**Exhibit 8-3**A random sample of 81 automobiles traveling on a section of an interstate showed an average speed of 60 mph. The distribution of speeds of all cars on this section of highway is normally distributed, with a standard deviation of 13.5 mph.4. Refer to Exhibit 8-3. The 86.9% confidence interval for μ is _____.- ○ 46.500 to 73.500- ○ 57.735 to 62.265- ○ 59.131 to 60.869- ○ 50 to 70**Explanation:**The image contains a question related to the computation of an 86.9% confidence interval for the mean speed (μ) of automobiles on a section of an interstate. Using statistical concepts, it seeks the interval estimate between which the true average speed of vehicles will fall 86.9% of the time.The question presents four potential ranges for this confidence interval. Determining the correct option involves statistical calculations using the given sample mean, standard deviation, and sample size.

4. The given sample size is 81, sample mean is 60 mph, and population standard deviation is 13.5…

Answered: 4. Refer to Exhibit 8-3. The 86.9%…

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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A sample of 9 adult elephants had an average weight of 12,625 pounds. The standard deviation for the sample was 24 pounds. Find the 90% confidence interval of the population mean for the weights of adult elephants. Assume the variable is normally distributed. Round intermediate answers to at least three decimal places. Round your final answers to the nearest whole number.
You measure the lifetime of a random sample of 64 tires of a certain brand. The sample mean is = 50 months. Suppose that the lifetimes for tires of this brand follow a Normal distribution, with unknown mean and standard deviation = 3.79 months A 99% confidence interval for is: a. 40.2 to 59.8. b. 48.39 to 51.61. c. 48.78 to 51.22. d. 49.8 to 50.2.
Step 3 The necessary value for Za/2 was determined to be 1.645. Recall the given information. Sample 1 n1 = 400 P1 = 0.56 P2 = 0.41 Substitute these values to first find the lower bound for the confidence interval, rounding the result to four decimal places. P₁(1-P₁) P₂(1-P₂) 1 72 lower bound = P₁-P₂-²a/2 = 0.560.41 Sample 2 n2 = 300 = 0.0559 = 0.56 0.41 = 02441 X Substitute these values to find the upper bound for the confidence interval, rounding the result to four decimal places. P₁(1-P₁) P₂(1-P₂) upper bound = P₁ P₂ + ²a/2√ n1 n2 X + ✔ - 1.645, + 0.56(1 0.56) 0.41(10.41) 300 +1.645. 400 0.56(1 0.56) + 400 + 0.41(1 0.41) 300
A researcher is interested in finding a 95% confidence interval for the mean number minutes students are concentrating on their professor during a one hour statistics lecture. The study included 133 students who averaged 36.1 minutes concentrating on their professor during the hour lecture. The standard deviation was 13.4 minutes. Round answers to 3 decimal places where possible. b. With 95% confidence the population mean minutes of concentration is between and minutes. c. If many groups of 133 randomly selected members are studied, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population mean minutes of concentration and about percent will not contain the true population mean minutes of concentration.
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4. Refer to Exhibit 8-3. The 86.9% confidence interval for µ is 46.500 to 73.500 57.735 to 62.265 59.131 to 60.869 50 to 70