Answer-Key-Module-23478.docx - Answer Key Module 2 1 2 3 2...

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Answer Key Module 2 1. f ( x i ;a i , 17,9 ) = ( 3 2 )( 2 1 )( 8 4 )( 4 2 ) ( 17 9 ) = 0.1037 2. f ( x = 1 ) = ( 5 1 ) ( 0.20 1 ) ( 0.80 4 ) = 0.4096 3. f ( x≤ 7 ; 0.6 ) = f ( x = 1 ; 0.6 ) + f ( x = 2 ; 0.6 ) + f ( x = 3 ; 0.6 ) + f ( x = 4 ; 0.6 ) + f ( x = 5 ; 0.6 ) + f ( x = 6 ; 0.6 ) + f ( x = 7 ; 0.6 f ( x≤ 7 ; 0.6 ) = ( 0.6 ) ( 0.4 ) 0 + ( 0.6 ) ( 0.4 ) 1 + ( 0.6 ) ( 0.4 ) 2 + ( 0.6 ) ( 0.4 ) 3 + ( 0.6 ) ( 0.4 ) 4 + f ( 0.6 ) ( 0.4 ) 5 + ( 0.6 ) ( 0.4 ) 6 f ( x≤ 7 ; 0.6 ) = 0.9984 4. f ( x = 9 ; 0.002 ) = ( 0.002 ) ( 0.998 ) 8 = 1.9682 x 10 3 5. f ( x = 45 ; 0.002 3 ) = ( 0.006 ) ( 0.998 ) 44 = 5.4941 x 10 3 6. f ( x≥ 1 ; 70,10,5 ) = f ( x = 1 ; 70,10,5 ) + f ( x = 2 ; 70,10,5 ) + f ( x = 3 ; 70,10,5 ) + f ( x = 4 ; 70,10,5 ) + f ( x = 5 ; 70,10,5 ) f ( x≥ 1 ; 70,10,5 ) = ( 5 1 )( 70 9 ) + ( 5 2 )( 70 8 ) + ( 5 3 )( 70 7 ) + ( 5 4 )( 70 6 ) + ( 5 5 )( 70 5 ) ( 70 10 ) = 1.0895 7. f ( x i ,a i , 200,9 ) = ( 80 3 )( 60 3 )( 20 1 )( 40 2 ) ( 200 9 ) = 0.0373 8. f ( x≤ 4 ; 0.6 ) = f ( x = 1 ; 0.6 ) + f ( x = 2 ; 0.6 ) + f ( x = 3 ; 0.6 ) + f ( x = 4 ; 0.6 ) f ( x≤ 4 ; 0.6 ) = ( 0.6 ) ( 0.4 ) 0 + ( 0.6 ) ( 0.4 ) 1 + ( 0.6 ) ( 0.4 ) 2 + ( 0.6 ) ( 0.4 ) 3 f ( x≤ 4 ; 0.6 ) = 0.9744 9. f ( x i ;a i , 232,000,10 ) = ( 52,200 2 )( 126,208 5 )( 53,592 3 ) ( 232,000 10 ) = 0.0749 10.A. x 500 = 0.01 100 = 0.05 B. f ( x = 1 ; 0.05 ) = ( e 0.05 ) ( 0.05 1 ) 1 ! = 0.0476 ¿ 1 0.0476 = 0.9524
Answer Key Module 3: 1. a. F ( x > 1.02 ) = 1.02 1.05 1 1.02 0.95 dx = 1 0.95 7 1 1.02 0.95 dx = 1 1.02 0.95 1.05 0.95 = 0.07 0.1 = 0.7 b. 0.90 = 1 x 0.95 1.05 0.95 0.01 =− x + 0.95 → x = 0.96 c. μ = 1.05 + 0.95 2 = 1 , σ 2 = ( 1.05 0.95 ) 2 2 = 5 x 10 3 2. a. μ = 374 + 380 2 = 377 , σ 2 = ( 380 374 ) 2 2 = 18 b. F ( x < 375 ) = 375 374 380 374 = 1 6 c. ( 377 375 ) $ 0.002 = $ 0.004 3. z = 6,250 6,000 100 = 2.5 a. P ( Z < 2.5 ) = 0.9938 z = 5,800 6,000 100 =− 2 z = 5,900 6,000 100 =− 1 b. P ( 2 < Z 1 ) = 0.1587 