Waiting times (measured in minutes) at a cashier counter in a supermarket are exponentially distributed with parameter 0.1. If 100 customers come to the counter in a day, approximate the probability that at least half of them must wait for more than 10 minutes. No continuity correction is required.
So here's what I did.
Let $X_i \sim Exp(0.1)$ be the waiting time of a customer where $i = 1,2,..,100$. So we know that $E[X_i] = 10$ and $Var(X_i) = 100$.
Let $X$ be the sum of all the $X_i$.
If half of them must wait for more than 10 minutes, that means that total waiting time of 100 customers has to be $X>1000$. So I am trying to evaluate $P[X>1000]$.
So now what I did was, to standardize it:
$P[\frac{\bar{X}-10}{10/\sqrt{100}}>\frac{1000/100-10}{10/\sqrt{100}}]$.
But this looks weird. Is is 1000? If not what's it suppose to be?
Edit:
$E[Y_i] = P[X_i>10] = e^{-0.1(10)} = e^{-1}$
$Var(Y_i) = e^{-1}(1-e^{-1})$
$Y = (\sum_{n=1}^{100} Y_i) \sim B(100,e^{-1}) \approx. N(100e^{-1},100e^{-1}(1-e^{-1}))$
Looking for $P(Y>50) = P(Z>\frac{50-100e^{-1}}{\sqrt{100e^{-1}(1-e^{-1})}})= P(Z>2.74)$
Correct?
Actually, if half of them (that's $50$) must wait at least 10 minutes, then all we know is $X>500$
Anyway, that's not the way. The event "at least 50 people waited more than 10 minutes" implies the event "$X>500$", but the reverse is not true, hence computing the latter probability would only give you a bound (and quite useless) on the former.
What you must do is : define $Y_i$ (indicator variable) such that $Y_i=1$ if $X_i>10$ ("success"), $Y_i=0$ otherwise. Compute the expectation and variance of $Y_i$. Let $Y=\sum Y_i$. Then notice that $Y$ is a binomial, which can be approximated by a Gaussian. Also notice that the event of interest is equivalent to $Y_i\ge N/2=50$
I hope you can go on from here.