I have a question like this.
There are $50$ people in a line. The time takes to serve a person has a mean of $5$ mins and standard deviation of $3$ mins. $5$ people can be served at a time. What is the probability that $50$ people can be served within an hour$?$
We can take $\mu = 5$ and $\sigma = 3$
So a normal distribution can be derived $N(5, \frac{3^2}{50})$
But the answer in the book says normal distribution is $N(250, 3\sqrt{50})$
How can I derive such an equation$?$
Let $t$ be the random variable which represent the time to serve a single customer. The problem tells us that the expectation value of $t$ is $5 \ \mathrm{min}$,
$$ E(t)=5\ \mathrm{min},$$
and it has a standard deviation of $3\ \mathrm{min}$, which means that the variance is $9 \ \mathrm{min}^2$,
$$ Var(t) = 9 \ \mathrm{min}^2.$$
We want to determine the probability distribution for the total time, which will be the sum of the individual times; this means that we should use the central limit theorem for sums rather than for the average value of a sample. The total time, which we will denote by $T$, is the sum of the time for $50$ customers,
$$ T = t_1 + t_2 + \cdots + t_{50}, $$
we can determine the expected value of $T$ by using the sum rule for expectations,
$$ E(T) = E(t_1+t_2+\cdots+t_{50}) \\ = E(t_1) + E(t_2) + \cdots + E(t_{50}) \\ = 5\ \mathrm{min} + 5\ \mathrm{min} + \cdots 5\ \mathrm{min} \\ = 50 \times 5\ \mathrm{min} \\ = 250 \ \mathrm{min},$$
similarly we can determine the variance of $T$ by using the sum rule for variances,
$$Var(T) = Var(t_1+t_2+\cdots t_{50})\\ = Var(t_1) + Var(t_2) + \cdots + Var(t_{50}) \\ 9 \ \mathrm{min}^2 + 9 \ \mathrm{min}^2 + \cdots + 9 \ \mathrm{min}^2 \\ = 50 \times 9 \ \mathrm{min}^2, $$
the standard deviation is just the square root of the variance so get,
$$ \sigma = \sqrt{50\times 9 \ \mathrm{min^2}} = 3 \sqrt{50} \ \mathrm{min}$$
So our distribution will be $N(250 \ \mathrm{min}, 3 \sqrt{50} \ \mathrm{min})$.
