BIOS 516D Final Fall 2009

BIOS 516D Final Fall 2009 - BIOS 516D Final Exam, Fall 2009...

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BIOS 516D Final Exam, Fall 2009 Name: 1. The mean of the normal distribution above is 1. 0 2. 20 3. 40 4. 60 5. 80 2. The approximate standard deviation of this normal distribution is 1. 0 2. 20 3. 40 4. 60 5. 80 3. The total cholesterol values for a certain population are approximately normally distributed with a mean of 100 mg/100 ml and a standard deviation of 20 mg/100 ml. Without using a table or a calculator, find the probability that an individual picked at random from this population will have a cholesterol value greater than 160 mg/100 ml. 1. virtually zero, as 160 mg/100 ml is three standard deviations above the mean 2. virtually 100%, as 160 mg/100 ml is three standard deviations above the mean 3. virtually 50%, as 160 mg/100 ml is three standard deviations above the mean 4. virtually 75%, as 160 mg/100 ml is the third quartile 5. virtually 75%, as 160 mg/100 ml is the first quartile
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BIOS 516D Final Exam, Fall 2009 Name: 4. We are given that P(Z > z) = 0.1345. We know that P(Z < −z) = 1. virtually zero 2. virtually 100% 3. 0.1345 4. 1 − 0.1345 5. −0.1345 5. Suppose we do not know the shape of the distribution of X. But we know that the mean of the distribution is 16 and the standard deviation is 2. What is the distribution of X for samples of size 64? 1. N(16, 2) according to the Central Limit Theorem 2. N(16, .25) according to the Central Limit Theorem 3. N(0, 1) according to the Central Limit Theorem 4. It is not possible to tell the shape, center and spread of the distribution; the sample size is not large enough for us to use the Central Limit Theorem. 6. If your score on a test is at the 60th percentile, you know that your score 1. lies below the first quartile 2. lies above the third quartile 3. could be anywhere 4. lies between the first quartile and the median 5. lies between the median and the third quartile 7. A researcher designs a study to test the effectiveness of a new hair growth formula in restoring hair to patients with male pattern baldness. The design consists of measuring each patient’s hair surface area (HSA) at baseline, then administering the new treatment for 8 weeks, whereupon the patient’s HSA is again measured and the change in HSA is noted. Which of the following are true (circle all that apply)?: 1. This is a one-sample design 2. This is a paired design 3. We have not discussed this specific design
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BIOS 516D Final Exam, Fall 2009 Name: 8. The globally averaged atmospheric carbon dioxide (CO 2 ) concentration is known
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BIOS 516D Final Fall 2009 - BIOS 516D Final Exam, Fall 2009...

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