sol1 - Problem 21/14 (a) (b) One of the following answers...

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Unformatted text preview: Problem 21/14 (a) (b) One of the following answers is fine: Mode: from the display, it is the stem 6, around 6.5. Median: 7. Mean: 7.7. (c) The display appears to be rather spread-out. (d) No, the data is positively skewed, or skewed to the right. (e) The last observation (18.9) appears to be an outlier. 1 Problem 22/20 (a) The stem and leaf plot could be either the one shown by JMP or one by hand (like the one above on the right). One of the following answers for a typical value is fine: From the display, somewhere in the low 2000s. Median: 2100. Mean: 2231.36. (b) The histogram is shown above. It could be said that there are two modes. From the JMP stem and leaf, the proportion of subdivisions with total length less that 2000 is (5+5+10+3)/47 = .489 = 48.9%. Similarly, the proportion of subdivisions with total length between 2000 and 4000 is (7+3+6+1)/47 = .361 = 36.1%. 2 Problem 24/28 After stacking the columns (Table > Stack) the above histogram was constructed. The distribution is unimodal and slightly positively skewed. The mean and median are 113.72 and 113, which corroborates the skewness of the distribution. 3 Problem 30/34 (a) Mean of U: 21.5454 Mean of F: 8.56 (b) Median of U: 17 Median of F: 8.9 The median of U is so di ff erent from its mean because the maximum value in that sample is an outlier (e.g see the box plot)....
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This note was uploaded on 09/09/2008 for the course ENGRD 2700 taught by Professor Staff during the Summer '05 term at Cornell.

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sol1 - Problem 21/14 (a) (b) One of the following answers...

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