chapter 1

Chapter 4: Continuous Random Variables and Probability Distributions73.( ) 16823.5+( 1.2) / 2( ) ( ) ( )a. E(X) = e = 68.0335; V(X) =2223.5 + 1.2 1.2e ⋅ e −1= 14907. ;σ x = 122.0949⎛ ln(250) − 3.5 ⎞ ⎛ ln(50) −3.5⎞b. P(50 ≤ X ≤ 250) = P ⎜ z ≤⎟ − P⎜z ≤ ⎟⎝ 1.2 ⎠ ⎝ 1.2 ⎠P(Z ≤ 1.68) – P(Z ≤ .34) = .9535 - .6331 = .3204.⎛ ln(68.0335) −3.5⎞c. P(X ≤ 68.0335) = P ⎜ z ≤⎟⎝ 1.2 ⎠distribution is not a symmetric distribution.= P(Z ≤ .60) = .7257. The lognormal74.a. .5 = F( µ ~ ⎛ ln( µ) ~ − µ ⎞) = Φ⎜⎟ , (where µ ~ refers to the lognormal distribution and µ and⎝ σ ⎠σ to the normal distribution). Since the median of the standard normal distribution is 0,ln( µ~ ) − µ= 0 , so ln(µ ~ ) = µ ⇒ µ ~ = e µ . For the power distribution,σµ ~ 3.5= e = 33. 12⎛ ln( X ) − µ⎜⎝ σb. 1 - α = Φ(z α ) = P(Z ≤ z α ) = ≤ z = P(ln(X)≤ µ + σz)= ( µ + σzααP X ≤ e ) , so the 100(1 - α)th percentile is . For the power distribution,α⎞⎟⎠µ σze +the 95 th 3.5+(1.645)(1.2) 5.474percentile is e = e = 238. 41α75.5+(.01) / 2 5.005a. E(X) = = e = 149. 15710+(.01) .01e ; Var(X) = e ⋅ ( e −1) = 223. 594b. P(X > 125) = 1 – P(X ≤ 125) =⎛ ln(125) − 5 ⎞= 1 − P ⎜ z ≤ ⎟ = 1− Φ − =⎝ .1 ⎠154( 1.72) . 9573⎛ ln(110) − 5⎞⎜ ⎟⎝ .1 ⎠c. P(110 ≤ X ≤ 125) = Φ( −1 .72) − Φ= .0427−.0013=. 0414d. µ ~ = e 5 =148. 41 (continued)e. P(any particular one has X > 125) = .9573 ⇒ expected # = 10(.9573) = 9.573f. We wish the 5 th 5+( −1.645)(.1)percentile, which is e = 125. 90