Solution to Homework 5

Transcription

1 Solution to Homework. [ -2] A system consisting of one original unit plus a spare can function for a random amount of time X. If the density of X is given (in units of months) by { Cxe x/2 x > f(x) = x what is the probability that the system functions for at least months? First we need to find the value of C. Since f(x)dx =, we have = Cxe x/2 dx = C = C (( 2xe x/2 ) = C ( 4e x/2 = 4C. xe x/2 dx ) 2e x/2 dx) integration by parts Therefore, C = /4. Then P(X ) = 4 xe x/2 dx = ) (( 2xe x/2 2e x/2 dx) 4 ) (e /2 4e x/2 integration by parts = 4 = 4 4 e /2 = [ -4] The probability density function of X, the lifetime of a certain type of electronic device (measured in hours), is given by f(x) = x 2 x > x (a) Find P(X > 2). (b) What is the cumulative distribution function of X? (c) What is the probability that, of 6 such types of devices, at least 3 will function for at least hours? What assumptions are you making?

2 (a) P(X > 2) = 2 x 2 dx = x 2 = 2 =.. (b) Since P(X x) = x dt = t2 t x = x, provided x >, the cumulative distribution function of X is given by x > F(x) = x x (c) For each device, P(X ) = x 2 dx = x = 2 3 =.667. Let Y be the number of devices which function for at least hours among those 6. If we assume that the devices works independently, then Y Binomial(6, 2/3). Therefore P(Y 3) = P(Y = 3)P(Y = 4)P(Y = )P(Y = 6) ( )( ) 3 ( 6 2 = ( )( ) 4 ( ) ( )( ) ( ( )( ) 6 ( ) = ) 6 4 ) [ -] A filling station is supplied with gasoline once a week. If its weekly volume of sales in thousands of gallons is a random variable with probability density function { ( x) 4 < x < f(x) = otherwise what must the capacity of the tank be so that the probability of the supplys being exhausted in a given week is.? Let c be the capacity of the tank and X be the weekly volume of sales. Then the event the supplys being exhausted in a given week is same as 2

3 {X c}. Thus we need to find c such that P(X c) =.. Since f(x) = whenever x, we may assume c <. Then P(X c) = Therefore, c =. =.69. c ( x) 4 dx = ( x) = ( c). 4. [ -6] Compute E[X] if X has a density function given by f(x) = x 2 x >. x c E[X] = xf(x)dx = = ln x =. x x 2 dx = x dx This is an example of infinite expected value. In this case, we say the random variable does not have finite expectation.. [ -2] A bus travels between the two cities A and B, which are miles apart. If the bus has a breakdown, the distance from the breakdown to city A has a uniform distribution over (,). There is a bus service station in city A, in B, and in the center of the route between A and B. It is suggested that it would be more efficient to have the three stations located,, and miles, respectively, from A. Do you agree? Why? Let X denote the distance to city A when the bus breaks down and Y denotethedistancetothenearestservicestation. ThenX Unif(,). If the three stations located,, and miles, respectively, from A, then X < X < Y = g (X) = X X. X < X < 3

4 Thus E[Y] = E[g (X)] = ( x) dx [ x dx = xdx ] ( x)dx [ ( = ) 2 (x 2 ) ] = 2.. ( x)dx (x 2 ) x dx (x )dx ( ) 2 x On the other hand, if the three stations located,, and miles, respectively, from A, then X < X < Y = g 2 (X) = X X X < X < Thus E[Y] = E[g (X)] = [ = = x dx ( x)dx (x )dx [ ( x 2 ( 2 x = 9.3. x dx (x )dx ( x)dx ) ( ) 2 x ) (x 2 ) x dx ( x)dx ] (x )dx (x 2 ) ( 2 x Therefore, the second plan for the service locations is better. ) ] 4

5 6. [ -8] Suppose that X is a normal random variable with mean. If P(X > 9) =.2, approximately what is Var[X]? Let σ denote the standard deviation of X. Then ( X.2 = P(X > 9) = P > 9 ) ( = P Z > 4 ), σ σ σ where Z is a standard normal random variable. From the standard normal table, we have P(Z >.84) =.2 (The table actually gives you P(Z.84) =.8). Therefore, σ is approximately 4/.84 = 4.76 and the variance is approximately = [ -22] The width of a slot of a duralumin forging is (in inches) normally distributed with µ =.9 and σ =.3. The specification limits were given as.9±.. (a) What percentage of forgings will be defective? (b) What is the maximum allowable value of σ that will permit no more than in defectives (meaning no more than %) when the widths are normally distributed with µ =.9 and σ? (a) Let X be the width of a randomly selected forging. Then this forging is defective if and only if X >.9 or X <.89. Thus P({a randomly selected forging is defective}) = P(X >.9 or X <.89) = P(X >.9)P(X <.89) ( X.9 = P.3 > = P(Z >.67)P(Z <.67) = 2P(Z <.67). =.9. ) P ( X.9.3 >.89.9 ).3 Here Z is a standard normal random variable. Thus 9.% of forgings will be defective. (b) Since. P({a randomly selected forging is defective}) = P(X >.9 or X <.89) = P(X >.9)P(X <.89) ( X.9 = P >.9.9 ) ( X.9 P σ σ σ = P(Z >.σ )P(Z <.σ ) = 2P(Z <.σ ), >.89.9 ) σ

6 and P(Z < 2.8) =.49 and P(Z < 2.7) =., we have σ./2.8 =.9. Therefore, the maximum allowable value of σ that will permit no more than in defectives (meaning no more than %) when the widths are normally distributed with µ =.9 and σ is [ -39] If X is an exponential random variable with parameter λ =, compute the probability density function of the random variable Y defined by Y = logx. The distribution function of Y is F Y (y) = P(Y y) = P(logX y) = P(X e y ) = e y y e e x dx = e x = exp( e y ), for all < y <. Therefore, the pdf of Y is f Y (y) = d dy F Y(y) = exp( e y )e y < y <. 9. [ -4] Find the distribution of R = Asinθ, where A is a fixed constant and θ is uniformly distributed on ( π/2,π/2). Such a random variable R arises in the theory of ballistics. If a projectile is fired from the origin at an angle α from the earth with a speed v, then the point R at which it returns to the earth can be expressed as R = (v 2 /g)sin2α, where g is the gravitational constant, equal to 98 centimeters per second squared. The distribution function of R is F R (r) = P(R r) = P(Asinθ r) = P(θ arcsin(a r)) arcsin(a r) = π dθ = θ arcsin(a r) π π/2 = arcsin(a r) π 2, π/2 for A r A and F R (r) = for r < A and r > A. Therefore, the pdf of R is f Y (y) = d dr F R(r) = Aπ A r A A 2 r2,. otherwise 6