Try this formula:
n = [(z-value)^2 * p * q]/E^2
Note: n = sample size needed; .5 for p and .5 for q are used if no value is stated in the problem. E = maximum error, which is .02; Z-value is found using a z-table (1.96 for 95% confidence as an example).
I'll let you take it from here.
Assume you want to estimate with the proportion of students who commute less than 5 miles to work within 2%, what sample size would you need?
2 answers
1-.02=2.33=z-value
plug into the equation
[(2.33)^2*.5*.5]/(.02)^2
3393.06
plug into the equation
[(2.33)^2*.5*.5]/(.02)^2
3393.06