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8 Suppose a customer service phone line gets an average of 2 calls every 5 minutes What is the probability that they they will gct exactly 3 calls in the next 10 mi...

Question

8 Suppose a customer service phone line gets an average of 2 calls every 5 minutes What is the probability that they they will gct exactly 3 calls in the next 10 minutes? WJL7S(6) What is the probability that they will get exactly 5 calls in the next 10 minutes?

8 Suppose a customer service phone line gets an average of 2 calls every 5 minutes What is the probability that they they will gct exactly 3 calls in the next 10 minutes? WJL 7S (6) What is the probability that they will get exactly 5 calls in the next 10 minutes?



Answers

Two telephone calls come into a switchboard at random times in a fixed one-hour period. Assume that the calls are made independently of one another. What is the probability that the calls are made a. in the first half hour? b. within five minutes of each other?

Figured the easiest way to show you how to do this problem is to use a graphing calculator. Because all you need to do is plug in the value they ask uh into t let's forget that. And for tea. Um So when you see Holy cow, I did not expect that to happen. Uh Okay, I'm just gonna keep happening, isn't it? Okay. So what I need to do is just show you how you can do this um with a couple different things. So first of all, just type in the equation and you can use Y equals, that's not a big deal. Uh one minus E to the negative. Okay, Okay. The flu threat key over three. And uh I'm not too concerned about the graph, but it should make sense as to zero being a probability of zero because it's impossible to have the probability happen immediately, like at zero minutes sometime has to elapse after you hang up whether it's a second or something like that. Um So what you can actually do is just create a table And make the values what you want. So the first value they ask for is one half. So that probability would be .154. The next probability they asked for is I think it was one minute, wasn't it? 02 minutes, two minutes. About .487. Next value they ask for is five minutes, hopes keep messing up. So at five minutes It's about .811. And they actually don't care about these other values. So let me delete them. Uh and you can see the correct answers here here and here for part a part B and part c. And what you could actually, I hope this answer makes sense because if you were to try Um typing in y equals one or that would be a guarantee. Well, there's, if you look closely there's no guarantee because that's a horizontal assent to it. So there's no guarantee that after 15 minutes a call will come in because that would be a probability probability of one is a guaranteed occurrence, but it's very very likely to happen. You know, you have a .99% Okay. Yeah, 99% probability I said that wrong. I don't know what I was thinking. Uh so anyway, using your graphing calculators, all you need to know how to do just plug in values for tea. Mhm, mm. Okay. Yeah.

All right, this question asks us, but a Passat industry shin with an average of 48 calls per hour. So party asks, what is the probability of getting three calls in five minutes? So let's first compute the expected amount of calls in that time interval. So they're 48 calls in 60 minutes, and we're dealing with five minutes and working that Oh, you'll see that the minute that the minutes cancel and we're left with calls, so we expect four calls. So now we can plug are expected value before in for the mean and three in for observed so probability that X equals three equals four to the third power times he to the negative forth all over three factorial. And that is equal to point 1954 All right, and then Part B asks for 10 calls in 15 minutes. So once again we expect 48 calls in 60 minutes times 15 minutes, minutes cancel. And in this case, we're left with 12 as our expected value. So we want to find the probability that there are 10 calls in this interval which is are expected value raised, Thio the observed times e to the negative of are expected all over our observed factorial, and that works out to be 0.10 48 Part See asks us for the expected amount of calls if the operator takes a five minute break. So once again, 48 calls for 60 minutes times five minutes, and that equals four waiting. And that is our expected. And it wants the probability that there are no calls so probability that X equals zero and that is our main. Two are observed times e to the negative of are mean all over our observe factorial and remember that zero factorial is one, and we end up with point zero 18 32 All right, and then Part D asks a similar question. But this case, if the operator takes a three minute break, what's the probability is no calls again? 48 calls in 60 minutes, times a three minute break. That's our new expected value, which becomes 2.4 calls, and it wants the probability that X equals zero and that is just equal to are mean to the observed times e to the negative of are mean all over our observed factorial. And that turns out to be 0.9 07

In this problem it is given that the probability get your call to a service line is answered in less than 30 seconds is 0.75. Assume that your calls are independent. It is asked what is the probability that you must call six times in order for two of your calls to be answered in less than 30 seconds. Let x denote the number of calls until obtaining two calls and certain less than 30 seconds. So x follows negative binomial distribution with r is equal to two And P is equal to 0.75. We know that if X follows negative paranormal distribution then probability of x equal to x is x minus one C. R minus one into one minus P raised two X minus are into P raised to our their exchanges from our art Chrisman artist to to infinity. We have to find the probability that we must call six times. That is we have to find the probability of X equal to six. This is equal to x minus one. That is 6 -1. Our is too So are my in this one, That is 2 -1. So this is 6 -1 c to -1 into one minus. He is 0.75 So 1 -0.75. Rest too. X -R. That is 6 -2 in two P PS 0.75 raised to our ari's too. So this is equal to 6 -1 is 5 to -1 is one. So 57 Into 0.25 raised to six minus 2, 30. 0.25 raised to four Into 0.75 raised to two. This is equal to zero point Zygo Ben 10 0.0110. So the probability that we must call six times in order for to offer our calls to be answered in less than 30 seconds is 0.0110 in the next part it Disaster. What is the mean number of calls To obtain two answers in less than 30 seconds. That is We have to find mean mu for the negative binomial distribution, we know that mean mu that is expected X is equal to uh huh, divided by P. There are is too. So this is equal to two divided by p. 0.75, two divided by 0.75. This is equal to multiplying and dividing the 100. We get 200 divided by 75. Now we know Dad, 25 into eight is 200 And 25 into 3 75. So this is equal to That is 200 x 75 is equal to eight divided by three. So The mean number of calls to obtain two answers in less than 30 seconds is eight x 3.

In this problem, it is given that the probability that your call to a service line is answered in less than 30 seconds is 0.75. Assume that your calls are independent. It is asked what is the probability that you must call four times to obtain the first answer? In less than 30 seconds. Let x denote the number of calls until we are answered in less than 30 seconds. So X is a geometric random variable with P is equal to 0.75. We know that if x is a geometry random variable, then probability of X equal to X is one minus P raised two x minus one in to pay. We are X ranges from one to up to infinity. So this is equal to 1 -0.75 is 0.25. So this is equal to 0.25 Raised two x -1 Into 0.75, we have to find the probability that we must call four times. So we have to find the probability of X equal to four, Probability of x equal to four is equal to 0.25, raised to X -1. But here X is four, so 4 -1, that is three Into 0.75. 0.25, rest 2 3 In 20.75. This is equal to zero point 01 56 25, 0.015, Into 0.75. This is equal to zero point 01 17, 0.0117. So the probability that we must call four times to obtain the first answer in less than 30 seconds is 0.0117. In the next part it is asked what is the mean number of calls until you are answered in less than 30 seconds. You know Dad for a geometry random variable X. I mean that is expected X is equal to one divided by p Mean is equal to one divided by P. Here P is equal to 0.75. So this is equal to one divided by zero point 75. Multiplying and dividing by 100. We get this is equal to 100 Divided by 75. So this is equal to we here 25 into 425 into three is 75. So this is equal to four divided by three. That is the mean number of calls until we are answered in less than 30 seconds is 4x3.


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