0.0228 = 0.1359 c. 0.95 = z = interpolationbetween 1.64 1.65 = 1.645 1.645 = x 6,000 100 x = 6,000 + 164.5 = 6,164.5 kg / cm 2 4. a. F ( x > 1 hr = 60 mins ) = 1 [ e 1 ( 60 ) 10 ( e 1 ( 0 ) 10 ) ] = 2.4788 x 10 3 b. F ( 60 < x < 70 ) : F ( x < 70 ) = [ e 1 ( 70 ) 10 ( e 1 ( 0 ) 10 ) ] = 0.9991 F ( 60 < x < 70 ) = 0.9991 0.0024788 = 0.9966 c. ln ( 0.10 )= ln ( 10 )∗ ln ( e 10 x ) 2.3026 = 2.3026 10 x
x = 0.4605 minutes Answer Key Module 4: 1. f xy ( x, y ) 1:1 ¿ 4 40 12 40 = 0.0300 1:2 ¿ 3 39 12 39 = 0.0237 1:3 ¿ 2 38 12 38 = 0.0166 1:4 ¿ 1 37 12 37 = 0.0088 Up to… 4:4 ¿ 1 37 9 37 = 0.0066 Y=Mechanical Engineering Majors X=Industrial Engineering Majors 1 2 3 4 Marginal Probability of Y 1 0.0300 0.0237 0.0166 0.0088 0.0791 2 0.0289 0.0217 0.0152 0.0080 0.0739 3 0.0277 0.0208 0.0139 0.0073 0.0696 4 0.0263 0.0197 0.0131 0.0066 0.0657 Marginal Probability of X 0.1129 0.0859 0.0589 0.0307 0.2883 (not equal to 1 because Z=electrica l eng’g. students are not included. f x ( x ) = 1:1 + 1:2 + 1:3 + 1:4 ¿ 0.0300 + 0.0237 + 0.0166 + 0.0088 = 0.0791
f y / 3 ( y ) Y=Mechanical Engineering Majors X=Industrial Engineering Majors 3 1 0.2818 2 0.2581 3 0.2360 4 0.2224 2. a. P ( x = 2 ) = 0.20 + 0.15 + 0.10 + 0.05 = 0.5 b. P ( x = 1 , y = 2 ) = 0.15 + 0.20 = 0.35 c. P ( z < 1.5 ) = 0.05 + 0.15 + 0.20 + 0.10 = 0.5 d. P ( x = 1 / y = 1 ) = 0.05 + 0.10 + 0.15 + 0.20 0.05 + 0.10 + 0.20 + 0.15 = 1 e. f x / yz ( x ) = 0.10 + 0.15 1 = 0.25 3. a. μ = 2 + 2 = 4 mm,s = 0.1 + 0.1 = 0.2 mm b. z = 4 4.3 0.2 =− 1.5 = P = 0.0668 4. Inverse solution x = ( y 10 ) 2 Jacobian J = 1 2 f y ( y ) = ( y 10 ) 2 18 1 2 f y ( y ) = y 10 72 , 10 ≤ y ≤ 26 Answer Key Module 7: Laboratory #7 1. Step 1: H 0 : μ = 1.5 inches (The mean inside diameter of a bearing used in an automotive is 1.5 inches.)
H 1 : μ≠ 1.5 inches (The mean inside diameter of a bearing used in an automotive is not 1.5 inches.) Step 2: ´ x = 1.4975 inches let σ = s = 0.01 inches n = 25 bearings = 0.01 (confidence level = 0.99)

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