# Probability of average given mean and variance

For the life of me I cannot find a way to solve this question. Any help would be appreciated!

From past experience, a professor knows that the test score of students taking a final examination is a random variable with mean 65. Suppose in addition the professor knows that the variance of a student's test score is equal to 30. How many students would have to take the examination so as to ensure, with probability at least 0.8, that the class average would be within 5 of 65?

• @Max It's not the sample of test scores that has a smaller variance, but the sample mean of them; and the variance of the mean would be $30/n$ not $30/\sqrt{n}$. And in any case, you don't need the CLT for the result about the variance of the average - just independence and two basic properties of variances. – Glen_b Feb 17 '13 at 23:27
• @Glen_b You are correct, sir. My comment is now deleted. – assumednormal Feb 17 '13 at 23:31
• You might like to construct an improved version of it as an answer. If you did so I would be inclined to upvote it. – Glen_b Feb 17 '13 at 23:33

The 'from past experience' bit should be taken as code for 'here are some population parameters'. The question gives the shared mean and variance (and implies identically distributed scores, though you don't actually need that for this question). You can probably safely assume independence.

The central issue (finding the variance of an average of independent random variables - and hence its standard error) follows from two facts:

1) $\operatorname{Var}(aX) = a^2\, \operatorname{Var}(X)$

2) The variance of a sum is the sum of the variances, if the variables are independent.

From this, you can obtain that the variance of the average of $n$ independent random variables with a common variance is $1/n\,$ times the variance of one of them.

Hence the standard deviation of the distribution of the average (the standard error of the average) is the standard deviation of the distribution of one of the variables divided by $\sqrt{n}$.

If the variance for one student is 30, what is the variance for the class average for a class of size $n$?
In either case, you can write down the width of an interval that has an 80% chance of including the class average using the properties of either the normal distribution or the Chebyshev inequality. Since you know the half-width (5), you can use this to solve for $n$.