# Review Exam Suppose that number of cars that passes through a certain rural intersection is a Poisson process with an average rate of 3 per day.

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1 Review Exam 2 This is a sample of problems that would be good practice for the exam. This is by no means a guarantee that the problems on the exam will look identical to those on the exam but it should give you an idea of the level of difficulty and the scope of the topics that will be covered. The problems are not meant to give you an exact approximation of the length of the exam. On your exam there will be space for you to complete the problems. Make sure that in each case where you define a random variable to state clearly what the random variable is, what type it is and what the associated parameters are. Suppose that number of cars that passes through a certain rural intersection is a Poisson process with an average rate of 3 per day. a What is the probability that exactly 4 cars will pass through the intersection in any 2 day period? Let X equal the number of cars passing through in a 2 day period. Since this is a Poisson process then X is Poisson random variable with λ = 2 3 = 6. We then wish to compute P X = 4 = e ! b What is the probability that at least two cars will pass through during any 8 hour period? For this problem we define a new random variable Y that is the number of cars passing through during an 8 hour period. Again this is a Poisson random variable but this time the parameter λ = 3/3 = Since 8 hours is one third of a day. Consequently we wish to find P Y 2 and one computes that as follows: P Y 2 = P Y = p p = e /! e / = 2 e c Some repairs need to be made to the pavement of the intersection. Suppose that exactly 5AM the a car passes through the intersection and immediately after the intersection is closed for 2 hours for repairs. What is the probability that the next car arriving at the intersection will arrive after the the intersection has been reopened at 5PM? For this problem we define the random variable Z =the elapsed time between two cars. Since this is a Poisson process, Z is an exponential random variable with λ = α = 3. Which means that the pdf for Z is given by fz = 3e 3z for z >. Our question then asks us to compute P Z >.5. Note that 2 hours is one half of the day To do this we set up the integral: P Z >.5 =.5 3e 3z dz = e 3z.5 = e 3.5 = e.5

2 2. A company purchases two identical components to be used in a particular machine. The first one is used until it fails where upon the second one immediately replaces it. Assume that the lifetimes of the components are independent and identical exponential random variables with the average lifetime of each component being 4 years. What is the probability that the company will not have to purchase another component in the next 6 years? Here we can define: X = {the time until the first component fails} Y = {the time until the second component fails} Since the random variable are identical, independent, exponential random variables with mean 4 then the marginal densities are f X x = 4 e.25x for x > and f Y y = 4 e.25y for y >. Note that the average is equal to /λ so λ = /4. Since the variables are independent the joint density is just the product of the two marginal densities or, fx, y = 6 e.25x+y for x > and y >. We are asked to compute the probability that the company will not have to purchase another component in the next 6 years. This is the equivalent to saying what is the probability that the sum of X and Y is greater than 6. To do this we must think of the region in the plane described by X + Y > 6. This is the region above the line Y = 6 X and also with X and Y positive. It is a little easier to think of this as the compliment where X + Y 6 The integral we set up is then: P X + Y > 6 = P X + Y 6 = Evaluating the first integral we have, 6 P X+Y 6 = or or 6 e.25x 4e.25y 6 x 6 6 x 6 dx = 6 e.25x e.25y dydx 6 e.25x 4 4e.256 x dx 6 P X + Y 6 = 4 e.25x 4 e.256 dx = e.25x.25xe.5 6 P X +Y 6 = [ e.5.5e.5 ] = [ +.5e.5 + ] =.5e

3 3. Five individuals from an animal population thought to be near extinction in a certain region have been caught, tagged, and released to mix into the population. After they have had an opportunity to mix, a random sample of of these animals is selected. Let X = the number of tagged animals in the second sample. If there are actually 25 animals of this type in the region. a What is the probability that X = 2? For this problem X is hypergeometric with parameters N = 25, M = 5 and n =. Thus the pmf would be equal to: 5 2 x Thus, px = x P X = 2 = p2 =, x 5 b What is the probability that X is at least 3? Using the same density function we have that, 2 7 P X 3 = p3 + p4 + p5 =

4 4. Suppose that 2 discrete random variables Y and Z have the following joint probability mass function py, z z py, z /2 /3 /4 /5 / y a Compute the marginal probability mass function for Y. To do this we recall that P Y y = py, z. In other words we add up the rows of the chart above: So: z P Y = z p, z = =.5 Similarly we get:.5 y =.35 y = 2 P Y y =.5 y = 4 otherwise b Find a formula for the cumulative distribution function for Y and give a rough sketch of the function. Recall that the cdf for Y is the function F y = P Y y. Since this is a discrete function this will look like a step like function. THe formula for that is: y <.5 y < 2 F y =.85 2 y < 4 4 y c Find the probability, P Y Z. To find this probability we add up all the cases where Y Z is less than or equal to. Namely: P Y Z = p, + p, /2 + p, /3 + p, /4 + p, /5 + p, /6 + p2, = =.55 d Find the expected value of Y Z. To compute the expected value we recall that: EY Z = y zpy, z y z One easy way to begin this problem is to make a chart for the values of y z.

