Chapter 5 HW Solutions

Chapter 5 HW Solutions - 0.167 0.375 0.125 Ex 5.10 First...

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Solution to HW # 5 Ex 5.2 (a): = 0.5*0.333+5*0.167+7*.375+11*.125 = 5 (0.5-5) 2 (0.333) + (5-5) 2 (0.167) + (7-5) 2 (0.375) + (11-5) 2 (0.125) = 12.74 The median can be located anywhere between 5 and 7. (b) = (0.5-5) 3 (0.333) + (5-5) 3 (0.167) + (7-5) 3 (0.375) + (11-5) 3 (0.125) = -0.344625 = 0.000057 = (0.5-5) 4 (0.333) + (5-5) 4 (0.167) + (7-5) 4 (0.375) + (11-5) 4 (0.125) = 304.551 = 1.8 (c) E[(h(X)] = {0.5 2 + 0.5 + 1}(0.333) + {5 2 + 5 + 1}(0.167) + {7 2 + 7 + 1}(0.375) + {11 2 + 11 + 1}(0.125) = 43.76 Prof. Barnes software can also be used to plot the distribution function and determine the aforementioned values. (d) When x=0.5, y=h(0.5)=1.75, p(y)=0.333 When x=5, y=h(5)=31, p(y)=0.167 When x=7, y=h(7)=57, p(y)=0.375 When x=11, y=h(11)=133, p(y)=0.125 Hence, the pdf of Y is as follows: y=h(x) 1.75 31 57 133 p(y) 0.333
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Unformatted text preview: 0.167 0.375 0.125 Ex. 5.10: First moment about mean = E{X-μ} = E(X) – E(μ) = μ – μ = 0. Or Ex 5.21 (a) If this is a valid pdf, then and f(x) ≥ 0 for all x. Both psi () and eta () are positive, and x is at least eta and therefore also positive, so the entire expression of f(x) will always be positive. This proves that f(x) is a valid pdf. (b) See jmp file (Run script of overlay plot to see the graph). Since jmp doesn’t provide direct formula for pareto distribution, we need to create the formula by inputting the pdf of pareto in the cell. By changing the value of and separately, we can find the effects show an exponential decay. controls the rate of decay. controls the starting point and the initial height. (c)...
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Chapter 5 HW Solutions - 0.167 0.375 0.125 Ex 5.10 First...

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