STAT515, Review Worksheet for Midterm 2 Spring 2019

Transcription

1 STAT55, Review Worksheet for Midterm 2 Spring 29. During a week, the proportion of time X that a machine is down for maintenance or repair has the following probability density function: 2( x, x, f(x The cost of this down time due to lost production and maintenance is given by C + 3X + 6X 2. Find the expectation of C. Solution: E(X xf(xdx x 2( xdx (2x 2x 2 dx (x 2 23 x3 3. E(X 2 Thus, x 2 f(xdx x 2 2( xdx (2x 2 2x 3 dx ( 2 3 x3 2 x4 6. E(C E ( + 3X + 6X 2 + 3E(X + 6E(X

2 2 2. Assume X is a normal random variable with mean 2 and variance 6, and Y is a Gamma random variable with parameters α 5 and β 2. In addition X and Y are independent. Construct a box with length L X, width W 2 X, and height H Y. Let V be the volume of the box. Calculate the expected value E(V. Solution: Since X N(µ, σ 2 N(2, 6, then E(X 2 V (X + [E(X] 2 σ 2 + µ Also, Y Gamma(α, β Gamma(5, 2, then E(Y αβ 5 2. By independence, we get E(V E(LW H E(2X 2 Y 2E(X 2 E(Y

3 3 3. If a is uniformly distributed over [, 2], what is the probability that the roots of the equation are both real? x 2 + ax + a + 5 Solution: The roots are both real if and only if the discriminant is nonnegative, that is, a 2 4(a + 5 (a (a + 6 a or a 6. Since a U(, 2, we get P ((a (a 6 P (a [, 6] [, 2] [( 6 ( ] + [2 ] 6 2 ( 32 2

4 4 4. A certain brand of cereal has an average shelf life of 7 months. Assume that based on the current technology used in the manufacturing and packaging processes the standard deviation from its average shelf life is σ month. (a Use Tchebysheff s Theorem to determine whether it is likely (probability of occurrence > 66% that the contents of a cereal box will stay fresh anywhere between 5 and 9 months. Solution: Let Y be the shelf life, then we only know the expected value µ 7 and standard deviation σ. By Tchebysheff s Theorem, P (5 Y 9 P ( Y µ 2σ 75% > 66%. 22 So it is very likely. (b What is the desired standard deviation if the percentage in (a needs to improve to at least 99%? Solution: We seek the value of σ such that P (5 Y 9 99%. Tchebysheff s Theorem, P (5 Y 9 P ( Y µ 2σ σ ( 2 2 σ2 4. σ By Then we just need σ2 4 99% σ2.4 σ.2.

5 5 5. Arrivals of customers at a checkout counter follow a Poisson distribution. It is known that, during a given 3-minute period, one customer arrived at the counter. Find the probability that the customer arrived during the last 5 minutes of the 3-minute period. Solution: Let Y be the arrival time of a customer during the given 3-minute period, then Y U(, 3, and thus P (Y 25 P (Y [25, 3]

6 6 6. The lifetime of a certain electric machine is modeled by an exponential distribution with mean β years. Let the random variable Y be the lifetime of the machine. (a Find the smallest value of β for which the probability that the product will last at least 4 years is greater than or equal to.6. Solution: Let Y be the lifetime of the electric machine, then Y exp(β. Its cdf for y > is given by F (y e y/β P (Y 4 F (4 e 4/β We require P (Y 4.6, then we need e 4/β.6 β 4 ln (b Now assume that the mean of Y is 5 (that is, β 5. If Y is the lifetime of the machine, then the machine will generate a total profit of e 2Y +3 dollars. Compute the expected profit over the lifetime of the machine. You should compute all integrals but you can leave your final answer in terms of exponentials. Solution: When β 5, the pdf of Y is given by 5 f(y e y/5, y >, Thus, E ( e 2Y So the expected profit is e 2y f(ydy 5 ( 5 e 2y 5 e y/5 dy e 5 y dy 5 e 5 y E ( e 2Y + 3 E ( e 2Y

7 7 8. The following questions concern a circular dartboard of radius r. Suppose that a dart player throws uniformly at random, that is, πr, f(x, y 2 x2 + y 2 r 2, (a Show that the x coordinate and the y-coordinate of the dart cannot be independent. Solution: Consider two special events A (X.8r and B (Y.8r. It is not hard to see that P (A > and P (B >. Also, for any (x, y A B, we must have x 2 + y r 2.28r 2 > r 2, then A B is outside the circle of radius, and therefore, P (A B. If X and Y are independent, so are events A and B. However, P (A B < P (AP (B. Contradiciton!. So X and Y must be dependent. (b What is the probability that the dart is closer to the bulls-eye than the outside of the board? Solution: The probability is ( P X2 + Y 2 r/2 X 2 +Y 2 r/2 πr 2 dxdy polar 2π dθ πr 2 ( πr 2π 2 2 ρ r/2 r/2 ρdρ 4. (c If the dart is closer to the bulls-eye than the outside, what is the probability that the dart at most r away from the bulls-eye? 4 Solution: The probability is ( P X2 + Y 2 r/4 X 2 + Y 2 r/2 P ( X 2 + Y 2 r/4 P ( X 2 + Y 2 r/2 /6 /4 4. where for the numerator, we apply similar calculations as part (b.

