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2008hw2_solutions

# 2008hw2_solutions - ST 314 1 Text p 80 exercise 2.64 HW#2...

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ST 314 HW #2 Solutions 1. Text, p. 80, exercise 2.64. As described in the assignment, the appropriate way of analyzing these data is to form a new column of numbers as follows: Difference = Operator #1’s time – Operator #2’s time One-Variable Analysis - DIFFERENCE Analysis Summary Data variable: DIFFERENCE 20 values ranging from -0.28 to -0.11 Stem-and-Leaf Display for DIFFERENCE: unit = 0.01 1|2 represents 0.12 1 -2|8 1 -2| 4 -2|544 8 -2|3333 (6) -2|111100 6 -1|99988 HI|-0.11 Box-and-Whisker Plot -0.29 -0.25 -0.21 -0.17 -0.13 -0.09 DIFFERENCE - 1 -

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Histogram for DIFFERENCE -0.29 -0.25 -0.21 -0.17 -0.13 -0.09 DIFFERENCE 0 2 4 6 8 10 frequency Comments : The distribution of the differences between the two operators is slightly skewed to the left and single-peaked, with one possible outlier. The range of the data is all below zero, so we the data strongly suggest that operator #1 has lower running times than does operator #2. 2. Text, p. 115, exercise 3.2. Let Y = the number of extruders down on a day. a. P(Y 3) = P(Y = 3) + P(Y = 4) = .05 + .05 = .10 b. P(Y < 1) = P(Y = 0) = 0.5 c. E(Y) = Σ y y p(y) = 0(.5) + 1(.3) + 2(.1) + 3(.05) + 4(.05) = 0.85 d. First find E(Y 2 ) = Σ y y 2 p(y) = 0(.5) + 1(.3) + 4(.1) + 9(.05) + 16(.05) = 1.95
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