Question

A small business collects surveys from 15 of their biggest customers. 75% indicated an interest in participating in their upcoming beta testing program. What’s the probability that over five weeks, the average number of emergency visits was greater than 42?

Answer 1

This is a binomial distribution question with
n = 75
p = 0.75
q = 1 - p = 0.25
This binomial distribution can be approximated as Normal distribution since
np > 5 and nq > 5
Since we know thatP(x > 42.0)=?
The z-score at x = 42.0 is,
z = 42.0-56.25/3.75
z = -3.8
This implies that
P(x > 42.0) = P(z > -3.8) = 1 - 7.234804392511998e-05
P(x > 42.0) = 0.9999