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 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.

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