Suppose the manager agrees to pay each employee a $50 bonus if they meet a certain goal. On a typical Saturday, the oil change facility will perform 45 oil changes between 10 Am and 12 PM. Treating this as a random sample, there would be a 10% chance of the mean oil-change time being at or below what value? (a) To compute probabilities regarding the sample mean using the normal model, what size sample would be required?OA. Any sample size could be used.OB. The sample size needs to be greater than or equal to 30.OC. The sample size needs to be less than or equal to 30.OD. The normal model cannot be used if the shape of the distribution is skewed right.(b) What is the probability that a random sample of n = 45 oil changes results in a sample mean time less than 15 minutes?The probability is approximately 0.0109(Round to four decimal places as needed.)(c) Suppose the manager agrees to pay each employee a $50 bonus if they meet a certain goal. On a typical Saturday, the oil-change facility will perform 45 oil changes between 10 A.M. and 12P.M. Treating this as a random sample, there would be a 10% chance of the mean oil-change time being at or below what value? This will be the goal established by the manager.There is a 10% chance of being at or below a mean oil-change time of minutes.(Round to one, decimal place as needed.)2000DO00F4F51HMacBook AirF6F7Vi L.▶11F8MoreF9Time Remaining: 01:10:09 Submit quizF10X

Central Limit Theorem for mean:If a random sample of size n is taken from a population having mean μ…

Suppose the manager agrees to pay each employee a $50 bonus if they meet a certain goal. On a typical Saturday, the oil change facility will perform 45 oil changes between 10 Am and 12 PM. Treating this as a random sample, there would be a 10% chance of the mean oil-change time being at or below what value?

Suppose the manager agrees to pay each employee a $50 bonus if they meet a certain goal. On a typical Saturday, the oil change facility will perform 45 oil changes between 10 Am and 12 PM. Treating this as a random sample, there would be a 10% chance of the mean oil-change time being at or below what value?

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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QUESTION 13 You are interested in seeing whether emotions impact decision making. You have three groups--a happy group, a sad group, and a neutral group. For the happy group there are 5 participants and the mean risky decision making score 5.0 with a standard deviation of 0.7. For the sad group there are 5 participants and the mean risky decision making score 5.4 with a standard deviation of 1.1. For the neutral group there are 5 participants and the mean risky decision making score 5.8 with a standard deviation of 2.8. The sum of squares between samples is equal to 1.6. The sum of squares within samples is equal to 38.0. You calculated the f-ratio above. The corresponding p-value for the f-ratio you calculated is p = 0.78. Do you accept or reject the null hypothesis? O You accept the null hypothesis because your p-value is greater than .05 and not significant. O You reject the null hypothesis because your p-value is greater than .05 and significant. O You accept the null hypothesis…
Question 8 Consider this scenario: A cereal-packing company where the content of cereal boxes must have a mean of 368 grams to satisfy the labeling on the box. Because it is expensive to weigh every box produced at the factory, a sample of boxes is used to estimate the actual mean. After analyzing the data sample, the operations manager must decide whether or not the manufacturing process is working as expected. Suppose a sample of 49 boxes is taken from thousands produced during an 8-hour shift and the standard deviation (σ) of the cereal-filling process is 15, what is the standard error of the mean? Use the equation, σX̅ = σ /(square-root of n), where n is the sample size. Enter your result as a value with one digit after the decimal point (e.g. 1.7).
Question three John and his friend were recently having a heated discussion about the number of men living in a western parish who possibly cared for children that are not biologically theirs during 2019. Assume that a survey of the parish’s population revealed that the average number of cases and the standard deviation was 5,000 and 6000 in 2019, respectively. John and his friend do not believe that the population average is correct. Assume that a survey based on a sample of the 800 people was conducted and found that the average number of cases in the western parish was 10,000 over the 12 months period. Conduct a hypothesis test, using a 5% significance level to determine if the average number of cases is actually higher based on the sample data.
EXPERIMENT 1: A researcher believes that on average women weigh less today than during the renaissance period around the 15th to 16th centuries. To investigate this belief, a random sample of 41 adult wômen is chosen and their weights are recorded. The mean weight of the sample is 111 pounds with a standard deviation of 12.4 pounds. Based on historical records and paintings of women from the renaissance, it is determined that the average weight of an adult women from that period would have been about 115 pounds.
Listed below are time intervals (min) between eruptions of a geyser. The "recent" times are within the past few years, and the "past" times are from 1995. Does it appear that the variation of the times between eruptions has changed? Use a 0.05 significance level. Assume that the two populations are normally distributed.
Question 4
QUESTION 7 You are interested in seeing whether emotions impact decision making. You have three groups--a happy group, a sad group, and a neutral group. For the happy group there are 5 participants and the mean risky decision making score 5.0 with a standard deviation of 0.7. For the sad group there are 5 participants and the mean risky decision making score 5.4 with a standard deviation of 1.1. For the neutral group there are 5 participants and the mean risky decision making score 5.8 with a standard deviation of 2.8. The sum of squares between samples is equal to 1.6. The sum of squares within samples is equal to 38.0. What is the degrees of freedom within samples? Please do the calculations on the handout provided.
QUESTION 9 The length of an adult Great White Shark is normally distributed with a mean of 3.4 meters and a standard deviation of 0.5 meters. What is the probability that a randomly selected adult Great White is more than 4.7 meters long? Click Save and Submit to save and submit. Click Save All Answers to save all answers.
QUESTION 6 The Corner Drug Store fills prescriptions for a popular children's antibiotic, amoxicillin. The daily demand for amoxicillin is normally distributed, with a mean of 200 kilograms and a standard deviation of 80 kilograms. The vendor for the pharmaceutical firm that supplies the drug calls the drugstore pharmacist every 30 days to check the inventory of amoxicillin. During a call, the druggist indicated that the store had 60 kilograms of the antibiotic in stock. The lead time to receive an order is 4 days. a) Determine the order size that will enable the drugstore to maintain a 95% service level.
Question 1 Suppose a waitress keeps track of her tips, as a percentage of the bill, for a random sample of 60 tables. Suppose that, unknown to the waitress, her population mean tip percentage is 14%, with a standard deviation of 3%. c. Describe the sampling distribution of her sample mean tip percentage. Round to 3 decimal places, when necessary. The sampling distribution has a mean of ___% and a standard deviation of ___% Answer for blank # 1: 14 Answer for blank # 2: 0.387 d. Between what two values is there a 68% chance that her sample mean tip percentage will fall? Enter your answers in increasing order, rounded to two decimal places. There is a 68% chance that her sample mean tip percentage will fall between ___% and ___% Answer for blank # 1: Answer for blank # 2
question 2b. The annual yield (expressed as a percentage) for all investments in a particular asset class in the latest financial year was normally distributed with a mean of 4.5% and a standard deviation of 1.25%.i. Represent the probability that a randomly selected investment in this class will have an annual yield greater than 6.00%on a diagram and determine this probability. ii. Represent the value for annual yield that only 10% of investments in this class exceed on a diagram and determine this value.
QUESTION 18 Find the z score that corresponds to P90. Show all work and upload it as a photo or pdf. For full marks, include a diagram of the distribution that indicates important points, regions, etc.