PracticeMT1_Ans.doc

# PracticeMT1_Ans.doc - BA 240 Winter 2006 W Li Practice Exam...

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BA 240 Winter 2006 W. Li 01/19/2006 Practice Exam 1 Solution Problem 1: Consider following sample of 16 quantitative measurements: 88, 53, 64, 59, 72, 91, 59, 68, 76, 55, 62, 65, 84, 62, 59, 77 Ordered Data: 53, 55, 59, 59, 59, 62, 62, 64, 65, 68, 72, 76, 77, 84, 88, 91 a) Calculate the three Measures of Central Tendency. Mean: 375 . 68 16 91 88 ... 59 55 53 n x x i Median: 5 . 64 2 65 54 M Mode: 59 b) Describe the shape (skew) using part a). Justify your answer. Skew to RIGHT since Mean > Median c) Calculate the Five-number Summary. Min Q1 Median Q3 Maximum 53 59 (4 th #) 64.5 77 (13 th #) 91 d) Calculate the standard deviation. 76931 . 11 15 75 . 2077 1 2 n x x s i (11.77) e) Find the z-score for the Minimum and the Maximum. 306 . 1 769 . 11 375 . 68 53 s x x z 922 . 1 769 . 11 375 . 68 91 s x x z f) Assume data is normal. Construct an interval that will include about 68% of the data set above. Consider using Empirical Rule: 68% means within the range of s x : 144 . 80 , 606 . 56 ) 769 . 11 ( 375 . 68 s x g) Count how many data actually fall into the interval in part f), and what is the percentage of the data? 11 data fall into (56.606, 80.144). Thus, 11/16 = 68.75% of data fall into the range. Problem 2 : Fred keeps track of his commute times. He has found that they have a normal shaped distribution with a mean of 46.2 minutes and a standard deviation of 11 minutes. a) What percentage of his commute times is between 13.2 minutes and 79.2 minutes? 11 , 2 . 46 s x s x 3 ) 11 ( 3 2 . 46 2 . 13 s x 3 ) 11 ( 3 2 . 46 2 . 79 From Empirical Rule, 99.7% (100%) of time is between 13.2 minutes and 79.2 minutes (within the range s x 3 ). Practice Exam 1 Solution 1

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BA 240 Winter 2006 W. Li 01/19/2006 b) What percentage of time does it take him at most 68.2 minutes to get to work?
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