95% CI for the population proportion

[okep]

Homework Statement

A 95% CI for the population proportion of professional tennis players who earn more than $2 millions a year is found to be [0.82, 0.88]. Given this information the sample size that was used was approximately A) 545 B) 382 C) 233 D) 378
Answer: A) 545

Homework Equations

Z = 1.96 for 95% CI
+/- Z[S/sqrt(n)] = 0.03

The Attempt at a Solution

+/- 1.96 [0.03)/sqrt(n) = 0.03
sqrt(n) = 1.96(2)/0.03
sqrt(n) = 130
n = 11.4 <-- this incorrect.

The answer should be A) 545. But where did i go wrong? How do i solve this problem? Thanks.

 

[rsa58]

first use the fact that .03 = 2 *(1.96)*sqrt(p(1-p)/n) now solve for n and plug in the two bounds for the CI given. you will find that n lies between 450 and 630 hence the answer.