posted by BJ
Suppose that you are in a class of 29 students and it is assumed that approximately 16% of the population is left-handed. (Give your answers correct to three decimal places.)
(a) Compute the probability that exactly five students are left-handed.
(b) Compute the probability that at most four students are left-handed.
(c) Compute the probability that at least six students are left-handed.
P(x) = (nCx)(p^x)[q^(n-x)]
x = 5
p = .16
q = 1 - p = 1 - .16 = .84
n = 29
Substitute and calculate for your probability.
x = 0,1,2,3,4
p,q,n stay the same.
Add each probability you calculate for the total probability.
Add together the probabilities you calculated for (a) and (b). Then subtract that value from 1. This will be your answer for (c).
You could also use a binomial probability table to find the above probabilities as well. This would be easier than calculating each by hand.
I hope this will help get you started.