2. (12 pts) Aaron needs to complete 30 tasks in sequence, with times required to

Question

2. (12 pts) Aaron needs to complete 30 tasks in sequence, with times required to do each of the tasks being inde to complete 34 tasks in sequence, with the times required for each task being independent with mean minutes and standard deviation 8 minutes, and independent of Aaron's times. Suppose Aaron started S minutes before Brian did. Approximate the probability that Brian co completes all his tasks. If you use any of the results covered in class, provide its name and justify its use. pendent random variables with mean 25 minutes and standard deviation 7 minutes. Brian needs 21 mpletes all his tasks before Aaron

Explanation / Answer

Here as number of tasks for both Aaron and Brian are greater than 30, we can apply central limit theroem, as we can say that mean time taken in completing the tasks can be taken as normally distributed.

Here mean time taken for Aaron to complete his tasks = 30 * 25 = 750 minutes.

Standard deviation of time taken for aaraon to complete his 30 tasks = 7 * sqrt(30) = 38.34 minutes

mean time taken for Brian to complete his tasks = 34 * 21 = 714 minutes.

Standard deviation of time taken for aaraon to complete his 34 tasks = 8 * sqrt(34) = 46.65 minutes

Here Aaron started five minutes before brian so we take mean time taken by brian can be taken as 719 minutes with no effect on standard deviation.

Now if X is the total time taken to completed Aaron's tasks and Y is the total time taken to complete Brian tasks.

So, we have to find

Pr(X < Y ) = Pr(X - Y < 0)

So, mean value of X - Y = E(X - Y) = 750 - 719 = 31 minutes

Varaince of (X-Y) = Var(X-Y) = Var(X) + Var(Y) = 30 * 7 + 34 * 8 = 482

std (X-Y) = sqrt(482) = 21.9545

So,

Pr(X - Y < 0) = Pr(X - Y < 0 ;31 ; 21.9545)

Z = (0 - 31)/ 21.9545 = -1.412

Pr(X - Y < 0) = Pr(X - Y < 0 ;31 ; 21.9545) = Pr(Z < -1.412) = 0.0790

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