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University of Chicago

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Problem 11 Medium Difficulty

An online retailer has determined that the average time for credit card transactions to be electronically approved is 1.6 seconds.
(a) Use an exponential density function to find the probability that a customer waits less than a second for credit card approval.
(b) Find the probability that a customer waits more than 3 seconds.
(c) What is the minimum approval time for the slowest $ 5 \% $ of transactions?

Answer

A. $-e^{-1 / 1.6}+1 \approx 0.465$
B. $e^{-3 / 1.6} \approx 0.153$
C. $\int_{0}^{b} f(t) d t=0.95$

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Video Transcript

and this problem were given that an online retailer has determined that the average time for credit card transactions to be Elektronik Clea approved is one point six seconds. And for a when you'd use an exponential density function to find the probability that a customer waits less than a second for a credit card approval. Exponential density functions are discussing examples two and three of this section and an example. Three. It tells us that the general form for an exponential density function his FF T equals zero T is less than or equal to zero and one divided by a mute times E to the minus. T divided by mu, where twenty is positive, where this mu is the average value of our random variable. The problem tells us that the average time for transaction approval is one point six seconds, so one point sixes are mute. So are exponential density function. Zero if he is less than or equal to zero and one divided by one point six times E to the minus T divided by one point six. If he is positive, we're going to use this in part B and C as well. But for Now let's finish part, eh? Which I says to find the probability that the wait time is less than or equal to one second. Of course, we can't wait negative time. So what we care about is the probability that our wait time is between zero and one second. We can evaluate this probability using an integral of our exponential density function. The bounds of the integral are the same is the bounds. In the previous lines, we have this lower bound of zero, his upper bound of one. This is the integral from zero one of one divided by one point six times e to the minus t divided by one point six TT. Because we only care about integrating above zero, we don't have to worry about the tea lesson or equal to zero part of our density function Santa derivative of this Inter grant is Maya's E to the minus t divided by one point six. So we need to evaluate this from zero to one. We get one minus e to the minus one divided by one point six, which is approximately zero point four six four or forty six point four percent. So this tells us at the probability of waiting less than one second is forty six point four percent in part B. We're asked to find the probability that a customer weights more than three second. That's the probability that the wait time is at least three seconds again we can express. This isn't in a roll of our probability density function. Our lower bound is three because we care about wait times that are at least three seconds. But in this part, we're not given an upper bound. We care about all wait times bigger than three seconds. That's all wait times between three and infinity, which gives us our upper bound. So this is an improper integral. The limit is B goes to infinity of the integral from three to be of our exponential density function one divided by one point six I've e to the minus. T divided by one point sixty t, which is the limit as B goes to infinity of minus e to the minus. Be divided by one point six minus minus e to the minus three divided by one point six. This first term got this exponential decay. There's minus being the exponents. So this goes to zero is B goes to infinity. So when all the dust settles, we're only left with E to the minus three, divided by one point six, which is approximately zero point one five three or fifteen point three percent. So the probability that the customer waiting more than three seconds is fifteen point three percent party's a little bit different were asked to find the minimum approval time for the slowest five percent of transactions. I think it's good to kind of think about what this looks like graphically. So let's draw are exponential density function. You've got this nice exponential decay happening. So the overall area under this curve from zero to infinity is one. So we care about there's little flavor at the end that represents the five percent of the longest transaction times, five percent of longest transactions. So there's this cut off value right here. Let's call it see, So the goal is the fine. See, you want to find this minimum transaction time for the longest five percent of transactions. So to do this, we need to set up an animal that we can solve. So what is this sea satisfy? To see satisfies that the integral from sea to infinity of our exponential density function is equal to zero point zero five. That's five percent of the area under this graph. Now this integral on the left is just improper integral, which we evaluated in part B. So when you evaluate this improper in a roll, you get e to the minus. C divided by one point six equals zero point zero five. We can solve this equation. Using the log rhythm minus C divided by one point six is the natural log of zero point zero five. So C is equal to negative one point six times the natural log of zero point zero five, which is approximately four point seven nine. So our final answer is that the minimum approval time for the slowest five percent of transactions is approximately four point seven nine seconds.