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Historically; 8 percent of a mail-order firm's repeat charge-account customers have an incorrect current address in the firm's computer database The numbe...

Question

Historically; 8 percent of a mail-order firm's repeat charge-account customers have an incorrect current address in the firm's computer database The number of customers out f 15 who have an incorrect address in the database is a binomial random variable with n = 15 and T = 0.08.(a) What is the probability that none of the next 15 repeat customers who call will have an incorrect address? (Round your answer to 4 decimal places )Probability(b) What is the probability that four customers w

Historically; 8 percent of a mail-order firm's repeat charge-account customers have an incorrect current address in the firm's computer database The number of customers out f 15 who have an incorrect address in the database is a binomial random variable with n = 15 and T = 0.08. (a) What is the probability that none of the next 15 repeat customers who call will have an incorrect address? (Round your answer to 4 decimal places ) Probability (b) What is the probability that four customers who call will have an incorrect address? (Round your answer to 4 decimal places:) Probability (c) What is the probability that five customers who call will have an incorrect address? (Round your answer to 4 decimal places:) Probability (d) What is the probability that fewer than six customers who call will have an incorrect address? (Round your answer to 4 decimal places:) Probability



Answers

People visiting video rental stores often rent more than one DVD at a time. The probability distribution for DVD rentals per customer at Video To Go is given in the following table. There is a five-video limit per customer at this store, so nobody ever rents more than five DVDs.
$$\begin{array}{|c|c|}\hline x & {P(x)} \\ \hline 0 & {0.03} \\ \hline 1 & {0.50} \\ \hline 2 & {0.24} \\ \hline 3 & {} \\ \hline 4 & {0.70} \\ \hline 5 & {0.70} \\ \hline 5 & {0.74} \\ \hline\end{array}$$
a. Describe the random variable $X$ in words.
b. Find the probability that a customer rents three DVDs.
c. Find the probability that a customer rents at least four DVDs.
d. Find the probability that a customer rents at most two DVDs.
Another shop, Entertainment Headquarters, rents DVDs and video games. The probability distribution for DVD rentals per customer at this shop is given as follows. They also have a five-DVD limit per customer.
$$\begin{array}{|c|c|}\hline x & {P(x)} \\ \hline 0 & {0.35} \\ \hline 1 & {0.25} \\ \hline 2 & {0.20} \\ \hline 3 & {0.10} \\ \hline 4 & {0.05} \\ \hline 5 & {0.05} \\ \hline\end{array}$$
e. At which store is the expected number of DVDs rented per customer higher?
f. If Video to Go estimates that they will have 300 customers next week, how many DVDs do they expect to rent next week? Answer in sentence form.
g. If Video to Go expects 300 customers next week, and Entertainment HQ projects that they will have 420 customers, for which store is the expected number of DVD rentals for next week higher? Explain.
h. Which of the two video stores experiences more variation in the number of DVD rentals per customer? How do you know that?