8 8 9. Suppose that X Uniform(,. Given the value X x, we generate Y X x Uniform(x,, that is, the conditional distribution of Y given X x is uniform over the interval [x, ]. (a What is the joint density of (X, Y. Solution: By the given information, we have, x, f X (x f Y (y X x, elsewhere, Then the joint density is f(x, y f X (x f Y (y X x (b What is the marginal density of Y? Solution: f Y (y f(x, ydx (c Find the expected value E(Y. Solution: E(Y yf(x, ydxdy y x x x y,, elsewhere, x x y, dx, y,, elsewhere. ln x xy x, y, ln( y, y, ( y x dy dx x 2 2 x ( 2 y2 x 2 x dx ( + xdx 4 ( + x y yx dx

9 9. Let Y and Y 2 have joint density function e (y +y 2 f(y, y 2, if y >, y 2 >,, elsewhere (a Find the marginal density functions for Y and Y 2. (b What is P ( < Y < 2.5? (c For any y 2 >, what is the conditional density function of Y given that Y 2 y 2? (d Using Part (c to answer: Are X and Y independent or not? Solution: (a f (y f(y, y 2 dy 2 e (y +y 2 dy 2, y >, elsewhere e y, y >, elsewhere f 2 (y 2 f(y, y 2 dy e (y +y 2 dy, y 2 >, elsewhere e y 2, y 2 >, elsewhere (b P ( < Y < (c and (d: If y 2 >, f (y Y 2 y 2 f(y, y 2 f 2 (y 2 f (y dy So Y and Y 2 are independent. 2.5 e y dy e y 2.5 e e 2.5 e (y +y 2, y e y >, 2 elsewhere e y, y >, elsewhere f (y

10 . Let Y and Y 2 have joint density function, if y C, y 2 f(y, y 2, elsewhere Find C and after that compute the probability: P ( Y 2 e Y Solution: As f(y, y 2 is a bivariate pdf, we have f(y, y 2 dy dy 2. The integral of LHS is equal to C dy dy 2 dy dy 2 C C [ ( ][ ( ] 4 C So 4 C and thus C 4. P ( Y 2 e Y 4 max, e y } y 2 min,e y } ( e y e y dy 2 dy dy dy 2 ( dy 2 dy 4 2e y dy dy 4 2ey e

11 2. The lifetime of lightbulbs that are advertised to last for 4 hours are normally distributed with a mean of 45 hours and a standard deviation of 5 hours. What is the probability that a bulb lasts longer than the advertised figure? Solution: let Y be the lifetime of lightbulb, then Y N(µ, σ 2 with µ 45 and σ 5. P (Y > 4 ( 5 2π exp (y 42 dy

12 2 3. Because of the relatively high interest rates, most consumers attempt to pay off their credit card bills promptly. However, this is not always possible. An analysis of the amount of interest paid monthly by a bank s Visa cardholders reveals that the amount is normally distributed with a mean of 25 dollars and a standard deviation of 8 dollars. What proportion of the bank s Visa cardholders pay more than 33 dollars in interest? Solution: let Y be the monthly interest paid, then Y N(µ, σ 2 with µ 25 and σ 8. P (Y > 33 ( 8 2π exp (y 252 dy

13 3 4. Wires manufactured for use in a system are specified to have lengths between 2 and 4 mm. The actual wires produced by company X have a normal distribution with mean length 3mm and standard deviation.5mm. (a What is the probability that a randomly selected wire from company X s production will meet specifications? Solution: Let Y be the length of the wires, then Y N(µ, σ 2 with µ 3 and σ.5. The probability to meet specifications is P (2 Y 4 P ( Y µ 2σ 95% (b If of these wires are used in each system and all are selected from company X, what is the probability that all ten in a randomly selected system will meet the specifications? Solution: Let Z be the number of wires which meet the specifications, then Z B(n, p with n and p.95. the probability that all ten will meet the specifications is ( P (Z.95 (

14 4 5. Suppose that X is the total time between a customer s arrival in the store and departure from the service window, Y is the time spent in line before reaching the window, and the joint densities of these variables is given by e x, y x, f(x, y (a Find the marginal densities f X (x for X and f Y (y for Y. Solution: f X (x f Y (y f(x, ydy f(x, ydx x e x dy, x,, elsewhere y e x dx, y,, elsewhere xe x, x,, elsewhere e y, y,, elsewhere (b What is the conditional density of Y given that X x? Be sure to specify the values of x for which this density is defined. Solution: f Y (y X x f(x, y f X (x e x xe, < y x x x (c Are X and Y independent? Please support your answer. Solution: No, since f Y (y X x f Y (y. (d Find the expected value E ( Y X. ( Y y E f(x, ydxdy X x ( x y x e x dy dx ( e x yx x 2 y2 dx 2 y xe x dx 2 Γ(2 2 2.