Q1. a) If X~N(25, 16), then find the following probabilities. i) P(X > 31) ii) P(16 < X < 34) b) Obtain 95% confidence interval for the population mean if the sample of size 49 gave mean 48 and the standard deviation 2.6
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- ) A random sample of size n = 6 is selected from a normal population, with 12) the sample variance s? = .18. Determine a confidence interval for the population variance o? with a confidence level of 98 percent.•If n=32, x=26.2, a 5.15,a=0.01:answer the following Two questions 021. The confidence interval for the population mean is C) (24.42, 27.56) A) (26.08, 26.32) B) (24.42, 27.98) D) (23.86, 28.54) Q22. The maximum error (the margin of error) of the estimation "E" is: C) 2.34 B) 0.78 A) 1.78 D) 0.62 A -0.637Prof. Gersch knows that the score on a standardized test is perfectly normal with unknown mean and a standard deviation of 50. He then takes a random sample of 25 tests and finds that the average of these is 200. How many exams should he sample so that the maximum margin of errorfor his 95% confidence interval is 10?: a. 98 b. 97 c. 100 d. 96 e. 10
- 113. The life in hours of a 75-watt light bulb is known to be normally distributed with σ = 25 hours. Arandom sample of 20 bulbs has a mean life of 1014 hours. Determine the lower limit of the 99%confidence interval on the mean life.23) The records of 100 postal employees who volunteered for a study, showed that the average time these employees had worked for the postal service was x = standard deviation of the population of time postal service employees have spent with the postal service is approximately 8 years. Assume that we know that the normal with standard deviation o 5 years. A 95% confidence interval for the mean time u the population of postal service employees have spent with the postal service is:
- A random sample of size 64 generated a sample mean of 80. The sample standard deviation is 16. a) Calculate the confidence interval (CI) for 95% confidence level (CL) for the population mean. b) What are the values of CI if CL changed to 90%?8 and 9Q3. A business man want to decide about the hair-cut prices and so he decided to know the mean time spet on hair-cuts. He has no record of time needed for hair-cut. He wanted a max error of +0.2. He took a sample of size 30 and found that s= 1.5 min. Find the sample size needed to achieve the 95% confidence level.
- 4) a. A group of doctors on a particular hospital identified that they attend an average of (15 + 18) patients per hour. Assume the standard deviation is 4. 18 random sample of 36 hours was chosen. Find the 95% confidence interval. b. Find the 90% confidence interval for the variance if a study of (9 + 18) students found the 6.5 years as standard deviation of their ages. Assume the variable is normally distributed.4#solve B plz