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Suppose $64$ students are taking an exam. The exam has $16$ questions in multiple choice format. Students must fill an oval for every one of the $16$ questions. They can't revise their answers. The time taken for a students to answer a question is exponentially distributed with mean $8$ minutes. So $f(t) = \frac{1}{8}e^{-t/8}$. The times for answering questions are independent. Using the normal approximation calculate the probability that the average time taken by the students is more than $132$ minutes.

So we want to find $P[T_1+T_2 + \cdots + T_{64} > 132(64)]$. What would be the distribution of a sum of exponential random variables? Would it be a gamma distribution? If we let $S = T_{1} + T_{2} + \cdots + T_{64}$, then $P[S > 132(64)] = 1-P[S \leq 132(64)]$.

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    The sum of k exponentials is gamma-distributed, as you've pointed out. But as k goes to infinity, the gamma distribution, appropriately rescaled, converges to normal.2010-11-14
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    just so you know, I corrected a typo in your last equation.2010-11-14

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Yes. It will be a Gamma distribution. But here as the problem says you can approximate it by a normal distribution. If you look at the variable $T = \displaystyle \sum_{k=1}^{64} T_i$, it has mean $64 \times 16 \times 8$ and the variance is again $64 \times 16 \times 8^2$.

(Note that $T_i$ has mean $16 \times 8$ and has variance $16 \times 8^2$).

Hence all you need now is to look at the random variable $\frac{T-64 \times 16 \times 8}{\sqrt{64 \times 16 \times 8^2}}$.

So, $P(T \geq 132 \times 64) = P(\frac{T-64 \times 16 \times 8}{32 \times 8} \geq \frac{132 \times 64 - 128 \times 64}{256}) \approx P(Z \geq 1) = 0.158655$

where $Z$ is the standard normal random variable with zero mean and unit variance. Look up the table for this value.

(Also, Kindly check my arithmetic calculations)

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The sum of i.i.d exponential random variables follows the Erlang distribution (which is a special case of the Gamma distribution). For details, check out http://en.wikipedia.org/wiki/Erlang_distribution