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Homework: Chapter Six Spring 2019 Save Score: 0.2 of 1 pt 2 of 8 (8 complete) HW Score: 70.73%, 5.66 of 8 p 6.210-T Question HelpA study discovered that Americans c...

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Homework: Chapter Six Spring 2019 Save Score: 0.2 of 1 pt 2 of 8 (8 complete) HW Score: 70.73%, 5.66 of 8 p 6.210-T Question HelpA study discovered that Americans consumed an average pf 13.2 pounds of chocolate per year Assume that the annual chocolate consumption follows the normal distribution with a standard deviation of 3.4 pounds_Complete parts a through e below a. What is the probability that a American will consume less than 9 pounds of chocolate next year?1075 (Round to four decimal plac

Homework: Chapter Six Spring 2019 Save Score: 0.2 of 1 pt 2 of 8 (8 complete) HW Score: 70.73%, 5.66 of 8 p 6.210-T Question Help A study discovered that Americans consumed an average pf 13.2 pounds of chocolate per year Assume that the annual chocolate consumption follows the normal distribution with a standard deviation of 3.4 pounds_Complete parts a through e below a. What is the probability that a American will consume less than 9 pounds of chocolate next year? 1075 (Round to four decimal places as needed ) b. What is the probability that an American will consume more than 11 pounds of chocolate next_ year? 7422 (Round to four decimal places as needed American will consume between 10 and 14 pounds of chocolate next_ 'year? c. What is the probability that an (Round to four decimal places as needed:) osof ge



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The number of chocolate chips in an 18 -ounce bag of Chips Ahoy! chocolate chip cookies is approximately normally distributed with a mean of 1262 chips and standard deviation 118 chips according to a study by cadets of the U.S. Air Force Academy. (Source: Brad Warner and Jim Rutledge, Chance, Vol. 12, No. $1,1999,$ pp. $10-14 .$ ) (a) What is the probability that a randomly selected 18 ounce bag of Chips Ahoy! cookies contains between 1000 and 1400 chocolate chips? (b) What is the probability that a randomly selected 18 ounce bag of Chips Ahoy! cookies contains fewer than 1000 chocolate chips? (c) What proportion of 18 -ounce bags of Chip Ahoy! cookies contains more than 1200 chocolate chips? (d) What proportion of 18 -ounce bags of Chip Ahoy! cookies contains fewer than 1125 chocolate chips? (e) What is the percentile rank of an 18 -ounce bag of Chip Ahoy! cookies that contains 1475 chocolate chips? (f) What is the percentile rank of an 18-ounce bag of Chip Ahoy! cookies that contains 1050 chocolate chips?

