CWA CH 11 FINAL_97_20

CWA CH 11 FINAL_97_20 - Chapter 11 PROBABILITY AND CALCULUS...

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Chapter 11 PROBABILITY AND CALCULUS 11.1 Continuous Probability Models 1. f ( x )= 1 9 x ¡ 1 18 ;[2 ; 5] Show that condition 1 holds. Since 2 · x · 5 ; 2 9 · 1 9 x · 5 9 1 6 · 1 9 x ¡ 1 18 · 1 2 : Hence, f ( x ) ¸ 0 on [2, 5]. Show that condition 2 holds. Z 5 2 μ 1 9 x ¡ 1 18 dx = 1 9 Z 5 2 μ x ¡ 1 2 dx = 1 9 μ x 2 2 ¡ 1 2 x ¶¯ ¯ ¯ ¯ 5 2 = 1 9 μ 25 2 ¡ 5 2 ¡ 4 2 +1 = 1 9 (8 + 1) =1 Yes, f ( x ) is a probability density function. 2. f ( x )= 1 3 x ¡ 1 6 ;[3 ; 4] Show that condition 1 holds. Since 3 · x · 4 ; 1 · 1 3 x · 4 3 5 6 · 1 3 x ¡ 1 6 · 7 6 : Hence, f ( x ) ¸ 0 on [3, 4]. Show that condition 2 holds. Z 4 3 μ 1 3 x ¡ 1 6 dx = μ x 2 6 ¡ x 6 ¶¯ ¯ ¯ ¯ 4 3 = 1 6 (16 ¡ 4 ¡ 9+3) =1 Yes, f ( x ) is a probability density function. 3. f ( x )= 1 21 x 2 ;[1 ; 4] Since x 2 ¸ 0 ;f ( x ) ¸ 0 on [1 ; 4] : 1 21 Z 4 1 x 2 dx = 1 21 μ x 3 3 ¶¯ ¯ ¯ ¯ 4 1 = 1 21 μ 64 3 ¡ 1 3 =1 Yes, f ( x ) is a probability density function. 4. f ( x )= 3 98 x 2 ;[3 ; 5] Since x 2 ¸ 0 ;f ( x ) ¸ 0 on [3, 5]. Z 5 3 3 98 x 2 dx = x 3 98 ¯ ¯ ¯ ¯ 5 3 = 1 98 (125 ¡ 27) =1 Yes, f ( x ) is a probability density function. 5. f ( x )=4 x 3 ;[0 ; 3] 4 Z 3 0 x 3 dx =4 μ x 4 4 ¶¯ ¯ ¯ ¯ 3 0 =4 μ 81 4 ¡ 0 =81 6 =1 No, f ( x ) is not a probability density function. 6. f ( x )= x 3 81 ; [0, 3] Z 3 0 x 3 81 dx = x 4 324 ¯ ¯ ¯ ¯ 3 0 = 1 4 6 =1 No, f ( x ) is not a probability density function. 716
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Section 11.1 Continuous Probability Models 717 7. f ( x )= x 2 16 ;[ ¡ 2 ; 2] 1 16 Z 2 ¡ 2 x 2 dx = 1 16 μ x 3 3 ¶¯ ¯ ¯ ¯ 2 ¡ 2 = 1 16 μ 8 3 + 8 3 = 1 3 6 =1 No, f ( x ) is not a probability density function. 8. f ( x )=2 x 2 ;[ ¡ 1 ; 1] Z 1 ¡ 1 2 x 2 dx = 2 3 x 3 ¯ ¯ ¯ ¯ 1 ¡ 1 = 2 3 (1 + 1) = 4 3 6 =1 No, f ( x ) is not a probability density function. 9. f ( x )= 5 3 x 2 ¡ 5 90 ; [ ¡ 1 ; 1] Let x =0 .Then f ( x )= f (0) = ¡ 5 90 < 0 : So f ( x ) < 0 for at least one x -value in [ ¡ 1 ; 1] : No, f ( x ) is not a probability density function. 10. f ( x )= 3 13 x 2 ¡ 12 13 x + 45 52 ; [0 ; 4] Let x =2 .Then f ( x )= f (2) = ¡ 3 52 < 0 : So f ( x ) < 0 for at least one x -value in [0 ; 4] : No, f ( x ) is not a probability density function. 11. f ( x )= kx 1 = 2 ;[1 ; 4] Z 4 1 kx 1 = 2 dx = 2 3 kx 3 = 2 ¯ ¯ ¯ ¯ 4 1 = 2 3 k (8 ¡ 1) = 14 3 k If 14 3 k =1 ; k = 3 14 : Notice that f ( x )= 3 4 x 1 = 2 ¸ 0 for all x in [1, 4]. 12. f ( x )= kx 3 = 2 ;[4 ; 9] Z 9 4 kx 3 = 2 dx = 2 k 5 x 5 = 2 ¯ ¯ ¯ ¯ 9 4 = 2 k 5 (243 ¡ 32) = 422 k 5 If 422 k 5 =1 ;k = 5 422 : Notice that f ( x )= 5 422 x 3 = 2 ¸ 0 for all x in [4 ; 9] : 13. f ( x )= kx 2 ;[0 ; 5] Z 5 0 kx 2 dx = k x 3 3 ¯ ¯ ¯ ¯ 5 0 = k μ 125 3 ¡ 0 = k μ 125 3 If k ¡ 124 3 ¢ =1 ; k = 3 125 : Notice that f ( x )= 3 125 x 2 ¸ 0 for all x in [0, 5]. 14.
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CWA CH 11 FINAL_97_20 - Chapter 11 PROBABILITY AND CALCULUS...

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