STAT 418HW16 SolutionsSections 8.1-8.41, 2, 7, 8, 13, 15, 17, 18, 19, 21, 22; TE 71. By Chebyshev’s inequality with k =√20,P (0 < X < 40) = 1 − P (|X − 20| > 20) ≥ 1 − 1/20 = 19/202.a. By Markov’s inequality,P (X ≥ 85) ≤E(X)85= 15/17.b. By Chebyshev’s inequality with k = 10/√25,P (65 ≤ X ≤ 85) = 1 − P (|X − 75| ≥ 10) ≥ 1 − 25/100 = .75.c. V ar(¯X) = 25/n. By Chebyshev’s inequality with k =5√25/n,P (|¯X − 75| > 5) ≤2525n.Therefore n = 10.7. Let Z ∼ N (0, 1). Let Xi= {lifetime of bulb i}. Given: Xiis exponential with mean 5and variance 25.P (100Xi=1Xi> 525) ≈ PZ >525 − 500√100 × 25= P (Z > .5) = .3085.8. Let Xidenote the life of bulb i and let Ribe the time to replace the bulb. FindP (P100i=1Xi+P100i=1Ri≤ 550). BecauseP100i=1Xi+P99i=1Rihas mean 100∗5+99∗.25 = 524.75and variance 2500 + 99/48 = 2502, it follows thatP (100Xi=1Xi+99Xi=1Ri≤ 550) ≈ PZ ≤550 − 524.75√2502= P (Z ≤ .505)= .693,where Z ∼ N (0, 1) and V ar(Ri) = V ar12Unif[0, 1]= 1/48.13. Let¯X denote the average in the smaller class and¯Y denote the average in the largerclass.a. P (¯X > 80) = P¯X−7414/5>157≈ P (Z > 2.14) = .0612.b. P (¯Y > 80) = P¯Y −7414/8>247≈ P (Z > 3.43) = .0003.1c. Note that SD(¯Y −¯X) =p196/64 + 194/25 = 3.3. ThenP (¯Y −¯X > 2.2) = P ((¯Y −¯X)/3.3 > 2.2/3.3) ≈ P (Z > .67) = .2514.d. Same as in (c).15.P (10,000Xi=1Xi> 2, 700, 000) ≈ P (Z ≥ (2, 700, 000 − 2, 400, 000)/(800 · 100))= P (Z ≥ 3.75) ≈ 0.17. Let Z ∼ N (0, 1).P (X > 85) = PX − 755>105≈ P (Z > 2) = .025.18. Let Yidenote the additional number of fish that need to be caught to obtain a newtype when there are at present i distinct types. Then Yiis geometric with parameter4−i4.E(Y ) = E3Xi=0Yi= 1 + 4/3 + 4/2 + 4 = 25/3V ar(Y ) = V ar(3Xi=0Yi) = 4/9 + 2 + 12 = 130/9.Hence,a.P |Y − 25/3| >253r13009!≤ 1/10,so we can take a =25−√13003and b =25+√13003.b. When a =√1170/3,PY −253> a≤130130 + 9a2=110P Y >25 +√11703!≤ .1.19. By Jensen’s inequality,a.f00(x) = 6x ≥ 0 ∀ x; E(X3) ≥ (EX)3= 253= 15, 625.b.f00(x) = −14x−3/2< 0 ∀ x; E(√X) ≤√EX = 52c.f00(x) = −x−2< 0 ∀ x; E(log X) ≤ log EX = log 25 = 3.22d.f00(x) = e−x≥ 0 ∀ x; E(e−x) ≥ e−EX= e−2521. No. Apply Jensen’s inequality again.22.a.p = P (X ≥ 26) ≤E(X)26=2026b.p = P (X ≥ 26) = P (X − 20 ≥ 26 − 20) ≤2020 + 36= 5/14c.p = P (X ≥ 26)≤ e−26tM(t), t > 0= e−26te20(et−1), t > 0d.p = P (X ≥ 26) ≈ PZ ≥25.5 − 20√20= P (Z ≥ 1.23) = .1093e. p = .112184TE 7.P 100Y1Xi≤ a100!= P 100X1log Xi≤ 100 log a!E log Xi=166Xi=1log i = 1.1V ar(log Xi) =166Xi=1(log i)2− 1.12= 1.59 − 1.12= .358P 100X1log Xi≤ 100 log a!≈ PZ ≤100 log a − 100 ∗ 1.1√.358 ∗