EMGT 571 - Engineering Statistics Midterm Exam

# EMGT 571 - Engineering Statistics Midterm Exam - D)= P(B...

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EMGT571 Engineering Statistics Midterm Examination Answer the following questions from Supplementary Problems in "Business Statistics For Contemporary Decision Making" by Black. Question # 2.22 Frequency 41 673 564 402 379 202 108 73 54 Percent 1.6 27.0 22.6 16.1 15.2 8.1 4.3 2.9 2.2 Cum % 100.0 27.0 49.6 65.7 80.8 88.9 93.3 96.2 98.4 Problem Other 9 5 3 10 4 8 6 1 2500 2000 1500 1000 500 0 100 80 60 40 20 0 Frequency Percent Pareto Chart of Problem 3.46 30 house holds 2 3 1 2 6 4 2 1 5 3 2 3 1 2 2 1 3 1 2 2 4 2 1 2 8 3 2 1 1 3 N for Variable Mean Q1 Median Q3 Range IQR Mode Mode C1 2.500 1.000 2.000 3.000 7.000 2.000 2 11

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4.34 30 20 45 55 40 95 a) P(E)= 40/95= 0.4211 b) P(B
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Unformatted text preview: D)= P(B)+P(D)-P(B ∩ D) = 30/95 + 55/95 – 15/95 = 0.7368 c) P(A ∩ E)= 20/95 = 0.3158 d) P(B|E)= 5/40 = 0.1250 e) P(A ∪ B)= P(A) + P(B) = 30/95 + 20/95 = 0.5263 f) P(B ∩ C)= 0.00 g) P(D|C)= 30/45 = 0.6667 h) P(A|B)= 0.00 i) Are variables 1 and 2 independent? Why or why not? P(B)=20/95 = 0.2105 P(D)=55/95 = 0.5789 No they are not independent because P(B|E) ≠ P(B) and P(D|C) ≠ P(D) 5.36 P(x)=λ x e-λ /x! a) P(x=4|λ=1.25) b) P(x≤1|λ=6.37) c) P(x>5|λ=2.40) 6.36 7.29 7.30 8.46 8.48 D E A 10 20 B 15 5 C 30 15...
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EMGT 571 - Engineering Statistics Midterm Exam - D)= P(B...

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