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Probability of Sample Mean.

A manufacturer of power tools claims that the average amount of time required to assemble their top-of-the-line table saw is fifty (50) minutes with a standard deviation of forty (40) minutes (the very large standard deviation is due to a variety of factors including a large variation in skills among the 'Do it yourself' home handyman which is traditionally one of the companies customer base). Suppose a random sample of 64 purchasers of this table saw is taken.
What is the probability that the sample mean will be less than 48 minutes __________?
Answer should be to four decimal places which is consistent with the number of decimal places listed in your appendix tables e.g. 0.1234 etc.

The next three questions refer to the following information:
International Pictures is trying to decide how to distribute its new movie 'Claws'. 'Claws' is the story of an animal husbandry experiment at the University of Southern Queensland that goes astray, with tragic results. An effort to breed meatier chickens somehow produces an intelligent, 200Â kilogram chicken that escapes from the lab and terrorises the campus. In a surprise ending the chicken is befriended by coach Tim Galvano, who teaches it how to play Rugby and help his team win State, National and World Championships. Because of the movie's controversial nature, it has the potential to be either a smash hit, a modest success, or a total bomb. International is trying to decide whether to release the picture for general distribution initially or to start out with a 'limited first-run release' at a few selected theatres, followed by general distribution after 3Â months. The company has estimated the following probabilities and conditional profits for 'Claws':
PROFITS (Millions of $)
Level of success Probability Limited release General distribution
Smash .3 22 12
Modest .4 9 8
Bomb .3 -10 -2

International can run sneak previews of 'Claws' to get a better idea of the movies' ultimate level of success. Preview audiences rate movies as either good or excellent. On the basis of past experiences, it was found that 90% of all smash successes were rated excellent (and 10% rated good), 75% of all modest successes were rated excellent (25% rated good) and 40% of all bombs were rated excellent (60% rated good). The cost of running sneak previews is not cheap. Currently, this stands at $1m.
What is the opportunity loss for a General Distribution for a Smash level of success?___________
Answer should be to whole numbers only. You do not need to put any units in your answer.
What is the posterior probability of a bomb given the sneak preview indicates excellent? ___________
Answer should be to four decimal places e.g. 0.1234, 0.2345 etc.
What is the maximum amount that should be paid for the sneak preview (i.e. what would be the expected value of sample information (EVSI))? Select the closest correct answer.
a. $1.04 million
b. $2.58 million
c. $7.2 million
d. $8.24 million

The next two questions refer to the following information:
(All calculations should be to at least three decimal places)
The tourist industry is subject to enormous seasonal variation. A hotel in North Queensland has recorded its occupancy rate for each quarter during the past 5 years. These data are shown in the accompanying table.
Table 1: Occupancy rate
2004 2005 2006 2007 2008
Quarter 1 0.561 0.575 0.594 0.622 0.665
Quarter 2 0.702 0.738 0.738 0.708 0.835
Quarter 3 0.800 0.868 0.729 0.806 0.873
Quarter 4 0.568 0.605 0.600 0.632 0.670
The trend line for this decomposition model has been calculated to be (at 3 decimal places):
Y = 0.650 + 0.004 T where T represents time.
What is the coefficient of determination (R2) for this trend line? (Select the closest correct answer.)
a. 0.0932 (9.32%)
b. 0.3448 (34.48%)
c. 0.4554 (45.54%)
d. 0.7882 (78.82%)
What would be the forecast in Quarter 3, 2009 using the trend line previously given (i.e. Y = 0.650 + 0.004 T) and the relevant adjusted seasonal index?_____________
Answer should be consistent with the data provided and be to three decimal places e.g. 0.123, 0.456 etc.


Solution Summary

The solution examines probability of sample means.