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2 Assume that the amount of time an Internal Revenue Service examiner is supposed to spend reviewing a randomly selected return is 55.1 minutes, with a standard dev...

Question

2 Assume that the amount of time an Internal Revenue Service examiner is supposed to spend reviewing a randomly selected return is 55.1 minutes, with a standard deviation of about 17.6 minutes If a sample of 35 such reviews is selected, what is the probability that the average review time is more than an hour?

2 Assume that the amount of time an Internal Revenue Service examiner is supposed to spend reviewing a randomly selected return is 55.1 minutes, with a standard deviation of about 17.6 minutes If a sample of 35 such reviews is selected, what is the probability that the average review time is more than an hour?



Answers

Suppose the mean length of time that a caller is placed on hold when telephoning a customer service center is 23.8 seconds, with standard deviation 4.6 seconds. Find the probability that the mean length of time on hold in a sample of 1,200 calls will be within 0.5 second of the population mean.

Yeah. Okay. This problem the variable is a. Can a uniform distribution From A. S. 1.5 piece 2.2. Just to uh the domain of the beautiful distribution. And then we can calculate the expected value. It should be A plus B over two. And the result is 1.85. This is the results for the expected value. And we have the viruses B minus a square over to. And we calculated again a precise point four. Is there a 3? Yeah. Probably. We have to probability that actually squared and 1.5. It's more than two. It should be 2 -1.5 times two months and 1.5. Uh this is 212. Sorry? Yeah. And we can calculate this would be 5-7 problem. See the community distribution function is it's a piecewise function. So first it is more than a. It's going to be zero. Otherwise it's gonna be uh 27 X -1.5. A base Career. Then 1.5 and smart and to want to. Otherwise it's gonna be one. So that's how this answers are

Okay, so we're given a normal distribution with a mean of 10 and a standard deviation of two. And what we're trying to find is the probability that the mean x value is at most 11 Over two different days. And so what I'm gonna do is cut this up into day one and day two and find the probabilities on day one and day two, and then Go from there. So on day one we have five people filling out the form, so and is equal to five. And what I'm gonna do is use something called a Z score, which basically just transforms our normal distribution into the standard normal distribution and transforms our Maximum value, or 11 into whatever that value would be on the standard distribution table, so that we can just look at that table to find our probabilities. So the Z score is the mean x value minus are you value divided by the standard deviation divided by the square root of the number of trials. And and so if I do this for 11, we would have 11 minus 10 divided by our standard deviation, which is too divided by squared five since we have five trials and this is equal to one over two, divided by the square to five. And now you can just simplify this two square to 5/2 and we can plug this into our probability. So we say the probability of Z being less than or equal to one, sorry, Square to 5/2. This is equal to our probability of our X men being less than equal to 11. So now we can just use the standard normal distribution table to find our probability. Um And so the last thing we're gonna want to do is Actually divide are square to five by two. So we have a value to look at for the standard distribution table. Standard normal distribution table, so this is squared of 5/2 is equal to one point 12 about. So if you look this up on the standard distribution table, we'd see that The probability of Z being less than or equal to 1.12 is eight 6.8686. And now what we want to do is find the probability on day two. So for day two we have N is equal to six and we're trying to find the same probability. So we're trying to find the probability of C sorry, of X being less than or equal to 11. So we can do the same thing. We can set our Z R. Z score equal to I mean X value minus you over the standard deviation divided by the squared of end. And if we do this for 11, again, going to see the only thing that changes is the number of trials or this squared of n. And so last time we had one over two divided by the square of five. This time we have 1/2 divided by the square of six. So that's just the squared of six over to And if we plug this into a calculator, we'll see that The square root of six divided by two is equal to around one 22 So now we can find the probability of RZ being less than or equal to 1.22, Which is it should be a little bit bigger than the other one. And it actually comes out to be .88 88 And now the last thing that we do is that we take these two probabilities .8888 10.8686. And since we're assuming that are probabilities are independent of each other, the two days are independent events. Where you can just say that the probability of, I'm gonna say of both days having a mean value of less than or equal to 11 is equal to the probability of the first day, multiplied by the probability of the second day. And we can do this because they're independent events. So This is equal 2.8. So the first day was point 8686, and the second day it was .88 88 And if we plug this into a calculator, You can see that this comes out to about .772.

Mhm. When this question, we're told that the main length of time is 47 days, standard deviation sixties. In the population we have a sample of size 50 for example, our mean is 47 our sample standard deviation is six over square root of 15, Which is .8485. And were asked for the probability that in this sample, the meantime will be more than 50 days. So that's the probability that C is more than 50 -47 over .84 five. Just the probability that Z is more than 3.54, That is 1- probability that sees less than 3.54. And from that table, this value is .9998, so the answer is find 0002.

For our problem were given that the average commute is 25 minutes and the standard deviation is 6.1 minutes. We first need to find the probability that a commute is greater than 30 minutes. So we're going to start by changing our value of 30 minutes to it. Z score or Z value. Using our formula C equals the value minus the mean over or divided by the standard deviation. So 30 minus 25 divided by 6.1, and this comes out to be approximately 0.8 two. We're going to round two decimal places here because our standard normal distribution table goes to two decimal places for our Z scores. So we're finding the probability that the Z score, or Z value, is greater than zero point 82 now for to draw sketch of what this would look like. You draw our standard normal distribution, the mean would be zero, and then we'd have one standard deviation away. Two standard deviations away three and you can also go in the other direction. Our Z value is going to be is 0.82 So we're in a sketch, be approximately here and We're looking for the probability that it's greater than that. So the area to the right will give us the probability that it's greater than 0.82 Now our standard normal distribution table gives us the area under the curve to the left of his E score. So since we're trying to find the area to the right, we need to do one minus. The value were given in the table. So if we go to our table and we look up the Z score 0.82 we're going to see the value 0.7 939 and we do one minus that value to give us the probability or area to the right. So that's going to be 0.2061 or we can write it as a percent so 20.61%. So there is a 20.61% chance that the commute would be longer than 30 minutes. Now, the problem also asked us to find the probability that a commute is less than 18 minutes and we're gonna follow the same steps is above starting by standardizing this value, finding the Z score so Z equals 18 minus 25 divided by 6.1. This would give us a Z score of approximately negative 1.15 So we are finding the probability that the Z score, or Z value is less than negative one 0.1 five. Now, if we were going to sketch this, there would be zero. We're going into the negatives here. You have to. So our Z score is negative 1.15 So approximately there, then we're finding the area under the curve to the left of that because we want to know less than that value. So because we're finding the area to the left of disease score, that's what our state a normal distribution table gives us. So we look up negative 1.15 as a Z score, and the area that the table gives us is going to be 0.1 to 51 or 12.51 percent. So there is a probability of 12.51% that a commute is less than 18 minutes


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