chapter 1

Chapter 4: Continuous Random Variables and Probability Distributions113.a. Y = -ln(X) ⇒ x = e -y = k(y), so k′(y) = -e -y . Thus since f(x) = 1,g(y) = 1 ⋅ | -e -y | = e -y for 0 < y < ∞, so y has an exponential distribution with parameter λ= 1.( y − µ )b. y = σZ + µ ⇒ y = h(z) = σZ + µ ⇒ z = k(y) =result follows easily.σand k′(y) =σ1 , from which thec. y = h(x) = cx ⇒ x = k(y) =cy and k′(y) =c1 , from which the result follows easily.114.a. If we let α = 2 and β = 2σ, then we can manipulate f(v) as follows:ν 2 22 2/ 2 2 / 2 2 2 1 ( / 2 ) 21 ( ν−νσ−νσ− − ν σ α α−− )( )2βf ν = e = νe= ν e = ν e ,222ασ2σ( 2σ)βwhich is in the Weibull family of distributions.−= 25 νν2−⎜− vF e800 d ν ; cdf:2 800F( ;2, 2 ) = 1−e⎛ ν⎟⎞⎝ σ ⎠ν σ= 1−e, so0400b. ( ν ) ∫F−625800( 25;2,2) = 1−e= 1−.458= . 542115.33651a. Assuming independence, P(all 3 births occur on March 11) = ( ) = . 0000000233651b. ( ) (365) = . 0000073c. Let X = deviation from due date. X∼N(0, 19.88). Then the baby due on March 15 was 4days early. P(x = -4) ˜ P(-4.5 < x < -3.5)⎛ −3.5⎞ ⎛ − 4.5 ⎞= Φ⎜⎟ − Φ⎜⎟ = Φ( −.18) − Φ( −.237) = .4286 −.4090= . 0196 .⎝19.88⎠ ⎝19.88⎠Similarly, the baby due on April 1 was 21 days early, and P(x = -21)⎛ −20.5⎞ ⎛ − 21.5⎞˜ Φ ⎜ ⎟ − Φ⎜⎟ = Φ( −1.03) − Φ( −1.08) = .1515 −.1401= . 0114 .⎝ 19.88 ⎠ ⎝ 19.88 ⎠The baby due on April 4 was 24 days early, and P(x = -24) ˜ .0097Again, assuming independence, P( all 3 births occurred on March 11) =. 0196 .0114 .0097 = .( )( )( ) 00002145d. To calculate the probability of the three births happening on any day, we could makesimilar calculations as in part c for each possible day, and then add the probabilities.172