5 z y z /2 /3 /4 /5 /6 - -/2 -/3 -/4 -/5 -/6 y 2 3/2 5/3 7/4 9/5 / /2 /3 5/4 9/5 23/6 Multiplying these values times the values in the table for py, z and adding them up gives: EY Z = y y zpy, z = 79/8 = Suppose that the cdf for a continuous random variable X is given by: x < F x = k + 2x x 3 x 2 x > 2 z a Find k so that this is a true cdf. Since F x here is a cdf we need that the cdf should be continuous. To make this work we must choose k so that F = and F 2 = and that the limit from either side of - and 2 is the same. The condition at is automatically satisfied and the we just must choose k so that k =. Thus k = /27. b Find the pdf fx associated to X and give a rough sketch of the graph of fx. To find the pfd from the cdf we differentiate, consequently we have that: x < fx = x 2 x 2 x > 2 c Find the expected value EX and the variance V X. To find the expected value we must recall that EX = xfxdx. For this problem this reduces to: 4 EX = 9 x 2 9 x3 dx = 9 x x4 = = To find the variance we begin with computing: EX 2 = 4 9 x2 4 9 x4 dx = 27 x x5 = =.6 45 Now to compute V X we must use the short cut formula that: V X = EX 2 EX 2 = = 43 8 =.5375

6 d Find the probability P X >. For this problem we set up the we just want to integrate over the region larger than. This reduces to: 4 4 P X > = fxdx = 9 9 x2 dx = 9 x x3 2 8 = = A certain detector is made up of 6 individual sensors. Each sensor is works 4% of the time independent of any other sensor. a What is the probability that exactly 3 of the 6 sensors fail? Give an exact answer. For this problem we assume that any sensor is independent of any other sensor. Thus we can define the random variable: X = { The number of defective sensors among 6} X is then binomial with n = 6 and p =.6, thus the pdf is px = 6 x.6 x.4 6 x for x 6. Consequently to find the probability that exactly 3 fail we have: 6 P X = 3 = p3 = b Suppose that the detector shuts off if more then 35 sensors fail. Find the approximate probability that the detector will shut off. To find this probability we must approximate X with a normal random variable. This is justified since np =.66 = 36 > and n p = 6.4 = 24 >. Before we proceed we note that µ = np = 36 and σ = np p = We now proceed to approximate P X > 35. We begin with a continuity correction by computing P X > 35.5: P X > 35.5 = P X > P Z >.38 = Φ =

7 7. Suppose that two plants share the same root system. Let X equal the height of the first plant in feet at a certain time and let Y be the height of the second plant in feet. Suppose that the joint probability density function for X and Y is given by: fx, y = { 6 25 x 2 y + 2x < x < 2, < y < otherwise a What is the marginal probability density associated to X? f X x = 6 3 fx, ydy = 25 x2 y + 2xdy = 25 x2 y xy = 3 25 x x = 3 25 x x, < x < 2 b What is the expected height of the first plant? EX = 3 x 25 x x dx = x x3 = =.57 c On average what is the difference in the heights of the plants? Notice that the first plant is always at least as tall as the second plant. The difference in height of the plants can be represented by X-Y, so the question asks us to compute the expected value of this quantity. EX Y = 6 25 x yx2 y+2xdydx = x3 y 2 3 x2 y 3 + 2x 2 y xy 2 = 6 25 = x4 9 x x3 2 x2 x 3 y x 2 y 2 + 2x 2 2xy dydx dx = x3 3 x2 + 2x 2 x dx 2 = 6 [ ].233 2

8 Additional Problems. A random variable has a normal distribution with σ =. If the probability that the random variable will take on a value greater than 82.5 is.788, what is the probability that it will take on a value greater than 58.3? 2. The joint density function of X and Y is given by { kx < x <, < y < x fx, y = otherwise a Determine if X and Y are independent. b Find P Y X > Suppose that p = Pmale birth =.5. A couple wishes to have exactly one female child in their family. They will have children until this condition is fulfilled. a What is the probability that the family has 6 children? b What is the probability that the family has at most five children? c How many children would you expect this family to have? 4. Suppose that the height of a randomly selected college student is normally distributed with a mean of 66 inches and a standard deviation of 3. a What is the probability that a randomly chosen students height is between 62.5 and 67.5 inches? b If a classroom has 3 college students what is the probability that there are exactly 4 that are between 62.5 and 67.5 inches?

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