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hw5 fall 07 solns

# hw5 fall 07 solns - 1 ISyE 3770 Fall 2007 Homework#5(Covers...

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1 ISyE 3770 – Fall 2007 Homework #5 (Covers Modules 16–18) — Solutions 1. (Hines et al., 4–1). A refrigerator manufacturer subjects his finished products to a final inspection. Of interest are two categories of defects: scratches or flaws in the porcelain finish, and mechanical defects. The number of each type of defects is a random variable. The results of inspecting 50 refrigerators are shown in the following joint p.m.f. table, where X represents the occurrence of finish defects and Y represents the occurrence of mechanical defects. Y \ X 0 1 2 3 4 5 0 11/50 4/50 2/50 1/50 1/50 1/50 1 8/50 3/50 2/50 1/50 1/50 2 4/50 3/50 2/50 1/50 3 3/50 1/50 4 1/50 (a) Find the marginal distributions of X and Y . Solution: Let’s re-write the table, this time including the marginals. Y \ X 0 1 2 3 4 5 f Y ( y ) 0 11/50 4/50 2/50 1/50 1/50 1/50 20/50 1 8/50 3/50 2/50 1/50 1/50 15/50 2 4/50 3/50 2/50 1/50 10/50 3 3/50 1/50 4/50 4 1/50 1/50 f X ( x ) 27/50 11/50 6/50 3/50 2/50 1/50 (b) Find the marginal distribution of mechanical defects, given that there are no finish defects. Solution: f ( y | X = 0) = f (0 , y ) f X (0) = f (0 , y ) 27 / 50 = 11 / 27 if y = 0 8 / 27 if y = 1 4 / 27 if y = 2 3 / 27 if y = 3 1 / 27 if y = 4 0 otherwise

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2 (c) Find the marginal distribution of finish defects, given that there are no me- chanical defects. Solution: f ( x | Y = 0) = f ( x, 0) f Y (0) = f ( x, 0) 20 / 50 = 11 / 20 if x = 0 4 / 20 if x = 1 2 / 20 if x = 2 1 / 20 if x = 3 , 4 , 5 0 otherwise 2. (Hines et al., 4–4). Consider a situation in which the surface tension and acidity of a chemical product are measured. These variables are coded such that surface tension is measured on a scale 0 X 2, and acidity is measured on a scale 2 Y 4. The probability density function of (
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