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exam2dsol
Course: SMAM 351, Fall 2009School: RIT
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351 SMAM 1 A. (1) Exam 2 d Solutions P(X = 5) = P(X 5)-P(X 4) = .193 - .090 = .103 (2) p = 0.7 P(X 15) = 1-P(X 14) = 1-.098 = .902 B. (1) P(X 4) = (.3)x -1(.7) = x=1 4 .7 - .7(.3)4 = 1- (.3)4 = .9919 1- .3 (2) 9 p[X = 10] = (.7)4 (.3)6 = .0147 3 2. EX = 0(0.1) +1(0.3) + 2(0.4) + 3(0.2) = 1.7 EX 2 = 02 (0.1 + 12(0.3) + 22 (0.4) + 32(0.2) = 3.7 ) 2 = 3.7 - 1.72 = .81 = 0.9 3. 4000k = 10000 k = 2.5 1...
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351 SMAM 1 A. (1)
Exam 2 d Solutions
P(X = 5) = P(X 5)-P(X 4) = .193 - .090 = .103
(2) p = 0.7 P(X 15) = 1-P(X 14) = 1-.098 = .902 B. (1) P(X 4) = (.3)x -1(.7) =
x=1 4
.7 - .7(.3)4 = 1- (.3)4 = .9919 1- .3
(2) 9 p[X = 10] = (.7)4 (.3)6 = .0147 3
2. EX = 0(0.1) +1(0.3) + 2(0.4) + 3(0.2) = 1.7 EX 2 = 02 (0.1 + 12(0.3) + 22 (0.4) + 32(0.2) = 3.7 ) 2 = 3.7 - 1.72 = .81 = 0.9 3. 4000k = 10000 k = 2.5 1 1- = .84 2.52
4. A. P[X = 6] = P(X 6)- P(X 5) = .130 - .067 = .063 B. =5 P(X > 5) = 1- P(X 5) = 1- .616 = .384
5. A. 4 6 x 3 - x f(x) = ,x = 0,1 ,2,3 10 3 B. 4 6 4 6 + 0 3 1 2 20 + 60 2 f(0) + f(1 = ) = = 10 120 3 3 6. = .005(1000) 5 = P(4 X 7) = P(X 7) -P(X 3) = .867 - .265 = .602 7. A 0 x<0 1 2 G(x) = 6 (x + x) 0 x 2 1 x>2 B.
1 G(c) = 6 (c 2 + c) = 1 3
c2 + c = 2 c2 + c - 2 = 0 (c + 2)(c -1) = 0 c =1 Re ject the other solution. It lies outside the domain. C. EX =
1 0 6
2
(2x 2 + x)dx = 1.222
E(3X + 2) = 3EX + 2 = 3(1.222) + 2 = 5.666 EX 2 =
1 0 6
2
(2x 3 + x 2 )dx = 1.778
2 = 1.778 - (1.222)2 ...
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