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.

## Answer this Question

## Related Questions

- Statistics(math) - (20.47 S-AQ) A sample survey interviews an SRS of 252 college...
- Algebra - If a researcher wanted to know the mean weight (the mean is the sum of...
- Statistics - For some reason I didn't get the entire questions submitted on the ...
- Business Statistics - Suppose that in the past, 94% of all Hispanic grocery ...
- college psych - Which of the following is true of the prevalance rate of ...
- Statistics - An organization is studying whether women have caught up to men in ...
- Statistics - Suppose that, in fact, the blood cholesterol level of all men aged ...
- STAT - A college faculty consists of 400 men and 250 women. The college ...
- math - A small college has 1250 students. There are 30 more women than men. How ...
- statistics - In the past a certain college has found that 35% of the students ...

More Related Questions