Question 6 of 6- /2 EView PoliciesCurrent Attempt in ProgressFast Auto Service provides oil and lube service for cars. It is known that the mean time taken for oil and lube service at this garage is 15minutes per car and the standard deviation is 2.4 minutes. The management wants to promote the business by guaranteeing amaximum waiting time for its customers. If a customer's car is not serviced within that period, the customer will receive a 50 %discount on the charges. The company wants to limit this discount to at most 4% of the customers. What should the maximumguaranteed waiting time be? Assume that the times taken for oil and lube service for all cars have a normal distribution.Round your answer to the nearest minute.The maximum guaranteed waiting time should be approximately iminutes.eTextbook and MediaSave for LaterAttempts: 0 of 5 usedSubmit Answer50e Type here to search

X: time taken for oil and lube service for al cars. X~N(15,2.4)

Answered: Fast Auto Service provides oil and lube…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Similar questions

Two friends, Karen and Jodi, work different shifts for the same ambulance service. They wonder if the different shifts average different numbers of calls. Looking at past records, Karen determines from a random sample of 32 shifts that she had a mean of 4.8 calls per shift. She knows that the population standard deviation for her shift is 1.4 calls. Jodi calculates from a random sample of 40 shifts that her mean was 4.2 calls per shift. She knows that the population standard deviation for her shift is 1.1 calls. Test the claim that there is a difference between the mean numbers of calls for the two shifts at the 0.01 level of significance. Let Karen's shifts be Population 1 and let Jodi's shifts be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.
The Michelin tires in average will last for 60000 miles. A technician claims that the deviation of mileage for wearied off tires is 25000 miles. If 16 such tires are tested, find the probability that for those tires the average wearied off mileage is between 55000 and 65000 miles.
Kathy and her brother Clay recently ran in a local marathon. The distribution of finishing timefor women was approximately normal with mean 259 minutes and standard deviation 32minutes. The distribution of finishing time for men was approximately normal with mean 242minutes and standard deviation 29 minutes. (a) The finishing time for Clay was 289 minutes. Calculate and interpret the standardized scorefor Clay’s marathon time. Show your work. (b) The finishing time for Kathy was 272 minutes. What proportion of women who ran themarathon had a finishing time less than Kathy’s? Show your work. (c) The standard deviation of finishing time is greater for women than for men. What does thisindicate about the finishing times of the women who ran the marathon compared to thefinishing times of the men who ran the marathon?
Jack's clothing sells on average, 25 shirts per week, sometimes a little more, sometimes a little less but typically a deviation of about 3. The cost per jacket is $65. Because Jack's is located in a sketchy part of town their insurance on inventory is 33% on all goods. The good news is free shipping to customers although UPS charges Jack's $8.50 for every delivery. It takes 2 weeks from the time jackets are ordered until they are received. Jack's supplier has said they only ship by pallet, and each pallet contains 75 jackets. If the reorder point is 50 what is the average flow time?
Greg is in his first semester at a university and is taking a calculus course with a large enrollment. He just took the first midterm exam and is nervous about his score. Among all the students in the course, the mean of the exam was 85 with a standard deviation of 15.Find that grag has obtain less then 95.
The chart to the right shows that a professor's grading distribution is bell shaped, or normally shaped. The mean of the distribution is 74 and the standard deviation is 9. Using a Continuous Variable Normal Bell Distribution Model, calculate the minimum test score needed to score in the top 5% of the class. Complete your work in the worksheet by listing the formula inputs, labels for the formula inputs and make your calculations with formulas. This test problem is similar to what you studied in video # 32, 33 and 34 and homework problems # 17, 21 and 23. Relative Frequency Frequency Professor looks at all test score for a particular test (this is population data), and observes: 0.5 0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 Mean = 74 Median = 74 Mode = 73 SD=9 0 0 0 up to 10 up 10 to 20 0 20 up to 30 0 30 up to 40 2 40 up to 50 29 50 up to 60 X = Score 112 60 up to 70 225 70 up to 80 111 80 up to 90 21 2 90 up 100 up to 100 to 110
You get hired as a server at a local restaurant, and the manage tells you that servers tips are $42 on average but vary about $12 (mean =42, standard deviation =12). You decide to track your tips to see if you make a different amount, but because this is your first job as a server, you don't know if you will make more or less in tips. After working 16 shifts, you find that your average mightly amount is $44.50 from tips test for a difference between this value and the population mean at the a=0.05 level of significance.
Let's say that the mean score on this exam turns out to be 73 with a standard deviation of 12 and is normally distributed. If I gave a B to everyone who had an 80 or above, what percent of the class would get B's (or better)?
According to the New York Daily News, March 5, 2014, people 65 years and older watch an average of 50.5 hours of TV per week. We will estimate the standard deviation to be 12.7 hours per week. Your town has many social programs for senior citizens which may reduce the amount of TV that older people watch. You want to find out if it is true that people 65 and older in your town watch less than 50.5 hours of TV per week. So, you sample 40 people 65 and older in your town. You find that the average time that the 40 people watch per week is 45.8 hours. Can the claim be supported to a level of significance of α = .02, test the hypothesis?
Two friends, Karen and Jodi, work different shifts for the same ambulance service. They wonder if the different shifts average different numbers of calls. Looking at past records, Karen determines from a random sample of 40 shifts that she had a mean of 5.1 calls per shift. She knows that the population standard deviation for her shift is 1.2 calls. Jodi calculates from a random sample of 31 shifts that her mean was 5.8 calls per shift. She knows that the population standard deviation for her shift is 1.4 calls. Test the claim that there is a difference between the mean numbers of calls for the two shifts at the 0.01 level of significance. Let Karen's shifts be Population 1 and let Jodi's shifts be Population 2. Step 3 of 3 : Draw a conclusion and interpret the decision.
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