In questions. 78. We have information about a video rental store DVD rental store called Video to Go and were given a probability model in which we have X to be zero through five there, representing the number of DVD Reynolds per day for a customer and then the probability of each of those things happening. Now the four has 40.70 beside it, and looking at that, that exceeds 100% and that would not be a legitimate probability model. And then we would not be able to answer questions. Um, following this. So I think that's supposed to be a 1000.7 I think that maybe a textbook errors I'm gonna use this, um, as 0.7 for the rest of the problem. Also noticed that because they're probably model should add upto one. There's a gap without three. So we're mento. We're supposed to fund that gap. That gap is 30.12 and now each of those probabilities when you add them together, that adds up to one. Let's answer party A says. Describe the random variable x inwards. So X is gonna be the number of DVD Reynolds from video to go that is per day per customer be found The probability that a customer rents three DVDs? Well, that was the gap in the probability model. We've already figured that out. Um, that is 0.12 See, find the probability that a customer rents at least four DVDs, which that means four or five so or means ad. So four is point of 75.4 Those together may get 0.11 d. We want to find the probability that customer rents at most two DVDs, so at most two means zero or one or two. We can find each of those probabilities and the distribution above 0.3 plus 0.5 plus 0.24 That gives us 0.77 as our total. And then we're given information about a new store entertainment headquarters were given their probability distribution zero through five for DVD Reynolds and all of their probabilities. The probabilities do add upto one or 100%. Now, um, to make things simpler, I'm gonna change the X and the p of X and this one toe why? And people why, just when I calculate, um, there's a difference between video to Go and entertainment headquarters and E. It says, At which store is the expected number of DVD Reynolds Higher? Well, we're gonna find to the expected value of X, which is videos to go from the top and then the video of why, which is our new distribution. So in order to find your expected value, you're gonna take each value of X and multiply bites probability and add those together for video to go. We get 1.82 and we're gonna do that same thing for entertainment headquarters. Each value zero times it's probability, plus one times it's probability and so on, and we get 1.4 for its expected value. So the expected number is higher at the video to go in F. It says a video to go estimates that they will have 300 customers next week. How many DVDs do they expect your rent next week? So we would salute the expected value of 1.82 Well, that's the number of videos that we would x number of DVD rentals would expect per customer. If we expect to see 300 customers would just multiply those two values together and get 546. So video to go should expect 546 DVD. Reynolds Total and G says a video to Go expects 300 customers next week and entertainment headquarters projects that they will have 420 customers. For which store is the expected number of DVD Reynolds for next week. Higher. So we've already calculated, um, the expected number of DVD rentals for video to go in the previous question. Let's look at entertainment headquarters. We haven't expected value of 1.4 with a expected number of customers to be 420. We're gonna multiply these two values. Together, we'll get 588. Eso entertainment headquarters should expect Mawr DVD rentals next week. Um, 588 being greater than 546 and h which of the two video stores experiences Mawr variation in the number of DVD rentals per customer. How do you know that? So we can find the variation. We could find the standard deviation. Either one of these one. Answer the question. So I'm gonna look at the variation of X and the variation of why side note here is that variation is gonna be the summation of every single value minus the mean when I figure out how far away far away it is from the mean and square that value and then we're going to multiply it. By its probability, we're gonna add all of those together. So in X, as you can see, we have zero minus the mean, which is the expected number squared times. It's probability we're gonna then add that toe one minus the mean square times It's probability something with 2345 together. That adds up to be 1.3476 So the variation is 1.34 76 doing the same thing with why which is entertainment headquarters. We're gonna take zero minus. It's mean square that value to get rid of the negative, multiplied by the probability and do that with 1234 and five and do that all the way across. We get variation of to 0.4 so the various you can see is greater for why than it is for X, and we'll say that here, Entertainment headquarters says more variation and DVD Reynolds per week since the variation of why was greater than X 2.4 being greater than 1.3476 so why had a greater variance?

