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20 marks] In the old days, there was probability 0.8 of success in any attempt to make long-distance telephone call Suppose 1800 calls came into the exchange: Use...

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20 marks] In the old days, there was probability 0.8 of success in any attempt to make long-distance telephone call Suppose 1800 calls came into the exchange: Use normal approximation of the binomial to calculate the probability that the number of long-( distance calls are between 1400 and 14502[20 marks] The mean water pressure in the main water pipe from & town well should be kept at 56 psi Anything less and several homes will have an insufficient supply; and anything greater could burs

20 marks] In the old days, there was probability 0.8 of success in any attempt to make long-distance telephone call Suppose 1800 calls came into the exchange: Use normal approximation of the binomial to calculate the probability that the number of long-( distance calls are between 1400 and 14502 [20 marks] The mean water pressure in the main water pipe from & town well should be kept at 56 psi Anything less and several homes will have an insufficient supply; and anything greater could burst the pipe: Suppose the water pressure is checked at 47 random times The sample mean is 57.1. (Assume 0 7.) Is there any evidence to suggest the mean water pressure is different from 562 Use & = 0.01. [10 marks] mnajor car manufacturer wants t0 test new engine to determine whether it meets new air- ~pollution standards. The mean emission of all engines of this type must be less than 22 parts per million of carbon_ 10 engines are manufactured for testing purposes and the mean and standard deviation of the emissions for this sample of engines are determnined to be I 18.6 parts per million, 3.5 parts per mnillion_ At & = 0.01 level of significance; do the data provide sufficient evidence to allow the manufacturer to conclude that this type of engine meets the pollution standard? Assume normality: 20 marksl Using ` least square method, find the line of best fit for the following set of 9 data points: (90, 94) , (67 , 75) , (50, 54) ; (89, 90) , (30, 60), (75, 82) ; (80, 90) ; (70, 85) , (83,88)



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For each scenario (a)-(j), state whether or not the binomial distribution is a reasonable model for the random variable and why. State any assumptions you make. (a) A production process produces thousands of temperature transducers. Let $X$ denote the number of nonconforming transducers in a sample of size 30 selected at random from the process. (b) From a batch of 50 temperature transducers, a sample of size 30 is selected without replacement. Let $X$ denote the number of nonconforming transducers in the sample. (c) Four identical electronic components are wired to a controller that can switch from a failed component to one of the remaining spares. Let $X$ denote the number of components that have failed after a specified period of operation. (d) Let $X$ denote the number of accidents that occur along the federal highways in Arizona during a one-month period. (e) Let $X$ denote the number of correct answers by a student taking a multiple-choice exam in which a student can eliminate some of the choices as being incorrect in some questions and all of the incorrect choices in other questions. (f) Defects occur randomly over the surface of a semiconductor chip. However, only $80 \%$ of defects can be found by testing. A sample of 40 chips with one defect each is tested. Let $X$ denote the number of chips in which the test finds a defect. (g) Reconsider the situation in part (f). Now suppose that the sample of 40 chips consists of chips with 1 and with 0 defects. (h) A filling operation attempts to fill detergent packages to the advertised weight. Let $X$ denote the number of detergent packages that are underfilled. (i) Errors in a digital communication channel occur in bursts that affect several consecutive bits. Let $X$ denote the number of bits in error in a transmission of 100,000 bits. (j) Let $X$ denote the number of surface flaws in a large coil of galvanized steel.

In this problem. We're told that the Fire Department is going to sample 25 homes. And if the sample indicates that less than 80% of all homes have a smoke detector, uh, then they will implement a programme. And we're told that the criteria is that if the number of homes with the detector is less than 15 then they then they will assume that less than 80% of homes actually have the smoke detector installed. So we can say that the number of homes out of a sample of 25 is a binomial random variable based on 25 trials and a probability of success of P. So the probability that the probability of deciding that less than 80% of all homes have a detector is the probability that the number in the sample that have a detector is at most 15. So for part A were told that probability of success is 0.8. So the probability of having 15 or fewer successes and this comes out to zero point 0173 and then for beware s what the probability of not rejecting the claim is when P is equal to 0.7, or 0.6. So not rejecting the claim means concluding that at least 80% of the houses in the population do have the detector. So that means in our sample, looking for the probability that at least 16 of the houses have the detector and that's equal to one, minus the probability that at most, 15 have it. So if P equals 0.7, then the answer is zero point 811 And if P equals 0.6, then the probability of getting at least 16 houses in the sample is equal to zero point 4 to 5. So now, for Part C were asked how the probabilities of error in Parts A and B would change if the value of 15 in the decision rule is placed, replaced with 14. So basically, the probability of rejecting the claim that at least 80% of homes have the smoke detector is the probability that in the sample we get at most 14 homes. So notice here that we're calling these answers from Part A and B the error probabilities because in part, a. If the true value is 0.8, then That means at least 80% of homes have the smoke detector. But if her sample has a less than 15 or at most 15 homes, then we would reject the claim, which would be an error and similarly in part D. Either of these p values for probability would indicate that less than 80% of all homes in the population have the smoke detector. Yet we would have more than 16 or at least 16 homes in the sample that had the smoke detector. So then, in acts in error, decide that at least 80% of the homes in the population have a smoke detector, even when only 70% actually dio and the same thing with probability of 0.6, then only 60% of the homes in the population have the smoke detector. Yet we're finding with the probability of 4.25 that we would suggests that the claim is that at least 80% and so it just refer to this is part a people with 0.8. So we're looking for the probability that they have, at most 14 homes in the sample with a smoke detector, comes out to zero point 00 56 and then for B. We would instead be looking for the probability of getting at least 15 homes in the sample that do not have smoke detectors that do have smoke detectors and calculating this for probability of success. Being 0.7, we get 0.90 to 2, and if the probability of success is 0.6, we get zero point 5858

