person A runs 100 meters in 10 seconds on average, and person B runs the same race in 12 seconds on average, let's suppose we model time of person A with an exponential distribution and we do the same with person B. How to find probability that the fastest person runs this race in less than (i.e $< 11$)?
I've interpreted the problem with minimum function, (i.e $Z = min(X, Y)$). by definition of exponential cumulative function, I know that $P(T>t) = exp(-\lambda*t)$, and then I can multiply first cumulative function with second cumulative function, such as: $$P(X>t) * P(Y>t)$$ and so I get: $$exp(-\frac{1}{10}*t) * exp(-\frac{1}{12}*t)$$ (it's better to write exponentials like this, because I think they're easier to read).
I've used minimum function, because minimum function finds the fastest person to consider later on with a normal distribution.
the result is: $exp(-0.1833*t)$. Now, I know $\lambda$, and this is equal to $0.1833$. I can calculate average, and variance, as follows:
- $E[T] = 1 / \lambda = 5.4556$
- $var(T) = 1 / \lambda^2 = 29.7629$
now, I can create Z = N(5.4556, 29.7629).
now, I've used central limit theorem, $$P(Z<11) = P(N(0, 1) < \frac{11-5.4556}{\sqrt(29.7629)}) = 0.81134 = 81\%$$
but result must be equal to $86\%$
You've used a normal approximation when no such approximation is necessary or appropriate.
If $Z = \min(X,Y)$ is the time that it takes for the faster of the two people to finish the race, then as you observed, $$\Pr[Z > t] = \Pr[A > t]\Pr[B > t].$$ This is because both racers must take longer than $t$ to finish if the first racer to finish takes longer than $t$, and each person's race time is independent of the other.
Consequently, as you noted, $$\Pr[Z > t] = e^{-t/10} e^{-t/12} = e^{-11t/60},$$ hence $Z$ is exponentially distributed with mean $\operatorname{E}[Z] = 60/11$. Then the exact cumulative distribution function for $Z$ is $$\Pr[Z \le t] = 1 - e^{-11t/60}$$ and the probability that $Z < 11$ is simply
$$\Pr[Z < 11] = 1 - e^{-11(11)/60} = 1 - e^{-121/60} \approx 1 - 0.133098 \approx 0.866902.$$
There is no purpose to using a normal approximation because the exact distribution of $Z$ is known and easily calculated.