Binomial Distribution HW Help?

Among employed women, 35% have never been married. Select 15 employed women at random.





(a) The number in your sample who have never been married has a binomial distribution. What are n and p?





n = ____


p = ____





(b) What is the probability that exactly 4 of the 15 women in your sample have never been married?





(c) What is the probability that 2 or fewer women have never been married?








Any help would be greatly appreciated. I am failing AP Statistics, and I can't seem to get the hang of it.

Binomial Distribution HW Help?
Let X be the number of employed women who have not been married. X has the binomial distribution with n = 15 trials and success probability p = 0.35





In general, if X has the binomial distribution with n trials and a success probability of p then


P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)


for values of x = 0, 1, 2, ..., n


P[X = x] = 0 for any other value of x.





The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.


Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.





X ~ Binomial( n , p )





the mean of the binomial distribution is n * p = 5.25


the variance of the binomial distribution is n * p * (1 - p) = 3.4125


the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 1.847295





The Probability Mass Function, PMF,


f(X) = P(X = x) is:





P( X = 0 ) = 0.001562069


P( X = 1 ) = 0.01261672


P( X = 2 ) = 0.04755531


P( X = 3 ) = 0.1109624


P( X = 4 ) = 0.1792469 ← answer to b


P( X = 5 ) = 0.2123387


P( X = 6 ) = 0.1905604


P( X = 7 ) = 0.1319264


P( X = 8 ) = 0.07103729


P( X = 9 ) = 0.02975066


P( X = 10 ) = 0.009611752


P( X = 11 ) = 0.002352527


P( X = 12 ) = 0.0004222484


P( X = 13 ) = 5.246873e-05


P( X = 14 ) = 4.036056e-06


P( X = 15 ) = 1.448841e-07








The Cumulative Distribution Function, CDF,


F(X) = P(X ≤ x) is:





x


∑ P(X = t) =


t = 0





P( X ≤ 0 ) = 0.001562069


P( X ≤ 1 ) = 0.01417878


P( X ≤ 2 ) = 0.0617341 ← answer to c


P( X ≤ 3 ) = 0.1726965


P( X ≤ 4 ) = 0.3519434


P( X ≤ 5 ) = 0.5642821


P( X ≤ 6 ) = 0.7548425


P( X ≤ 7 ) = 0.8867689


P( X ≤ 8 ) = 0.9578062


P( X ≤ 9 ) = 0.9875568


P( X ≤ 10 ) = 0.9971686


P( X ≤ 11 ) = 0.9995211


P( X ≤ 12 ) = 0.9999434


P( X ≤ 13 ) = 0.9999958


P( X ≤ 14 ) = 0.9999999


P( X ≤ 15 ) = 1








1 - F(X) is:





n


∑ P(X = t) =


t = x





P( X ≥ 0 ) = 1


P( X ≥ 1 ) = 0.998438


P( X ≥ 2 ) = 0.9858212


P( X ≥ 3 ) = 0.9382659


P( X ≥ 4 ) = 0.8273035


P( X ≥ 5 ) = 0.6480566


P( X ≥ 6 ) = 0.4357179


P( X ≥ 7 ) = 0.2451575


P( X ≥ 8 ) = 0.1132311


P( X ≥ 9 ) = 0.04219384


P( X ≥ 10 ) = 0.01244318


P( X ≥ 11 ) = 0.002831425


P( X ≥ 12 ) = 0.0004788981


P( X ≥ 13 ) = 5.664967e-05


P( X ≥ 14 ) = 4.180941e-06


P( X ≥ 15 ) = 1.448841e-07