Posted by dais on .
A sample survey interviews an SRS of 263 college women. Suppose (as is roughly true) that 75% of all college women have been on a diet within the past 12 months.
Using the Normal approximation, what is the probability (±0.001) that 82% or more of the women in the sample have been on a diet?
Use the binomial approximation to a normal distribution.
mean = np = 263 * .75 = ?
standard deviation = √npq = (√(263 * .75 * .25) = ?
Note: q = 1 - p
Finish the calculation.
Next, use z-scores:
z = (x - mean)/sd
x = .82
Use mean and sd calculated above.
Finally, use a z-table to determine your probability (remember the question is asking "82% or more" when looking at the table).
I hope this will help get you started.