Probability - normal distributions The time it takes for a calculus student to answer all the questions on certain 
exam is an exponential random variable with mean 90 minutes. If all 100 
students of a calculus class are taking that exam, approximate the probability 
that more than 40 of them complete the exam in less than 90 minutes.
 A: We assume that the completion times are independent. This is not necessarily realistic. Two close friends who sit together and surreptitiously collaborate are likely to finish at roughly the same time. And members of a study group may have comparable finish times. Exponential distribution is also for various reasons an implausible model of test completion times. But in what follows, we hold our noses and calculate.  
If $X$ is the completion time of a randomly chosen students, then the probability that $X\lt 90$ is 
$\int_0^{90}\frac{1}{90}e^{-x/90}\,dx$. This is $1-e^{-1}$.
Use $p$ as an abbreviation for $1-e^{-1}$. Then the number $Y$ of students who complete in less than $90$ minutes has binomial distribution with parameter $100$ and $p$. 
The random variable $Y$ has mean $\mu=100p$ and variance $\sigma^2=100p(1-p)$.
The probability that $Y\le y$ is approximated by the probability that a normal with the same mean and variance is $\le y$. 
Let $y=40$ and use the normal approximation to estimate $\Pr(Y\le 40)$. Your final answer will then be $1-\Pr(Y\le 40)$.  
