Five students are waiting to talk to the ta when office...

"Five students are waiting to talk to the TA when office hours begin. The TA talks to the students one at & time, starting with the first student and ending with the fifth student, with no breaks between students Suppose that the time taken by the TA to talk to a student has a normal distribution with a mean of & minutes and a standard deviation of 2 minutes, and suppose that the times taken by the students are independent of each other_ (a) What is the probability that the total time taken by the TA to talk to all five students is longer than 45 minutes? (6) Suppose that the time that elapses between when the TA starts talking to the first student, and when the TA starts to have a headache, has a normal distribution with a mean of 28 minutes and a standard deviation of 5 minutes, which is independent of the times taken to talk to the students What is the probability that the TA's headache starts at a time after the TA has finished talking to the third student?"

Transcript

00:01 So i'm going to assume that the mean amount of time that it takes is, we know it's supposed to be a normal distribution, and you have an ampersand, and i'm going to assume that that is an 8 minutes, that that's the average amount of time, and we have 2 minutes as a standard deviation.

00:16 And we have 5 students, and they're supposed to be independent.

00:19 So the mean of, and if i let t stand for the total amount of time, then the mean of the total would end up being that 8 added on 5 times, or 8 times 5, which is going to be 40 minutes.

00:36 And then the standard deviation for the total time will be equal to the square root of, and then we add the variances of those 5 times.

00:51 And so this would end up being 4 times 5, it's going to end up being the square root of 20.

00:57 So we want to know what is the likelihood that the total time is more than 45 minutes.

01:04 And we could also do this by finding what the mean time is.

01:07 But let's convert that to a z value, and we'd have the 45 minus the 40, divided by that square root of 20.

01:16 And that gives us 5 divided by the square root of 20, which is a z value of 1 .1, and it's going to round to 2.

01:26 And so looking up in my table, the area above 1 .12 is the same as that below negative 1 .12.

01:35 Let me quick look at that.

01:37 Negative 1 .12 corresponds with .1314.

01:44 Now, we're dealing with his headache starting.

01:51 And we know that the headache has a normal distribution, that on the average starts 28 minutes after he starts working with the first student and has a standard deviation of 5.

02:03 And we are trying to find, we're looking at him finishing the third student.

02:09 So let's find what the mean for finishing with the, and i'll call it just the sub 3 for the third student, would be the, we know the mean is 8, at least i believe that's what it is.

02:21 So 24 would be the mean amount of time it takes him to finish with those three students.

02:27 And the standard deviation for those three students would be the square root of, again, 2 squared plus 2 squared plus 2 squared.

02:35 And so that's going to be 4 times, so that's going to be the square root of 12 minutes.

02:41 So we want to know what's the likelihood that the headache starts after...

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