Wednesday

October 1, 2014

October 1, 2014

Posted by **Kelsey** on Friday, June 7, 2013 at 12:46pm.

Using Normal approximation, what is the probability (±0.001) that 77% or more of the women in the sample have been on a diet? .

- Statistics(math) -
**Kuai**, Friday, June 7, 2013 at 3:19pmn = 252

phat = 0.77

p = 0.71

q = 1-p = 0.29

standard deviation = sqrt((pq/n))

= sqrt(0.7)(0.29)/252)

= 0.02858

z score = (phat -p)sd/sqrt(n)

z = (0.77-0.71)/0.02858

z = 2.10

P(phat > 0.77) = p(z < 2.10)

1- 0.9821 = 0.0179

**Answer this Question**

**Related Questions**

statistics - A sample survey interviews an SRS of 263 college women. Suppose (as...

Statistics - For some reason I didn't get the entire questions submitted on the ...

math - A small college has 1250 students. There are 30 more women than men. How ...

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

college psych - Which of the following is true of the prevalance rate of ...

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

STAT - A college faculty consists of 400 men and 250 women. The college ...

Math/statistics - a recent study indicated that 29% of the 100 women over age 55...

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

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