Statistics(math)
posted by Kelsey .
(20.47 SAQ) A sample survey interviews an SRS of 252 college women. Suppose (as is roughly true) that 71% of all college women have been on a diet within the past 12 months.
Using Normal approximation, what is the probability (±0.001) that 77% or more of the women in the sample have been on a diet? .

n = 252
phat = 0.77
p = 0.71
q = 1p = 0.29
standard deviation = sqrt((pq/n))
= sqrt(0.7)(0.29)/252)
= 0.02858
z score = (phat p)sd/sqrt(n)
z = (0.770.71)/0.02858
z = 2.10
P(phat > 0.77) = p(z < 2.10)
1 0.9821 = 0.0179