } { 6 } + \frac { 5 } { 6 } \right) ^ { 24 }$
• D. $\left( \frac { 1 } { 2 } + \frac { 1 } { 2 } \right) ^ { 32 }$

Answer 1

Answer: (D)

$\left( \frac { 1 } { 2 } + \frac { 1 } { 2 } \right) ^ { 32 }$

Answer 2

Given m + s² = 24, ms² = 128

\ m = 16 or 8

m = 16 Ž s² = 8

and m = 8 Ž s² = 56

Case 1 : np = 16, npq = 8 give

n = 32, p = q = $\frac { 1 } { 2 }$ Case 2 : np = 8 and npq = 56 give

q = 7 which is not feasible