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