SD probability problem involving chi-square distribution Preparing for an exam in statistics, I have been pondering the following problem:

Given that in country X 14% of people hold a university degree, find the probability that for a random sample of size 200 the standard deviation is less than 20.

Trying to compute this resulted in ridiculously high values for the chi-square-distribution.
Any input would be warmly welcomed.
 A: Sometimes on exams you just have to answer them literally as written.  It helps to explain and justify your interpretation.  In this case presumably the question refers to the university degree as coded with a binary ($0/1$) variable.  Let's assume it really does ask about samples with a standard deviation less than $20$.
There are just $201$ possible samples, parameterized by their total, $k$.  The mean of such a sample is $k/200$ and therefore its variance (without assuming any bias correction factor) is the average squared deviation from this mean, equal to
$$V(k) = \frac{1}{200}\left(k(1 - k/200)^2 + (200-k)(0 - k/200)^2\right) = k - k^2/200.\tag{1}$$
This attains its maximum when $k=100$, at which point it equals
$$V(100) = 100 - 100^2/200 = 50.$$
Since a standard deviation of $20$ corresponds to a variance of $20^2 = 400$, it is certain that the standard deviation will be less than $20$.  The answer, then, is $1$.

Using $1/199$ instead of $1/200$ as the factor in formula (1), to correct for bias, will not change this answer.
