2Let X follows a binomial distribution with mean equals 055

(2)Let X follows a binomial distribution with mean equals 0.55 and standard deviation equals 0.7228. Find n and p.

for binomial, mean = n*p

standard deviation = square root (n*p*(1-p))

so, 0.55 = n*p.

0.7228 = square root of (n*p(1-p)

Substituting 0.55 for n*p into the second equation:

0.7228 = square root of (.55*(1-p)

squaring both sids: (0.7228)^2 = .55(1-p)

Divide both sides by .55: (0.7228)^2/.55 = 1 - p

p = 1 - (.7228)^2/.55 = 0.05

n = .55/.05 = 11

So n = 11 and p = .05