Feb 21 Homework Solutions Math 151, Winter 2012 Chapter …

1 Feb 21 Homework SolutionsMath 151, Winter 2012 Chapter 5 Problems (pages 224-227)Problem 5A filling station is supplied with gasoline once a week. If its weekly volume of sales inthousands of gallons is a random variable with probability density functionf(x) ={5(1 x)40< x <1,0otherwisewhat must the capacity of the tank be so that the probability of the supply s being exhaustedin a given week want to find the capacityaso thatP{X a} 1 .01 =.99. For 0 a 1, wehaveP{X a}= a05(1 x)4dx= 1 (1 a)5,Hence we needP{X a} .99= 1 (1 a)5 .99= (1 a)5 .01= a 1 (.01)1/5 the capacity of the tank must be at least 6019 6 ComputeE[X]ifXhas a density function given by(a)f(x) ={14xe x/2x >00otherwiseE[X] = xf(x)dx= 014x2e x/2dx=( 12x2 2x 4)e x/2 0= 4.}}

2 (b)f(x) ={c(1 x2) 1< x < [X] = xf(x)dx= 1 1cx(1 x2)dx= 0sincef(x) is an even function. Note that the solution does not depend (c)f(x) ={5x 2x >50x 5E[X] = xf(x)dx= 55x 1dx= 5 log(x) 5= .Problem 10 Trains headed for destination A arrive at the train station at 15-minute intervals startingat 7 , whereas trains headed for destination B arrive at 15-minute intervals startingat 7:05 (a)If a certain passenger arrives at the station at a time uniformly distributed between7 and 8 and then gets on the first train that arrives, what proportion of thetime does he or she go to destination A?The passenger takes the train to destination A if he arrives between 7:05 and 7 , between 7:20 and 7:30 , between 7:35 and 7:45 , or between 7:50 and8 Thus the passenger takes the train to destination A with probability (10 +10 + 10 + 10)/60 = 40/60 = 2/3.}}

3 In other words, the passenger takes the train todestination A two-thirds of the time.(b)What if the passenger arrives at a time uniformly distributed between 7:10 and 8 passenger takes the train to destination A if he arrives between 7:10 and 7 , between 7:20 and 7:30 , between 7:35 and 7:45 , between 7:50 and 8 ,or between 8:05 and 8:10 Thus the passenger takes the train to destinationA with probability (5 + 10 + 10 + 10 + 5)/60 = 40/60 = 2/3. In other words, thepassenger still takes the train to destination A two-thirds of the 11A point is chosen at random on a line segment of lengthL. Interpret this statement, andfind the probability that the ratio of the shorter to the longer segment is less than1 point is chosen at random on this line means that we can represent the chosen pointby a continuous random variableXtaking values between 0 andL.

4 SoXis the distancefrom a fixed endpoint of the line segment to the chosen point. The random variableXhas a uniform probability density functionf(x) = 1/Lfor 0 x L. If we chose thepoint atX, the line is divided into two pieces of lengthXandL X. The ratio of theshorter to the longer segment is the smaller number ofX/(L X) or (L X)/X. WehaveX/(L X)<1/4 X < L/5 and (L X)/X <1/4 X >4L/5. Note that theeventsX < L/5 andX >4L/5 are disjoint. Hence the probability that the ratio of the2shorter to the longer segment is less than 1/4 isP{X/(L X)<1/4 or (L X)/X <1/4}=P{X < L/5 orX >4L/5}=P{X < L/5}+P{X >4L/5}= L/50(1/L)dx+ L4L/5(1/L)dx= 1/5 + 1/5= 2 15 IfXis a normal random variable with parameters = 10and 2= 36, compute(a)P{X >5}.Recall thatP{X a}= ((a )/ ) = ((a 10)/6) where (x) is the cumulativedistribution function of the standard normal random variable.

5 Using Table onpage 201, we haveP{X >5}= 1 P{X <5}= 1 ( 5/6) = (5/6) (.83) .7967.(b)P{4< X <16}.P{4< X <16}=P{X <16} P{X 4}= (1) ( 1) = 2 (1) 1 2(.8413) 1 =.6826.(c)P{X <8}.P{X <8}= ( 1/3) = 1 (1/3) 1 (.33) 1 .6293 =.3707(d)P{X <20}.P{X <20}= (5/3) ( ) =.9525.(e)P{X >16}.P{X >16}= 1 P{X <16}= 1 (1) 1 .8413 =. 16 The annual rainfall (in inches) in a certain region is normally distributed with = 40and = 4. What is the probability that, starting with this year, it will take over 10years before a year occurs having a rainfall of over 50 inches? What assumptions are youmaking?Assume the annual rainfall in different years are independent. LetXdenote the annualrainfall in a given year and recall thatP{X a}= ((a )/ ) = ((a 40)/4) where (x) is the cumulative distribution function of the standard normal random probability that it will take over 10 years before a year occurs having a rainfall ofover 50 inches is(P{X 50})10= ( ((50 40)/4))10= ( ( ))10 (.)

