### Exploring Oil Change Time DistributionThe shape of the distribution for the time required to get an oil change at a 20-minute oil-change facility is skewed right. Records indicate a mean time of 21.8 minutes with a standard deviation of 4.2 minutes. Let's explore this through the following tasks:#### View Distribution Tables- [Standard normal distribution table (page 1)](URL)- [Standard normal distribution table (page 2)](URL)### Tasks#### (a) Required Sample Size for Normal ModelTo compute probabilities regarding the sample mean using a normal model, consider the following:**Choose the appropriate statement:**- **A.** The sample size needs to be greater than 30.- **B.** The sample size needs to be less than 30.- **C.** Any sample size could be used.- **D.** The normal model cannot be used if the shape of the distribution is skewed right.#### (b) Probability for Mean Time Less than 20 MinutesCalculate the probability that a random sample of $ n = 45 $ oil changes results in a sample mean time less than 20 minutes.- **Probability is approximately:** $ 0.0021 $- *(Round to four decimal places as needed.)*#### (c) Manager's Goal for Employee BonusSuppose each employee can earn a $50 bonus for meeting a certain goal. For a typical Saturday, the facility performs 45 oil changes between 10 A.M. and 12 P.M. Using this as a random sample, determine the mean oil-change time for a 10% chance of being at or below.- **Mean time for a 10% chance:** $ \_ \_ \_ \_ \_ \_ $ minutes- *(Round to one decimal place as needed.)* **Standard Normal Distribution Table (Page 1 and Page 2)**The standard normal distribution table is used to find the probability that a normally distributed random variable Z, with mean 0 and standard deviation 1, is less than or equal to a given value.**Diagrams:**Each page features a bell curve representing the standard normal distribution. The area under the curve to the left of a vertical line at a specified Z value represents the cumulative probability for that Z score.**Table Layout:**- The tables include Z scores in increments of 0.01.- The leftmost column lists the integer and first decimal of Z values.- The topmost row lists the second decimal point of Z values.- The intersecting cells display the cumulative probability.### Page 1#### Z Scores and Probability Values:- **Range: −3.4 to −0.1** - Example: - For Z = −1.5 and 0.06: The probability is 0.0630. - For Z = −2.8 and 0.05: The probability is 0.0026.### Page 2#### Z Scores and Probability Values:- **Range: 0.0 to 3.4**- Example: - For Z = 1.2 and 0.03: The probability is 0.8907. - For Z = 2.5 and 0.08: The probability is 0.9924.**Usage Note:**When using the table, locate the Z score's integer and first decimal in the leftmost column, then find the second decimal in the topmost row. The cell where these intersect provides the cumulative probability.These tables are essential for statistical analysis and hypothesis testing involving normal distributions.

The mean is 21.8 minutes and the standard deviation is 4.2 minutes. Let x be the mean oil-change…

Answered: (c) Suppose the manager agrees to pay…

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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Lab 1.1: Data Collection ExperimentStudent Learning Outcomes• The student will demonstrate the systematic sampling technique.• The student will construct relative frequency tables.• The student will interpret results and their differences from different data groupings.Movie Survey: Below are the number of movies seen in the past month from 48 randomly selected individuals.4 14 7 3 9 3 6 10 1 0 6 3 2 7 11 13 5 7 8 4 5 3 7 9 12 5 1 0 2 8 6 4 5 9 12 10 4976469455771. Order the Data: Complete the two frequency tables below using your class data.Number of Movies Frequency Relative Frequency Cumulative Relative Frequency0 1 2 3 4 5 6 7+Number of Movies Frequency Relative Frequency Cumulative Relative Frequency0–1 2–3 4–5 6–7+Table 1.19 Frequency of Number of Movies Viewed1. Using the tables, find the percent of data that is at most two. Which table did you use and why?2. Using the tables, find the percent of data that is at most three. Which table did you use and why?3. Using the tables,…
Question 4 a) A medical researcher wants to investigate the amount of time it takes for patients' headache pain to be relieved after taking a new prescription painkiller. She wants to estimate the mean of the population of relief times. She believes that the population is normally distributed with a population standard deviation of 25 minutes. How large a sample should she take to estimate the mean time to within 5 minutes with 95% confidence? b) An epidemiologist is worried about the prevalence of the flu in East Vancouver and the potential shortage of vaccines for the area. She will need to provide a recommendation for how to allocate the vaccines appropriately across the city. She takes a simple random sample of 320 people living in East Vancouver and finds that 40 have recently had the flu.Suppose that the epidemiologist wants to re-estimate the population proportion and wishes for her 95% confidence interval to have a margin of error no larger than 0.05. How large a sample should…
A grocer finds that the cantaloupes in his store have weights with mean 0.7 kg and standard deviation 53 grams. If this is true, then 99.7% of cantaloupes will have weights between how many grams?
The duration (in minutes) for a sample of 20 flight-reservation telephone calls is shown in the following table: 2.1 10.4 4.8 5.5 4.5 2.8 5.9 10.5 4.8 3.3 5.8 6.6 7.5 4.8 5.5 3.5 5.3 3.6 7.8 6.0 NOTE: Ex=111 and x = 709.82 Assuming the population has a normal distribution, find a 95% confidence interval for the population mean time.
If the average low temperature of a winter month in Rochester, NY is 18∘18∘ and the standard deviation is 3.4∘3.4∘, then according to Chebyshev's theorem, the percentage of average low temperatures in Rochester, NY between 12.56∘12.56∘ and 23.44∘23.44∘ is at least %.
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).
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.
Copy each table to the Word document and highlight theappropriate p-value. Under each table type the hypotheses, decision and summary, using the resultsfrom Excel, again you should not be doing the calculations by hand! A random sample of 16 students selected from the student body of a large university was selected. We wantto determine if the average age of all the students at the university is significantly different from 24. Assumethe distribution of the population of ages is normal. t-Test: Paired Two Sample for Means Variable 1 Variable 2 Mean 22.8125 0 Variance 10.5625 0 Observations 16 16 Pearson Correlation #DIV/0! Hypothesized Mean Difference 24 df 15 t Stat -1.461538462 P(T<=t) one-tail 0.082249947 t Critical one-tail 1.753050356 P(T<=t) two-tail 0.164499893 t Critical two-tail 2.131449546 I am struggling with the explanation. How would I write this out? Thanks!
2. Problem 3, page 455. Checkout time at a supermarket is monitored using a mean and range chart. Six samples of n=20 observation have been obtained and the sample means and ranges computed: Sample Mean Range Sample Mean Range 1 3.06 0.42 4 3.13 0.46 2 3.15 0.5 3.06 0.46 3 3.11 0.41 3.09 0.45 a. Using the factors in the table, determine UCL and LCL for mean & range chart. b. Is the process under control? c. (Not in the book) If a new sample is collected and the mean is 3.02 and the range is 0.5. Is the process under control?
QUESTION 8 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 total? Please do the calculations on the handout provided.
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.
2. Ms. Palac, a sociologist, made a study on the age in years of a woman at the time of her first marriage. She randomly selected a sample of 24 and 21 married women from Village A and Village B, respectively. The following data are the mean and standard deviation for each group: Village A: x = 18.63 and s = 5.54 Village B: x = 24.90 and s = 5.54 Test the hypothesis that the women in Village A get married younger than women in Village B. Use 0.10 significance level. Step 1: Null and Alternative Hypothesis Ho: Ha: Step 2: Level of Significance and appropriate test statistic Step 3: Determine the Critical Region Degrees of Freedom (df): Critical Region: Step 4: Compute the z or t value Computations: Step 5: Decide whether to accept or reject Ho Step 6: Conclusion

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