The amount of time needed for a printer to print a file is a random variable with mean $E(X_i)=20$ minutes and variance $var(X_i)=4$ minutes$^2$. The times needed for difference file are independent from each other. Find the probability that the printer prints less than 15 files in 5 hours.
I have thought for a hour to answer this question. I'm not have an idea to do it. I cannot find the probability if given mean and variance. What the hint to answer this question?
The probability that fewer than $\ 15\ $ files are printed in $\ 5\ $ hours is $$ P\Big(\sum_{i=1}^{15}X_i>5\times60\Big)\ . $$ You're probably expected to appeal to the central limit theorem to deduce that $\ \sum_\limits{i=1}^{15}X_i\ $ is approximately normally distributed with mean $\ 15\times20\ $ minutes and variance $\ 15\times4\ $ minutes${}^2$. However, since $\ 15\times20=$$ 5\times 60=$$\,300\ $, you only really need for $\ X_i\ $ to be symmetrically distributed about its mean to deduce that $$ P\Big(\sum_{i=1}^{15}X_i>5\times60\Big)=\frac{1}{2}\ . $$