# HW7 - Chapter 4 Continuous Variables and Their Probability...

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59 Chapter 4: Continuous Variables and Their Probability Distributions 4.1 a. < < < < = = 4 1 4 3 9 . 3 2 7 . 2 1 4 . 1 0 ) ( ) ( y y y y y y Y P y F b. The graph is above. 012345 0.0 0.2 0.4 0.6 0.8 1.0 y F(y) 4.2 a. p (1) = .2, p (2) = (1/4)4/5 = .2, p (3) = (1/3)(3/4)(4/5) = 2., p (4) = .2, p (5) = .2. b. < < < < < = = 5 1 5 4 8 . 4 3 6 . 3 2 4 . 2 1 2 . 1 0 ) ( ) ( y y y y y y y Y P y F c. P ( Y < 3) = F (2) = .4, P ( Y 3) = .6, P ( Y = 3) = p (3) = .2 d. No, since Y is a discrete random variable.

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60 Chapter 4: Continuous Variables and Their Probability Distributions Instructor’s Solutions Manual 4.3 a. The graph is above. 01 y F(y) q = 1-p b. It is easily shown that all three properties hold. 4.4 A binomial variable with n = 1 has the Bernoulli distribution. 4.5 For y = 2, 3, …, F ( y ) – F ( y – 1) = P ( Y y ) – P ( Y y – 1) = P ( Y = y ) = p ( y ). Also, F (1) = P ( Y 1) = P ( Y = 1) = p (1). 4.6 a. F ( i ) = P ( Y i ) = 1 – P ( Y > i ) = 1 – P (1 st i trials are failures) = 1 – q i . b. It is easily shown that all three properties hold. 4.7 a. P (2 Y < 5) = P ( Y 4) – P ( Y 1) = .967 – .376 = 0.591 P (2 < Y < 5) = P( Y 4) – P ( Y 2) = .967 – .678 = .289. Y is a discrete variable, so they are not equal. b. P (2 Y 5) = P ( Y 5) – P ( Y 1) = .994 – .376 = 0.618 P (2 < Y 5) = P ( Y 5) – P ( Y 2) = .994 – .678 = 0.316. Y is a discrete variable, so they are not equal. c. Y is not a continuous random variable, so the earlier result do not hold. 4.8 a. The constant k = 6 is required so the density function integrates to 1. b. P (.4 Y 1) = .648. c. Same as part b. above. d. P ( Y .4 | Y .8) = P ( Y .4)/ P ( Y .8) = .352/.896 = 0.393. e. Same as part d. above.
Chapter 4: Continuous Variables and Their Probability Distributions 61 Instructor’s Solutions Manual 4.9 a. Y is a discrete random variable because F ( y ) is not a continuous function. Also, the set of possible values of Y represents a countable set. b. These values are 2, 2.5, 4, 5.5, 6, and 7. c. p (2) = 1/8, p (2.5) = 3/16 – 1/8 = 1/16, p (4) = 1/2 – 3/16 = 5/16, p (5.5) = 5/8 – 1/2 = 1/8, p (6) = 11/16 – 5/8 = 1/16, p (7) = 1 – 11/16 = 5/16. d. P ( Y 5 . φ ) = F ( 5 . φ ) = .5, so 5 . φ = 4. 4.10 a. F ( 95 . φ ) = φ 95 . 0 ) 1 ( 6 dy y y = .95, so 95 . φ = 0.865. b. Since Y is a continuous random variable, y 0 = 95 . φ = 0.865. 4.11 a. [] 1 2 2 / 2 0 2 0 2 = = = c cy cydy , so c = 1/2. b. 4 0 2 2 ) ( ) ( y y t y dt dt t f y F = = = , 0 y 2. c. Solid line: f ( y ); dashed line: F ( y ) 0.0 0.5 1.0 1.5 2.0 0.0 0.2 0.4 0.6 0.8 1.0 y d. P (1 Y 2) = F (2) – F (1) = 1 – .25 = .75. e. Note that P (1 Y 2) = 1 – P (0 Y < 1). The region (0 y < 1) forms a triangle (in the density graph above) with a base of 1 and a height of .5. So, P (0 Y < 1) = 2 1 (1)(.5) = .25 and P (1 Y 2) = 1 – .25 = .75.

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62 Chapter 4: Continuous Variables and Their Probability Distributions Instructor’s Solutions Manual 4.12 a. F (– ) = 0, F ( ) = 1, and F ( y 1 ) – F ( y 2 ) = 2 1 2 2 y y e e > 0 provided y 1 > y 2 . b. 2 3 . 1 ) ( 3 . φ = e F = .3, so 3 . = ) 7 ln(. = 0.5972. c. 2 2 ) ( ) ( y ye y F y f = = for y 0 and 0 elsewhere. d. P ( Y 200) = 1 – P ( Y < 200) = 1 – P ( Y 200) = 1 – F (2) = e –4 .
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HW7 - Chapter 4 Continuous Variables and Their Probability...

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