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Spring 36-201 2008 Homework 7 KEY This document available on: http://www.stat.cmu.edu/~gordonw/spring2008stat201.html Stat 201 Spring 2008 Homework 7 GRADING KEY Page 1 of 7 Due: Wednesday, March 5, at the beginning of lecture. Homework considered late beyond the first 10 minutes of lecture. Late Homework Policy: Late homework may be turned-in for a maximum of half credit, up to 4:00 PM on the same Wednesday the homework is due. Late homework should be placed in the manila envelope which will be posted by room 328-C, Facilities Management Building (FMS). Late homework should be labeled with the course as well as your name. Please do not turn in any homework to any mailboxes. Please include the top-page coversheet , filled-out and stapled to your homework. Show all work. This homework covers section 6.2 in the course textbook. Point Distribution: Exercise 1 Exercise 2 Exercise 3 effort & neatness TOTAL 32 11 37 20 100 NOTE: To save paper, you are encouraged to print double-sided ( duplex mode). 1. (Normal Distribution fundamentals) For each of the following, use the Standard Normal Distribution Table ( Table A in the appendix of the book, and also available on the instructor s web page) as appropriate: This document available on: http://www.stat.cmu.edu/~gordonw/spring2008stat201.html Stat 201 Spring 2008 Homework 7 GRADING KEY 1 (a) Page 2 of 7 Find the probability of an observation on the Standard Normal Distribution satisfying: Z < .38 ; include a sketch, with the axis appropriately labeled, the mean appropriately marked, the position in question appropriately labeled, and the appropriate area shaded and labeled. Solution: The probability is .3520 (or equivalently, 35.2%) ; the picture is as follows: To the students: In probability notation, we would write this result as: P ( Z < .38) = .3520 1 (b) Find the score, z , on the Standard Normal, so that the proportion of observations below z is .6950; include a sketch, with the axis appropriately labeled, the mean appropriately marked, the position in question appropriately labeled, and the appropriate area shaded and labeled. Solution: The z score is .51 ; the picture is as follows: To the students: In probability notation, we would write this result as: P ( Z < .51) = .6950 This document available on: http://www.stat.cmu.edu/~gordonw/spring2008stat201.html Stat 201 Spring 2008 Homework 7 GRADING KEY 1 (c) Page 3 of 7 Find the score, z , on the Standard Normal, so that the proportion of observations above z is .2643; include a sketch, with the axis appropriately labeled, the mean appropriately marked, the position in question appropriately labeled, and the appropriate area shaded and labeled. Solution: The z score is .63 (note that it is positive); the picture is as follows: To the students: In probability notation, we would write this result as: P ( Z > .63) = .2643 1 (d) Suppose a population of scores follows a Standard Normal Distribution. Based on the Normal Distribution Table in the text, the top 3% of scores have at least what value? Include a sketch, with the axis appropriately labeled, the mean appropriately marked, the position in question appropriately labeled, and the appropriate area shaded and labeled. Solution: The z score is 1.88 (note that it is positive); the picture is as follows: To the students: In probability notation, we would write this result as: P ( Z > .63) = .2643 This document available on: http://www.stat.cmu.edu/~gordonw/spring2008stat201.html Stat 201 Spring 2008 Homework 7 GRADING KEY 2. Page 4 of 7 Alice and Bob are comparing their college math entrance exam scores. Alice took the SAT, whose math portion has a Normal distribution with mean 480 and standard deviation 100; while Bob took the ACT, whose math portion has a Normal distribution with mean 18 and standard deviation 5. Alice scored 660 on the math SAT; while Bob scored 28 on the math ACT. Who did better ? [Hint: Consider the standardized scores.] Show appropriate calculations and state your conclusion. Solution: Bob did better. Alice s standardized score: z Alice = Bob s standardized score: z Bob = x 3. , and the Standard Normal Distribution Table ( Table A in the appendix of the text, and also available on the instructor s web page), as appropriate: 3 (a) Among developers