6 9938)10 . 23 One thousand independent rolls of a fair die will be made. Compute an approximationto the probability that the number 6 will appear between 150 and 200 times the number 6 appears exactly 200 times, find the probability that number 5 will appearless than 150 the number of times the number 6 appears. ThenXis a binomial randomvariable withp= 1/6 andn= 1000. By the DeMoivre-Laplace limit theorem,P{150 X 200}=P{150 np np(1 p) X np np(1 p) 200 np np(1 p)}=P{150 1000/6 1250/9 X np np(1 p) 200 1000/6 1250/9} P{ X np np(1 p) } ( ) ( ) .9977 (1 .9207)=. that the number 6 appears exactly 200 times, the number 5 appears less than150 times if it does so in the remaining 800 trials. Represent the number of times thenumber 5 appears in the remaining 800 trials by a binomial random variableYwithp= 1/5 andn= 800, since in the remaining 800 trials only the numbers between 1 and5 appear.

7 Thus, given that the number 6 appears exactly 200 times, the probability thatthe number 5 will appear less than 150 times isP{Y <150}=P{0 Y <150}=P{0 np np(1 p) Y np np(1 p) 150 np np(1 p)}=P{0 160 128 Y np np(1 p) 150 160 128}=P{ Y np np(1 p) .8839} ( .8839) ( ) (1 .8106) 0=. 34 Jones figures that the total number of thousands of miles that an auto can be driven be-fore it would need to be junked is an exponential random variable with parameter1 has a used car that he claims has been driven only 10,000 miles. If Jones purchasesthe car, what is the probability that she would get at least 20,000 additional miles out of4it? Repeat under the assumption that the lifetime mileage of the car is not exponentiallydistributed but rather is (in thousands of miles) uniformly distributed over(0,40).

8 LetXdenote the total number of thousands of miles that the auto is driven beforeit needs to be junked. We want to computeP({X 30}|{X 10}). AssumingXis anexponential random variable with parameter 1/20, we haveP{X 30}= 30120e x/20dx= 3/2e tdt=e 3/2,P{X 10}= 10120e x/20dx= 1/2e tdt=e 1/2,P({X 30}|{X 10}) =P{X 30}/P{X 10}=e 3/2/e 1/2=e 1 . that we could have saved some work here by using the memoryless property of theexponential random uniformly distributed over (0,40), we haveP{X 30}= 4030140dx= 1/4,P{X 10}= 4010140dx= 3/4,P({X 30}|{X 10}) =P{X 30}/P{X 10}= (1/4)/(3/4) = 1 35 The lung cancer hazard rate (t)of at-year-old male smoker is such that (t) =.027 +.00025(t 40)2fort that a 40-year-old male smoker survives all other hazards, what is the probabilitythat he survives without contracting lung cancer to(a) age 50?

9 LetXdenote the life-time of the smoker. ThenP({X >50}|{X >40}) =1 F(50)1 F(40).We know from equation ( ) on page 213 thatF(x) = 1 exp( x0 (t)dt).5 Hence the conditional probability that he survives to age 50 without cancer isP({X >50}|{X >40}) =1 F(50)1 F(40)=exp( 500 (t)dt)exp( 400 (t)dt)= exp( 5040 (t)dt)= exp( 5040(.027 +.00025(t 40)2)dt)= exp( .027 10 .00025(10)33) .7023.(b) age 60?Using the same steps as in part (a), the conditional probability that he survives toage 60 without cancer isP({X >60}|{X >40}) =1 F(60)1 F(40)=exp( 600 (t)dt)exp( 400 (t)dt)= exp( 6040 (t)dt)= exp( 6040(.027 +.00025(t 40)2)dt)= exp( .027 20 .00025(20)33) . 5 Theoretical Exercises (page 227-229)Problem 5 Use the result that, for a nonnegative random variableY,E[Y] = 0P{Y > t}dtto show that, for a nonnegative random variableX,E[Xn] = 0nxn 1P{X > x} lettingY=Xn, we know thatE[Xn] = 0P{Xn> t} the change of variablest=xnanddt=nxn 1dx, we get thatE[Xn] = 0nxn 1P{Xn> xn} unis an increasing function ofu 0, the events{Xn> xn}and{X > x}areidentical.

10 So we ve shownE[Xn] = 0nxn 1P{X > x}dx= 0nxn 1P{X > t}dt,as 8 LetXbe a random variable that takes on values bewteen0andc. That is,P{0 X c}= 1. Show thatVar(X) c2 a random variable that takes on values bewteen 0 andc, we know thatf(x) = 0 ifx 0 orx c. ThusE[X2] = c0x2f(x)dx c0cxf(x)dx=cE[X].HenceVar(X) =E[X2] E[X]2 cE[X] E[X]2=c2( 2),where =E[X]/c. We know 2attains its maximum value of 1/4 at = 1/2, soVar(X) c2( 2)| =1/2=c2(1/2 1/4) =c2 10 Letf(x)denote the probability density function of a normal random variable with mean and variance 2. Show that and + are points of inflection of this is, show thatf (x) = 0whenx= orx= + .Recall thatf(x) =1 2 e (x )2/2 the derivative with respect toxtwice, we getf (x) = (x ) 2 3e (x )2/2 (x) = 2+ (x )2 2 5e (x )2/2 (x) = 0 2+ (x )2= 0 x= orx= + , as claimed.