Tony’s Pizza Company finds that 65% of the general population likes pepperoni pi
ID: 3150987 • Letter: T
Question
Tony’s Pizza Company finds that 65% of the general population likes pepperoni pizza. I buy pizza for 56 of my intro stats students and it turns out that only 30 of these students like pepperoni pizza. Using the normal approximation to the binomial distribution, what is the probability of getting 30 or fewer students who like pepperoni pizza in a randomly selected group of 56 students? Enter your answer in decimal form rounded to the 4th decimal place.
I think I'm having a hard time trying to figure out this question. My teacher constantly ask us to draw out the graph and I did that. However, I think I may have been imputting my data in the wrong parts for the formula. Here is what I've tried (the teacher's way):
X: 30
N:56
p?: X/N = 0.54
P: 0.65
Using the TI-83 calculator: normalcdf(-1e99,0.54, 0.65, 0.64) = 0.4318 <---------------------- That answer was wrong. I can't figure out where I went wrong.
I then retried this problem with following my book directly and still got my answer wrong:
P: 0.65
n: 56
x: 30
µ: n * P = 36.4
P(x ? 30) = P(z ? -1.78) = 0.0375 <-----------------------------This answer was wrong as well. All I know is that the probability is supposed to be below the benchmark value of 0.05.
Explanation / Answer
p = 0.65 and q = 0.35
n =56
As np >5 and nq >5 we can approximate to normal with mu = 36.4 and sigma = 3.569
std error = 3.569/rt n = 0.477
P(X<=30) = P(Z<-6.4/0.477) = P(Z<-13.41)
= 0