the of statistics was the mathematician and social scientist Adolphe Quetelet* (1796 1874), who was the first person to use the Normal Distribution as a For each of the following, use the standardization formula, z = This document available on: http://www.stat.cmu.edu/~gordonw/spring2008stat201.html x = 660 480 100 = 1.8 Bob s standardize score is larger, indicating that he scored higher, relative to his distribution. = 28 18 5 =2 x Stat 201 Spring 2008 Homework 7 GRADING KEY Page 5 of 7 model of real data. For instance, based on information provided by the Scottish Army, Quetelet found that the chest measurements of male soldiers closely followed a Normal distribution, with mean = 40 inches, and standard deviation = 2 inches (according to An Introduction to Mathematical Statistics and its Applications, 2nd ed, by Larsen and Marx, p. 213). Based on this distribution, what proportion of male soldiers had chest measurement above 41 inches? Include relevant distribution sketches, with the axes appropriately labeled, the means appropriately marked, the position in question appropriately labeled, and the appropriate area shaded and labeled. Also show all calculations. * See http://www-groups.dcs.st-andrews.ac.uk/~history/Mathematicians/Quetelet.html Solution: The proportion is .3085 (or equivalently, 30.85%) the picture and resulting calculations are as follows: To the students: In probability notation, we would write this result as: P ( X > 41) = P ( Z > .5) = .3085 This document available on: http://www.stat.cmu.edu/~gordonw/spring2008stat201.html Stat 201 Spring 2008 Homework 7 GRADING KEY 3 (b) Page 6 of 7 Suppose you are working as a quality control investigator, testing circuit boards. Suppose circuit boards thinner than 10 mm are at risk of breaking. Suppose the distribution of thickness of circuit boards produced by Intel is supposed to be Normal, with mean = 12 mm, and standard deviation = 0.7 mm, if the manufacturing process is working correctly. (i) Intel claims to you that the process is working correctly. To test this claim, you select one circuit board at random. If the claimed distribution is true, how likely should it be for that one randomly-selected circuit board to be thinner than 10 mm? Include relevant distribution sketches, with the axes appropriately labeled, the means appropriately marked, the position in question appropriately labeled, and the appropriate area shaded and labeled. Also show all calculations. Answer: The probability is .0021 (or equivalently, 0.21%) ; picture and calculations are as follows: To the students: In probability notation, we would write this result as: P ( X < 10) = P ( Z < 2.86) = .0021 (ii) [Critical Thinking] If, in fact, the board you select does turn out to be thinner than 10 mm, do you think the manufacturing process is working correctly, as claimed? [Say yes or no.] Answer: No. Since a selection was obtained that should have had a very low probability, it is reasonable to be suspicious of the claim. I suspect that the process is not working correctly. [I.e., either the true mean number is higher than claimed; or the standard deviation is larger than expected; or the distribution is not really Normal (or any combination)]. This document available on: http://www.stat.cmu.edu/~gordonw/spring2008stat201.html Stat 201 Spring 2008 Homework 7 GRADING KEY 3 (c) Page 7 of 7 Suppose X represents the weights, in pounds, of sacks of potatoes at the Original Hot Dog Shop, and suppose X varies Normally with = 65 pounds and = 8.1 pounds. If I take a part-time job loading sacks of potatoes but I agree to lift only the lightest 10% of the sacks, approximately what weights will I be lifting? Include relevant distribution sketches, with the axes appropriately labeled, the means appropriately marked, the position in question appropriately labeled, and the appropriate area shaded and labeled. Also show all calculations. Solution: I will be lifting less than about 54 and 2 pounds (around 54.6 or 54.5, depending on z score used) Picture and calculations as follows: 1 To the students: In probability notation, we would write this result as: P ( X < 54.632 ) = P ( Z < 1.28) = .10 [end of homework 7] This document available on: http://www.stat.cmu.edu/~gordonw/spring2008stat201.html

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