In this problem. We're told that the Fire Department is going to sample 25 homes. And if the sample indicates that less than 80% of all homes have a smoke detector, uh, then they will implement a programme. And we're told that the criteria is that if the number of homes with the detector is less than 15 then they then they will assume that less than 80% of homes actually have the smoke detector installed. So we can say that the number of homes out of a sample of 25 is a binomial random variable based on 25 trials and a probability of success of P. So the probability that the probability of deciding that less than 80% of all homes have a detector is the probability that the number in the sample that have a detector is at most 15. So for part A were told that probability of success is 0.8. So the probability of having 15 or fewer successes and this comes out to zero point 0173 and then for beware s what the probability of not rejecting the claim is when P is equal to 0.7, or 0.6. So not rejecting the claim means concluding that at least 80% of the houses in the population do have the detector. So that means in our sample, looking for the probability that at least 16 of the houses have the detector and that's equal to one, minus the probability that at most, 15 have it. So if P equals 0.7, then the answer is zero point 811 And if P equals 0.6, then the probability of getting at least 16 houses in the sample is equal to zero point 4 to 5. So now, for Part C were asked how the probabilities of error in Parts A and B would change if the value of 15 in the decision rule is placed, replaced with 14. So basically, the probability of rejecting the claim that at least 80% of homes have the smoke detector is the probability that in the sample we get at most 14 homes. So notice here that we're calling these answers from Part A and B the error probabilities because in part, a. If the true value is 0.8, then That means at least 80% of homes have the smoke detector. But if her sample has a less than 15 or at most 15 homes, then we would reject the claim, which would be an error and similarly in part D. Either of these p values for probability would indicate that less than 80% of all homes in the population have the smoke detector. Yet we would have more than 16 or at least 16 homes in the sample that had the smoke detector. So then, in acts in error, decide that at least 80% of the homes in the population have a smoke detector, even when only 70% actually dio and the same thing with probability of 0.6, then only 60% of the homes in the population have the smoke detector. Yet we're finding with the probability of 4.25 that we would suggests that the claim is that at least 80% and so it just refer to this is part a people with 0.8. So we're looking for the probability that they have, at most 14 homes in the sample with a smoke detector, comes out to zero point 00 56 and then for B. We would instead be looking for the probability of getting at least 15 homes in the sample that do not have smoke detectors that do have smoke detectors and calculating this for probability of success. Being 0.7, we get 0.90 to 2, and if the probability of success is 0.6, we get zero point 5858

So we have a normal distribution, This is for individual weights, uh the cars, uh and they would hold 75 tons and the standard deviation is 0.8 tons. And we're told us approximately normally distributed. So in question, a what's the probability that we choose a random car? And it holds less than 74.5 tons. And so we need to convert that to a Z value. Excuse me, 74.5 minus 75 divided by the standard deviation. And when we do that, we find out that that Z value or the number of standard deviations below the mean, that that value is is negative 0.6 to five. And I used my normal CDF, we could round this to negative 0.63 and then look it up in a table that I used my normal CDF and I got that probability is about 26 27%. Now we have a different question. We're sampling 20 Yeah, cars finding a mean of those 20. And we want to know what's the probability that the mean. And so our new sampling distribution no longer has that same samples. Standard deviation. We still would have our means of X bars centered at 75 but our standard deviation is reduced. 2.8 divided by the square root of 20. So what is the likelihood of getting an X. Bar? That is less than just strictly, less than 74.5. So now we convert that to get the statistic and we take the and that should not be just strictly less than 74.5 minus 75 divided by the standard deviation. That is for that sampling distribution for means of size 20. And when I get that Z value that z value comes out to be negative 2.795 And again, I used my normal C D f. I could round that to negative 2.80 and look it up in the table and I get a very very small value. So it asked in part C. To do an interpretation. And if you found one car that had a value that was less than 74.5. That's not unusual. I wouldn't seo look hey the mean is not 75. You guys are wrong. But if you found 20 cars and you got the mean to be less than 74.5 we would say. I don't think 75 is the mean. I don't think so. We would think it's lower than that because this is really really unusual to happen. To get a mean of 20 cars to be this low or lower.


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