STAT 418HW10 SolutionsSections 5.5-5.719, 21, 23, 27, 31, 33, 35; TE 9, 1319. Let Z = (X − 12)/2, then Z is a standard normal. Find c such that 0.10 = P [Z >(c − 12)/2]. From Table 5.1, P [Z < 1.28] = .90 and so(c − 12)/2 = 1.28 ⇒ c = 14.5621. Let X = {Height}.P (X > 74) = PX − 712.5>74 − 712.5= P (Z > 1.2) = 1 − Φ(1.2) = 1 − .8849 = .1151P (X > 77|X > 72) =P (X > 77)P (X > 72)=P (Z > (77 − 71)/2.5)P (Z > (72 − 71)/2.5)=P (Z > 2.4)P (Z > .4)=1 − .99181 − .6554= .023823. Let X6= {number of times a 6 appears in 200 rolls} and X5= {number of times a 5appears in 200 rolls}.P (150 ≤ X6≤ 200) = P (149.5 < X6< 200.5) ≈ P 149.5 − 1000/6p1000/6 ∗ 5/6< Z <200.5 − 1000/6p1000/6 ∗ 5/6!= Φ 200.5 − 166.67p5000/36!− Φ 149.5 − 166.67p5000/36!= Φ(2.87) − Φ(−1.46) = .9258.P (X5< 150|X6= 200) = P (X5< 150 in 1000-200=800 rolls) ≈ P Z <149.5 − 800/5p800/5 ∗ 4/5!= Φ 149.5 − 160p3200/25!= 1 − Φ(.93) = 1 − .8238 = .1762.27.X ∼ B(10000, 5)P (X > 5799.5) ≈ PX − 5000√2500>5799.5 − 5000√2500= P (Z > 15.99) ≈ 0,where Z ∼ N(0, 1).131.a.E (|X − a|) =ZAa(x − a)dxA+Za0(a − x)dxA=A2−a −a2AddaE (|X − a|) =2aA− 1 = 0 ⇒ a =A2.dda2E (|X − a|) =2A> 0, thereforeA2is a minimum.b.E (|X − a|) =Za0(a − x)λe−λxdx +Z∞a(x − a)λe−λxdx= a(1 − e−λa) + ae−λa+e−λaλ−1λ+ ae−λa+e−λaλ− ae−λa= a + 2e−λaλ−1λ.ddaE (|X − a|) = 1 − 2e−λa.dda2E (|X − a|) = 2λe−λa> 0.Differentiation yields that the minimum is attained at a, wheree−λa=12or a =log 2λ33. Let X = {the number of years a radio functions}.P (X > 8) = 1 − P (X ≤ 8) = 1 −(1 −e−λ8) = e−188= e−1= .3679.35. Let X = {survival time}.a.P (X > 50) = 1 − F (50) = e−R5040λ(t)dt= e−R5040.027+.00025(t−40)2dt= e−[.027t+.00025(t−40)3/3]|5040= e−.35= .7047b.P (X > 60) = 1 − F (60) = e−R6040λ(t)dt= e−R6040.027+.00025(t−40)2dt= e−[.027t+.00025(t−40)3/3]|6040= e−1.21= .29822TE 9. The final step of parts (a) and (b) use that −Z is also a standard normal randomvariable.a. P (Z > x) = P (−Z < −x) = P (Z < −x).b. P (|Z| > x) = P (Z > x) + P (Z < −x) = P (Z > x) + P (−Z > x) = 2P (Z > x).c. P (|Z| < x) = 1 −P (|Z| > x) = 1 −2P (Z > x) = 1 −2(1 −P(Z < x)) = 2P (Z < x) −1.TE 13.a. f(x) =1b−afor all x ∈ (a, b); therefore the mode is all values in (a, b).b.f(x) =1√2πσ2e−(x−µ)2/2σ2, x ∈ (−∞, ∞)log(f(x)) = log(1√2πσ2) − (x − µ)2/2σ2ddxlog(f(x)) = −(x − µ)σ2= 0 ⇒ x = µ.ddx2log(f(x)) = −1σ2< 0.c.f(x) = λe−λx, x ≥ 0ddxf(x) = −λ2e−λx< 0Because f(x) is a strictly decreasing function on x ≥ 0, f(x) is maximized at x =