25 Refer to Exhibit 3. What is the probability that a randomly selected individu

Question

25 Refer to Exhibit 3. What is the probability that a randomly selected individual with an will get a starting salary of at least $47,s007 0.4332 0.9332 0.0668 0.5000 b. d. 26. Refer starting salaries of $34,000 to $46,000? 38.49% 38.59% a. b. 50% d. 76.98% 27. Refer to Exhibit 3. Suppose we take a sample of 100 MBAs, the sample average of their starting salary is $50,000. What is the standard error of the mean? when the confidence level is 95%, find za/2 : 28, 29, when the confidence level is 99%, find za/2 30. The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. If a sample of 25 fish yields a mean of 2.8 pounds, what is the Z-score for this sample mean? 31. Continue with the above question, what is the probability that the average weight of these 25 fish is less than 2.8 pounds? Continue with the above question, what is the probability that the average weight of these 25 fish is more than 3.2 pounds? 32. The fact that the sampling distribution of the sample mean can be approximated by a normal probability distribution whenever the sample size is large is based on the a. central limit theorem b. fact that there are tables of areas for the normal distribution 33. c. assumption that the population has a normal distribution d. All of these answers are correct. 34. A population has a mean of 53 and a standard deviation of 21. A sample of 49 observations will be taken. The probability that the sample mean will be less than 57.95 is a. 0 b. 0495 c. 9505 35. A random sample of 225 people was taken from a very large population. The population proportion of females is 0.8·The standard error of the sample proportion of females is

Explanation / Answer

28. When the confidence level is 95% then Za/2 = 1.96

29. When the confidence level is 99% then Za/2 = 2.56

30. Mean = 3.2
Stdev = .8
n=25
z = (SampleMean - Pop.Mean)/(Stdev/sqrt(n)) = (2.8-3.2)/(.8/sqrt(25)) = -2.5

31. P(Z<-2.5) = .00621

32. P(X>3.2) = P(Z> 3.2-3.2/(.8/sqrt(25)) = P(Z>0) = .50

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