Question

The mean of a binomial distribution is 5 and standard deviation is 2. Determine the distribution

Answer 1

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) = nc_xp^xq^n-x 

i.e P(X = x) = `[(25),(x)] (1/5)^x (4/5)^(25 - x)`