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Question
The mean of a binomial distribution is 5 and standard deviation is 2. Determine the distribution
Solution
In a binomial distribution
Mean np = 5 → (1)
Standard deviation `sqrt("npq")` = 2
Squaring on body sides
npq = 4 → (2)
Equation ÷ Equation
⇒ `"npq"/"np" = 4/5`
∴ q = `4/5`, p = `1 - "q"`
= `1 - 4/5 = (5 - 4)/5`
p = `1/5`
Substitute p = `1/5` in equation (1)
`"n"(1/5)` = 5
n = 5 × 5
⇒ n = 25
∴ The binomial distribution is
P(X = x) = ncxpxqn-x
i.e P(X = x) = `[(25),(x)] (1/5)^x (4/5)^(25 - x)`
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