Question

Determine the binomial distribution for which the mean is 4 and variance 3. Also find P(X=15)

Answer 1

In a binomial distribution

Mean np = 4 → (1)

variance npq = 3 → (2)

Equation (2) ÷ Equation (1)

⇒ `"npq"/"np" = 3/4`

q = `3/4` p = `1 - "q" = 1 - 3/4`

p = `(4 - 3)/4`

∴ p = `1/4`

Substitute p = `1/4` in equation (1)

n = `(1/4)` = 4

n = 4 × 4

⇒ n = 16

The binomial distribution is p(X = x)nc_xp^xq^n-x 

P(X = x) = `(16/x)(1/4)^x(3/4)^(16 - x)`

P(X = 15) = `16"c"_15 (1/4)^15 (13/4)^(16 - 15)`

= `16"c"_1 (1/4)^15 (3/4)^1`

= `16 xx 1/4^15 xx 3/4`

= `(16 xx 3)/4^16`

- `(16 xx 3)/(4^2 xx 4^14)`

= `(16 xx 3)/(16 xx 4^14)`

P(X = 15) = `3/4^14`