Posted by dais on Thursday, May 2, 2013 at 10:48pm.
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?

statistics  MathGuru, Friday, May 3, 2013 at 6:55pm
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 zscores:
z = (x  mean)/sd
x = .82
Use mean and sd calculated above.
Finally, use a ztable 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.
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