Were asked to answer questions but the plane for an executive Travelers club. We're told that this plan has been developed by an airline with the premise. 10% of its current customers would qualify for the membership in part A. We're told that among 25 randomly selected customers were asked for the probability that between two and six inclusive these customers qualify for this membership. Well, we have the X. First of all is a binomial distribution with a sample size of 25 and then also probability of 0.1, since there's a 10% chance they will qualify. So you have the probability that X lies between two and six is going to be be of six parameters 25 1 minus B of tu minus one. Because you want to include to in this, so be of one parameters 25 and 0.1. This is approximately 0.991 minus 0.271 and this is 0.720 in part B. We're asked the expected number of customers who qualify and the standard deviation of the number of qualified in a random sample, this time of 100 current customers. We have the expected value. X. This is going to be the sample size n times The probability P This is 25 times 0.1 just 2.5. Well, here's a mistake. Actually, Instead of being 25 times 250.1, this should be 100 times point. One comes out to be 10 and we have that. The standard deviation of X. This is square root of n times P times probably a failure. Que so this is the square root of 100 times 0.1 times one minus 10.1, which is 0.9. So this comes out to be the square root of 10 times. 100.9 is nine, which is three. So we expect a standard deviation here of three in court. See, we're told that X is the number in a random sample of 25 current customers who qualify for membership. We're told that we should reject the company's premise in favor of the claim that P is greater than 0.1. If X is greater than or equal to seven were asked to find the probability that the company's premise is rejected when it is actually valid. So here we have, the company's premise will be rejected, and it's actually valid if we have probability that X is greater than or equal to seven and P is equal 2.1. Well, this is equal to in Sex is by normally distributed one minus be of and here we want to include seven. So be a seven minus one is six the parameters of B. What we have that this is a random sample of 25 customers, so the sample size is 25 and we have that P was 0.1. This is approximately equal to one minus 10.991 which is equal to 0.9 so rather small percentage less than 1%. And in Part D. We're using the same decision rule introduced in part C so that the company's premises rejected in favor of the claim that he is greater than 0.1. Effects is greater than or equal to seven. Interesting. The problem. We asked the probability of the company's premises, not reject, even though P is equal to 0.2, this is a probability that premises not rejected, even though premises that invalid in this case, so 20% to qualify when we have. This is the case when we have that probability that X is going to be less than or equal to six and we have The P is equal to 0.2. So this probability because excess binomial distributed this is equal to B F six and again, this is a sample size of 25 p is point to and using a calculator, we approximate this to be 0.780 Notice that this is about 78% which is much larger than the result from C.

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

Well this problem, we are told that we have a shipment of 10 items with two defective an eight non effect. You have 10 total items. Two of them are defective, means your P. Value 0.2 is 20%. Now notice that we have a binomial distribution here because the china has chosen independently and there are only two options and see that's going to be effective or in effect, that's what's what's your own party On part of a We were told that we have a sample of three items. We want the probability that the shipment will be rejected and it's rejected if the defective item is found now here in this three. And so we want the probability that one of the three. Mhm. Again we are taking a sample three. We want the probability that at least one is effective, probably at least one is able to one minus the probability of zero. So since this is a binomial distribution, this is one minus 3 to 0 time. 0.220 0.8 to the third. And so then we evaluate this and this is one minus 10.8 to the third, that's 0.488 And so the probability is rejected this 0.488 That's what that sample of three. It's on me. Yeah. We want to sample four items to be selected. We want to know the probability that it is rejected. And so it means with a sample of four, that means we want the probability at least one of the four, I'm sorry, should still be excess rather than one. We want at least one of the four. But in his for now this is still the same as one minus the probability the next zero. And since it is for this is now one minus 4 to 0 time. 0.220 0.8 to the fourth. And so evaluating this, this gives us 0.5904 So that's the probability it is rejected here. Let's see. We have fighter five items. Okay. Yeah. Yeah. And so this is the probability that act is greater than or equal to one still. But in is now five. This is one minus 5 to 0, 0.220 0.8 to the fifth. So we evaluate this, this is 0.67232 Yeah, in part D. We want the problem. We want management wants there to be a 90% chance that this batch is rejected. We want to know how many we need to choose. Yeah. We still want the probability that acts is greater than or equal to one. We just don't know what it is once this is still one minus the probability that x zero. Mhm. And so this is one minus in 20 time 0.220 time 0.8 to the end. And we want this to be 0.90 This is one minus 10.8 to the in 0.90 So negative 0.8 to the end is negative 0.1. Yeah, so 0.8 to the end is 0.1. And so what this tells us is that in is april to the log base 0.8 of zero point more. The definition of war, that's all we can evaluate this for him. Whenever we do this gives us in is approximately 10.3. This this means we would need a sample size about least. Mhm. A what?


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