Wednesday

July 30, 2014

July 30, 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

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