### 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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The price of a share of stock divided by the company's estimated future earnings per share is called the P/E ratio. High P/E ratios usually indicate "growth" stocks, or maybe stocks that are simply overpriced. Low P/E ratios indicate "value" stocks or bargain stocks. A random sample of 51 of the largest companies in the United States gave the following P/E ratios†. 11 35 19 13 15 21 40 18 60 72 9 20 29 53 16 26 21 14 21 27 10 12 47 14 33 14 18 17 20 19 13 25 23 27 5 16 8 49 44 20 27 8 19 12 31 67 51 26 19 18 32 (a) Use a calculator with mean and sample standard deviation keys to find the sample mean x and sample standard deviation s. (Round your answers to four decimal places.) x = s = (b) Find a 90% confidence interval for the P/E population mean ? of all large U.S. companies. (Round your answers to one decimal place.) lower limit upper limit (c) Find a 99% confidence interval for the P/E population mean ? of all large U.S.…
A researcher believes that 52.5% of people who grew up as the only child have an IQ score over 100. However, unknown to the researcher, this figure is actually 50%, which is the same as in the general population.To attempt to find evidence for the claim, the researcher is going to take a random sample of 400 people who grew up as the only child. Let p be the proportion of people in the sample with an IQ score above 100. Answer the following: a) Find the mean of p: b)Find the standard deviation of p: c) Compute an approximation for P(p ≥ 0.525), which is the probability that there will be 52.5% or more people with IQ scores over 100 in the sample. Round your answer to four decimal places.
Question 22: A requirement for business graduate school is the GMAT. The GMAT is normally distributed with a mean of 527 and a Standard deviation of 112. The max score on the GMAT is 800. Typically high scorers on the GMAT will get scholarships for graduate school. A score of 720 or higher is a competitive range. Calculate the probability a test taker scores between 720 and 800. (Round to the hundredths)
The shape of the distribution of the time required to get an oil change at a 15-minute oil-change facility is unknown. However, records indicate that the mean time is 16.6 minutes, and the standard deviation is 4.9 minutes. Complete parts (a) through (c). (a) To compute probabilities regarding the sample mean using the normal model, what size sample would be required? O A. Any sample size could be used. B. The sample size needs to be greater than or equal to 30. C. The normal model cannot be used if the shape of the distribution is unknown. D. The sample size needs to be less than or equal to 30. (b) What is the probability that a random sample of n = 40 oil changes results in a sample mean time less than 15 minutes? The probability is approximately: (Round to four decimal places as needed.)
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 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.
Diamond Pricing. In a Singapore edition of Business Times, diamond pricing was explored. The price of a diamond is based on the diamond’s weight, color, and clarity. A simple random sample of 18 one-half-carat diamonds had the following prices, in dollars. 1676 1442 1995 1718 1826 2071 1947 1983 2146 1995 1876 2032 1988 2071 2234 2108 1941 2316 a. Apply the t-interval procedure to these data to find a 90% confidence interval for the mean price of all one-half-carat diamonds. Interpret your result. (Note: — x = $1964.7 and s = $206.5.)b. Obtain a normal probability plot, a boxplot, a histogram, and a stem-and-leaf diagram of the data.c. Based on your graphs from part (b), is it reasonable to apply the t-interval procedure as you did in part (a)? Explain your answer.
Read the following prompt and complete 3.a and 3.b
QUESTION 2 An environmentalist wanted to determine if the mean acidity of rain differed among Ipoh, Johor Bahru and Kuantan. He randomly selected six rain dates at each city and obtained the following data: Table 2 Ipoh 5.10 5.00 4.91 5.12 4.78 5.20 Johor Bahru 4.80 4.15 4.42 4.65 4.87 4.10 Kuantan 5.45 6.23 5.55 5.10 5.42 5.25 Answer the following questions without using any output from any software. a) Construct a complete Analysis of Variance table. b) At <= 0.05, test whether the mean acidity of rain is different among the three cities. c) Based on the answer in (a) and (b), write a brief conclusion on the mean acidity of rain among Ipoh, Johor Bahru and Kuantan.
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.
The Law of Large Numbers states that as the number of observations in a sample increases, the mean of this is sample becomes a. exactly equal to the mean. b. closer and closer to the population mean. c. close to 0. d. further and further from the population mean.
The average number of cavities that thirty-year-old Americans have had in their lifetimes is 4. Do twenty-year-olds have a different number of cavities? The data show the results of a survey of 11 twenty- vear-olds who were asked how many cavities they have had. Assume that the distribution of the population is normal. 2, 5, 3, 2, 5, 3, 5, 2, 2, 2, 4 What can be concluded at the a = 0.05 level of significance? a. For this study, we should use Select an answer b. The null and alternative hypotheses would be: ns Ho: ? vSelect an answer v re H: ? v Select an answer c. The test statistic 2 v = (please show your answer to 3 decimal places.) urse d. The p-value = e. The p-value is ? va f. Based on this, we should Select an answer v the null hypothesis. g. Thus, the final conclusion is that... (Please show your answer to 4 decimal places.) O The data suggest that the population mean number of cavities for twenty-year-olds is not significantly different from 4 at a = 0.05, so there is…

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