Statistics calculated from a sample of 17 observations are:\[\sum_{i=1}^{17} x_i = 1277\]\[\sum_{i=1}^{17} x_i^2 = 103225\]Provide the following, rounding your answers to two decimal places.(a) Find the sample mean. [Input Box](b) What is the sample standard deviation? [Input Box](c) Assume that the population distribution is normal. Find a 95% confidence interval for the population mean. (\[ \, \], \[ \, \]) [Input Boxes] **The waiting time to get food at a fast-food restaurant is 4.5 minutes on average with a standard deviation of 2 minutes. Suppose that a sample of 45 customers is randomly selected.**---**A. What are the mean and standard deviation of the sample mean waiting time of 45 customers?**- **Mean =** 4.5 minutes- **Standard Deviation** (standard error of the mean) = 0.29814 minutes - *Round standard error to 5 decimal places.*---**B. What is the probability that the average waiting time of 45 customers is less than 4.3 minutes?**- **Answer:** - *Round to 4 decimal places.*---**C. What is the probability that the average waiting time of 45 customers is greater than 4.7 minutes?**- **Answer:** - *Round to 4 decimal places.*---**D. What is the probability that the average waiting time of 45 customers is between 4.3 and 4.7 minutes?**- **Answer:** - *Round to 4 decimal places.*---There are no graphs or diagrams in this document. The content focuses on calculations related to the statistics of waiting times in a fast-food restaurant context.

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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 grade point averages (GPA) of 15 randomly selected college students are 2.3, 3.3, 2.6, 1.8, 0.2, 3.1, 4.0, 0.7, 2.3, 2.0, 3.1, 3.4, 1.3, 2.6, 2.6 1. The sample mean is (Round to one decimal place) 2. The sample standard deviation is (Round to one decimal place) 3. The sample size is 4. The degree of freedom for the Studentized version of the sample mean is 5. The point estimate for the mean GPA of ALL students at this college is 6. The t-value for a 99% confidence interval is t = 7. The Margin of error for a 99% confidence interval for this sample is (Round to one decimal place) O Th 00 .0/ ASUS VivoBaok EA home 19 end f10 f11 Pgup f12 Pgdn & 3 6. 7. 8 9.
Claim: The standard deviation of pulse rates of adult males is less than 12 bpm. For a random sample of 144 adult males, the pulse rates have a standard deviation of 10.8 bpm. Find the value of the test statistic. The value of the test statistic is ???
Pulse Rates of Males Refer to Data Set 1 “Body Data” in Appendix B and use the pulse rates of males. a. Find the mean and standard deviation, and verify that the pulse rates have a distribution that is roughly normal. b. Treating the unrounded values of the mean and standard deviation as parameters, and assuming that male pulse rates are normally distributed, find the pulse rate separating the lowest 2.5% and the pulse rate separating the highest 2.5%. These values could be helpful when physicians try to determine whether pulse rates are significantly low or significantly high.
Calculate the SEM (standard error of the mean) for the following data:a. Population standard deviation = 15b. Sample size = 25
1.
You work for a soft-drink company in the quality control division. You are interested in the standard deviation of one of your production lines as a measure of consistency. The product is intended to have a mean of 12 ounces, and your team would like the standard deviation to be as low as possible. You gather a random sample of 17 containers. Estimate the population standard deviation at a 95% level of confidence. 11.94 11.99 11.97 11.95 11.85 12.1 12.07 12.16 12.13 12.1 12.01 12.1 11.95 12 11.93 11.96 12.1 (Data checksum: 204.31) Note: Keep as many decimals as possible while making these calculations. If possible, keep all answers exact by storing answers as variables on your calculator or computer. Round final answer to 4 decimal places. a) Find the sample standard deviation: b) Find the lower and upper x² critical values at 95% confidence: Lower: Upper: c) Report your confidence interval for o: (
3565 6781 11366 5656 10024 9101 9286 10376 7290 15558 6657 20795 4027 19100 16223 14502 2830 5785 9070 7560 D) Calculate the mean and standard deviation of your data set. You may use technology such as Google Sheets, Excel, or a TI-83/TI-84. Round to the nearest hundredth. Mean 9777.6 Standard Deviation 5090.73 E) Using a significance level a = 0.05, find the critical value using the following steps. 1. What are the degrees of freedom? n D20 d. f.= n- 1=19 = 19 2. Will this be a one-tailed or two tailed test? two-tailed 3. Using table 5 from the back of your textbook (page A18 in Appendix B), what is the critical value? 1.729
Given Samples A and B below Sample A: 2.0 Sample B: 4.3 2.8 4.5 O a. Calculate the mean and standard deviation for each sample. Sample A: XA = Sample B: XB = ; SB = Round to two decimal places if necessary b. Calculate the coefficient of variation for each sample. Sample A: CVA = ; SA = _% 4.2 4.2 4.3 4.0 Sample B: CVB Round to two decimal places if necessary c. Which sample is more variable? o Sample A Sample B Neither sample is more variable than the other 4.4 4.3 3.7 4.9 _% 3.6 2.6 2.8 2.2 2.3 4.1
Determine the standard error. Complete solution
Assume that intelligence follows a normal distribution and that on a standard test for intelligence the true mean (μ) is 100 with a standard deviation (σ) of 15. What test score do you need for your intelligence to be in the top 1% of the population?
Consider the following sample data. Sample A: 11, 22, 33 Sample B: 81, 92, 103 Sample C: 1,100; 1,111; 1,122 (a) Find the mean and standard deviation for each sample. Sample A: Sample B: Sample C: Mean Sample Standard Deviation (b) What does this exercise show about the standard deviation? multiple choice The idea is to illustrate that the standard deviation is not a function of the value of the mean. The idea is to illustrate that the standard deviation is a function of the value of the mean.
You are interested in estimating the the mean weight of the local adult population of female white-tailed deer (doe). From past data, you estimate that the standard deviation of all adult female white-tailed deer in this region to be 17 pounds. What sample size would you need to in order to estimate the mean weight of all female white-tailed deer, with a 96% confidence level, to within 7 pounds of the actual weight? Sample Size:

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