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home / study / math / statistics and probability / questions and answers / question 1 the aptitude test scores of applicants ... Your question has been answered! Rate it below. Let us know if you got a helpful answer. Question Question 1 The aptitude test scores of applicants to a university graduate program are normally distributed with mean 500 and standard deviation 60. Applicants need a test score higher than 530 to be admitted into the graduate program. What proportion of applications qualify? [Marks 2] If the university wishes to set the cutoff score for graduate admission so that only the top 10% of applicants qualify for admission, what is the required cutoff score? [Marks 2] What percentage of applicants have test scores within two standard deviations of the mean? [Marks 3] Question 2 Government officials in Canberra have recently expressed concern regarding overruns on military contracts. These unplanned expenditures have been costing Australians millions of dollars every year. The prime minister impanels a committee of experts to estimate the average amount each contract costs the government over and above the amount agreed upon. The committee has already determined that the standard deviation in overruns is $17.5 million, and that they appear normally distributed. i.If a sample of 25 contracts is selected, how likely is it the sample will overestimate the population mean by more than $10 million? [Marks 3] ii.The prime minister will accept an error of $5 million in the estimate of µ. How likely is he to receive an estimate from the committee within the specified range? [Marks 4] Question 3 You have just graduated with a post graduate degree in business and have obtained a position with a large manufacturing firm. The director of marketing has asked you to estimate the mean time required to complete a particular unit of the manufacturing process. A sample of 600 units yields a mean of 7.2 days. Since the population standard deviation is unknown, the sample standard deviation of s=1.9 days must be used. Calculate and interpret the 90 percent interval for the mean completion time for the manufacturing process. If this mean time is estimated to be in excess of 7 days, a new process will be implemented to reduce production costs. [Marks 4] A construction firm was charged with inflating the expense vouchers it files for construction contracts with the federal government. The contract states that a certain type of job should average $1,150. In the interest of time, the directors of only 12 government agencies were called on to enter court testimony regarding the firm’s vouchers. If a mean of $1,275 and a standard deviation of $235 are discovered from testimony, would a 95 percent confidence interval support the firm’s legal case? Assume voucher amounts are normal. [Marks 4]
Explanation / Answer
Question 1 The aptitude test scores of applicants to a university graduate program are normally distributed with mean 500 and standard deviation 60. Applicants need a test score higher than 530 to be admitted into the graduate program. What proportion of applications qualify? [Marks 2]
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 530
u = mean = 500
s = standard deviation = 60
Thus,
z = (x - u) / s = 0.5
Thus, using a table/technology, the right tailed area of this is
P(z > 0.5 ) = 0.308537539 [ANSWER]
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If the university wishes to set the cutoff score for graduate admission so that only the top 10% of applicants qualify for admission, what is the required cutoff score? [Marks 2]
First, we get the z score from the given left tailed area. As
Left tailed area = 1 - 0.10 = 0.9
Then, using table or technology,
z = 1.281551566
As x = u + z * s,
where
u = mean = 500
z = the critical z score = 1.281551566
s = standard deviation = 60
Then
x = critical value = 576.8930939 [ANSWER]
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What percentage of applicants have test scores within two standard deviations of the mean? [Marks 3]
Hence, from 380 to 620.
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 380
x2 = upper bound = 620
u = mean = 500
s = standard deviation = 60
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -2
z2 = upper z score = (x2 - u) / s = 2
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.022750132
P(z < z2) = 0.977249868
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.954499736 = 95.4499736% [ANSWER]
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