Question

A survey revealed that 44% of movie goers will eat some amount of popcorn during a movie. If the next 17 movie goers are asked if they will eat popcorn during the next movie they attend, what is the probability that at least 8 will say they will eat popcorn during the next movie?

Answer 1

p eat = .44
p not eat = .56

p that k say yes (binomial)
P(17/k) = C(17,k) .44^k .56(17-k)

where
C(17,k) = 17!/[k!(17-k)!]

do that for k = 8 through 17 and add

OR
approximate binomial expansion with normal distribution

mean = n p = .44 * 17 = 7.48
sigma^2 = 17(.44)(.56) = 4.19 = n p(1-p)

use david lane tables http://davidmlane.com/hyperstat/z_table.html

Answer 2

First write out what you know in order to get the formula:

x = # of movie goers
44% of movie goers eat popcorn
so 44% of movie goers = movie goers who eat popcorn

Out of 17 movie goers how many will eat popcorn?
44% of x =
44% of 17 = 7 movie goers who will eat popcorn

But the question is asking for the percentage if the # of movie goers who eat popcorn = 8 so you know the answer has to be higher than 44%

The formula becomes:
y% of 17 = 8
y% x 17 = 8
y% = 8/17
y = (8/17)*100 = 47.06%

First, 47.06% is higher than 44% (logical verification)
Second, try it out to prove it:

47.06% x 17 = 8.0002