Okay. So in this case, the number of chocolate chip cookies is approximately normally distributed with a mean of 1 to 6 to with men of 1262 Okay, with the mean of 1 to 6 to. And what is the standard deviation? 118 sigma is 118 Yeah. What is the probability that a randomly selected bag contains between 1000 and 1400 contains between 2000 is somewhere around here and 1400 somewhere around here. So what I can do is I know the overall area under the normal curve. When I find the Z statistics for 1000 and 1400 right? The rural area under the normal curve will be one. So what I can do is I can find a P value for 1000, which will be the area and the stale P value for 1400 which will be an area in this stage. And I can subtract both of those areas from one. Right. So this is 1000. This is 1400. So let's just look at this. We are using the Z statistic where X minus mu by sigma gives me the value of C. Okay, this is part of a for X is equal to 1000. What will be the Z statistic for 1000? Italy, 1000 minus 1 to 6 to if I use a calculator. Okay, so this is 1000 minus 1262 divided by 118 which is Sigma. So this is minus 2.2. This is minus 2.2. Now, how do I find a probability for this? The probability for the number of chocolate cookies or a number of the cookies in the bag being less than 1000. It will be given by the P Value for this. How do you find the P value? We can use a set statistic table, A Z table, or and online to a lot of statistical package like Excel. So this is what I'm doing here. I put in the value of 2.22 03 I had enter and my P values 0.13 So this is 0.13 Right. So this area is zero point zero and three. We are going to do the same thing for 1400. So what is the Z statistic for 1400? This is going to be 1400 minus one, 262 and divided by 118 which is 1.1694 This is 1.1694 What is the corresponding P value? 1.1694 Okay, I put in this value. This is 1.1694 they had entered. And what the value is 0.1 to 11 This is 0.1211 So this area is 0.1211 Now, I want the area in between these two. That is this particular area. So I'm going to subtract both of these p values from one. This is going to be one minus 0.13 minus 0.1211 All right, one minus 0.0 and three. Minus zero point 1211 This is 12110.8659 So what is the probability? 0.8659 This is the probability that I want probability that this is between mhm 1000 and 1400. Okay? This is the probability. Let's make this a little neat. Okay? Okay. This is the probability. Now, both of these are inclusive. They say it doesn't really matter much, right? This is the beauty of the set table. Whether they're inclusive or exclusive, the probability will remain the same. Moving on to Part D. What is the probability that it contains fewer than 1000 chocolate chips? Now, we have already calculated the P value for 1000. Right? Which is this 0.13 So what is the probability that are Ex has less than 1000 chips? It is going to be 0.13 Okay, then moving on to part. See? Yeah. Uh, what proportion contains more than 1200 chocolate chips? Okay, so what we want is more than 1200 which means that if I look at the normal curve, this is 1262 right. 1262 This is the mean. We want more than to learn the government. It will be somewhere around here, right? More than means we want this area. So what I can do is I can find the p value for 1200 which will be this lesser area area and the tail, and I can subtract it from one. So let's find the sadistic. For 1200. This will be 1200 minus 1262 divided by 118 which is minus 1180.5 to 54 Negative 0.5254 And what is the P value for this? Okay, negative 0.5 to 54 I hit. Enter. This is 0.2996 point 2996.2996 Now I have to subtract this from one because I want the area on the bigger side. Right on the larger side, this side. So this is going to be one minus 10.2996 or this can be point 704 4.7004 Let's find out 0.7004 actually 0.7004 If so, this is the probability that the number of chips is greater than 1200. Okay, now moving on the part D What proportion of bags contain fewer fewer than 1125 ships contain fewer than 1125 Let's find the Z value for 1125 This is going to be 1125 minus 1262 divided by 118 which is minus 1.16 minus 1.16 Now, what is the corresponding P value? Minus 1.1 six minus 1.16 we had entered. This is 0.1230 Okay, so corresponding P value is 0.1230 What does this say? This is that probability that the number of chips is less than 1125 is 0.1230 Okay, moving on apart. Eat. What is the percentile rank for? 1475 chocolate chips. It was a percentile rank for 1475 chocolate chips. So let's find the Z 24 1475 What is this going to be? 1475 minus 1262 Divided by 118 which is 1.8 1.805 Okay, 1.805 means it is somewhere around here. So what I have to do is I have to find the p value is subtracted from one which will give me this. This is the side of the person time that we want. So what is the P value for this? 1.805 1.8. Do five hit. Enter. So this is 0.355 point 0355 This is the P value, this shaded region, and we also supply this from one. So this is one minus 0.355 So this is 0.9645 8.96 for five. Or I can say that this is 96% time. So this is 96 percentile. Okay, How about back if what is the percentile that contains 1050 chocolate chips? So what we are going to do is something similar against their value for 1050 which happens to be 1050 minus 1262 divided by 118 This is minus 1.7966 minus 1.7966 I will again find the P value for this one point 7 19 6. This is 0.362 0.362 So this is this is 0.36 to 0.362 So I would say that this is the fourth percentile, right? This happens to be the fourth was indicted. These are our answers.

