Monday

September 22, 2014

September 22, 2014

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

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 6:55pmUse 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...

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 ...

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 ...

Statistics - Among employed women, 25% have never been married. Select 10 ...

statistics - Find the indicated binomial probabilities. Round to the nearest 3 ...

Lincoln College Of New En - The percentage of physicians who are women is 27.9% ...