Question 973393
The mean of the sampling distribution is the point estimate, which is 0.38 or 342 adults from 900.

The standard deviation is
sqrt {[(p)(1-p)]/(900)}

This is sqrt {(0.38)*(0.62)/900}

=sqrt (0.0002618)=0.0162.  This is the standard deviation of the sampling distribution.

The z-value for the random sample is {0.36-0.38)/0.0162; it is the sample-the postulated mean all divided by the standard error of the sampling distribution.

This is (-.02/0.0162), and we want the area on the normal distribution to the left of z< -1.234

We would expect with a large sample to have a fairly small probability of being 2% away.  As a rough guide, the error is 1/sqrt (n), here about 3%.  But when the probability differs from 50%, which it does, the error is less.  

Probability is 0.1085.   A z-score of -1.28 is the 10th percentile, so this is reasonable.