In this problem. We're looking at an instructor who has a class of students and is grading papers from the information that the problem provides. We know that these stand simple size is 40 students and that the mean grading time for each paper is six minutes, and the standard deviation for the grading time is also six minutes. Because the sample size 40 is greater than 30 we know that the central limit theorem also applies, and that means that the total time spent creating the papers, let's call it Big T is an approximately normal distribution. And because of that, we can then calculate the mean and the standard deviation for the total time distribution, which will be, well, Call me a T, and if I t. This is based on the sample size mean and standard deviation, so the equations will be and times the sample mean and the square root of n times the sample standard deviation, which, when you plug in the numbers, comes out to be 240 minutes, along with approximately a standard deviation of 37.9 five. So for the first part of the problem, we are trying to determine the probability that in grading all these papers and starting at 6 50 um, he will be able to watch his show, which will be at 11 o'clock if we transferred this or Trans, convert this to numbers or convert this two minutes. We are essentially trying to find the probability that the total time spent grading t is going to be less than 250 minutes, which is the time between 6:50:11 p.m. Which is when he would be grading and before the TV show starts. And using the central limit theorem and Z a standard normal random variable. We can then approximate this probability to be the probability that Z is going to be less than T minus the total meat divided by the total standard deviation. And once we plug in the numbers, these numbers, um, we have the probability of Z being less than 250 minus 2 40 which is the mean, um, time spent creating total time spent grading and 30 divided by 37.9 five. This comes out to be zero 0.2 six cool, and if we refer to the Z score table that the probability that we get from there is zero point 60 to 6, or approximately 60 0.3%. This means that the probability that the lecture or the instructor is going to be able to finish grading before his show starts is approximately 60.3% for the second problem. We are looking at the probability that he will miss time or essentially miss part of his show, and from the second part tells us that the show starts at 11 10 PM rather than 11 p.m. And based on that and converted to numbers, we are trying to find the probability that T will be greater than 260 minutes, which is the time between 6 50 11 10 PM, which then approximates to the probability that Z a random, a standard normal random variable will be greater than T minus mut divided by by T and plugged in that we get probability that Z is less sorry. Greater than 260 minus 2 40 divided by 37 0.95 Since we're looking at greater than we are looking at the right side of the normal distribution, which means that with respect to disease score, we're going to have one minus the probability for that Z score which calculated out, which is this value is going to be zero point five two seven and looking to the Z score table and looking at 0.53 on the Z score table, we find that this equates to one minus 0.7019 which comes out to be 0.2 98 one, which means that there is approximately a 29.8% chance that the instructor will miss part of his show while grading all of his papers.

In this problem. We are looking at an instructor who has a class of students and it's creating papers from the problem. We know that the sample size is 40 students so and equals 40. And we also know that the mean grading time spent on each paper or mu is six minutes. Finally, we know that the standard deviation for the mean creating time is also six minutes. Because the sample size 40 is larger than a 30. We know that the central limit theorem is applicable, which means that the total time spent creating let's call it Big T is has an approximately normal distribution. And from that we also can find the mean and the standard deviation for that for big T, which is, um, you t and fi t these are not. These are directly related to the sample means and standard sample standard deviations, and we find them using the equations for the mean total time spent grading being the samples, times the sample mean and for the standard deviation. We have the square root of these number of samples times the sample standard deviation, and when we calculate these out, he's come out to be 240 minutes for the sample. Ah, for the mean total Times bank rating and we also have approximately 37.95 for the standard deviation of the time spent total time spent reading in the first part of the problem. We're trying to determine the probability that creating 40 papers the instructor will be done by 11 PM for the news and translating this two minutes in actual numbers were essentially trying to figure out the probability that the total time spent grading or big T is going to be less than 250 minutes, which is the time before 6 50 between 6, 50 and 11 o'clock, which is the time that ah, he starts grading and the time that he hopes to be done grading and using the central Limit theorem. And with Z being a standard normal random variable, we can approximate this probability to the probability that Z is less than the target time where the probability of time that we're examining, which is 2 50 minus the total time spent ratings mean divided by the standard deviation and using the numbers that we have, we get the probability that Z it's less than yeah, zero point 264 or which then we can use the Z score to figure out the corresponding probability and from the Z score table at the end of the book, we can find out that this is the score of 0.264 is correlates with a probability of 0.602 six, which means that there is approximately a 60.3% chance that the instructor will be able to finish grading before his show for the second part of the problem. Similarly, we're trying to determine the probability that he will actually miss part of the show, and we find out that the specific segment of the show he wants to watch starts at 11 10. So translated to numbers, we are looking at the probability that big T, the total time spent grading is going to be greater than 260 minutes and similarly to the previous part. We use the same equation with Z being an approximately normal random variable, but we replace the number that we were looking at 2 50 with 260 minutes instead which is the probability that the is going to be greater than 0.5 to 7. Since we're looking at the right side of the distribution, we have to subtract the probability from one so trans this translates to one minus the probability associated with disease score of zero point five 27 When we refer to disease Score table, we then find out that this becomes one minus 0.7 oh 19 which is 0.2981 or approximately 29 point heat percent, which means that the instructor has a 29.8% chance of missing his at least some time of his part of the show that he wants to watch.


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