SOLUTION: Suppose in fact 60% of all Filipino adults between the ages of 30 and 39 exercises at least 3 days per week. If we take a random sample of 100 Filipino adults in their 30s, what is

Question 1161992: Suppose in fact 60% of all Filipino adults between the ages of 30 and 39 exercises at least 3 days per week. If we take a random sample of 100 Filipino adults in their 30s, what is the probability that at least 70% of them exercise at least 3 days per week?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
p = .6
q = 1 - .6 = .4
n = 100
mean of population = .6
standard deviation of distribution of sample means = sqrt(.6 * .4 / 100) = .04899 rounded to 5 decimal places.
z = (.7 - .6) / .04899 = 2.04 rounded to 2 decimal places.
area to the left of that z-score is equal to .97932, based on the table that can be found at https://www.math.arizona.edu/~rsims/ma464/standardnormaltable.pdf
area to the right of that z-score is equal to 1 - .97932 = .02068

.02068 should be your answer.

p is the probability that the population ratio is as stated.
q is the probability that the population ratio is not as stated.
that's why you get p = .6 and q = .4
the standard distribution of sample means is equal to sqrt (p*q/n) which makes it equal to sqrt(.6 * .4 / 100) which gets you .04899 rounded to 5 decimal places.
the z-score formula is z = (x - m) / s
z is the z-score.
x is the raw score
m is the raw mean
s is the standard deviation of the distribution of sample means.
in this problem, x is equal to .7 and m is equal to .6 and s is equal to .04899.
that gets you the z-score of 2.04 rounded to 2 decimal places.
it's rounded to 2 decimal places because that's the accuracy of the z-score table used.
the table gives you the area to the left of that z-score which is the probability of getting that z-score or less.
the area to the right of that z-score is equal to 1 minus the area to the left of that z-score.
that tells you the probability of getting that z-score or more.
since the problem states at least, then you want the area to the right of that z-score, which is what was calculated above.

if you had used a calculator, you would have used unrounded numbers.
in that case:
s = sqrt(.6 * .4 / 100) = .0489897949
z = (.7 - .6) / that = 2.041241452.
p(z > 2.041241452) = .0206133484.
that's pretty close to the result from the table of .02068.
.02068 is only off from .0206133484 by 0.32% rounded to 2 decimal places.
that's considerably less than a 1% error.

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