MAE Probability and Statistical Methods for Engineers - Spring 2016 Final Exam, June 8

Transcription

1 MAE 18 - Probability and Statistical Methods for Engineers - Spring 16 Final Exam, June 8 Instructions (i) One (two-sided) cheat sheet, book tables, and a calculator with no communication capabilities are allowed, (ii) No cellphones, tablets, or laptops, (iii) You have 17 minutes, (iv) Do not get stuck! Move on and come back later if you have time, (v) Justify all your answers, (vi) Do not forget to write your name and student number in your exam. Questions 1. (1 points) The lifetime T of a laptop battery until it loses 8 percent of its capacity is distributed as a lognormal random variable with a mean of years and a coefficient of variation of.5. (i) What proportion of laptop batteries have lifetimes between 1.5 and years? (ii) What is the probability that a battery has a lifetime greater than one standard deviation from the mean? (iii) Six laptops are chosen at random. What is the probability that at least a laptop battery has a lifetime greater than one standard deviation from the mean? (i) The lifetime T has µ = years and δ =.5. These descriptors are related to the parameters of the lognormal as follows: ζ = δ =.5, (as δ is smaller than.3) λ = ln µ ζ = ln 1.65 = = (+ points: one per parameter) The proportion of batteries with lifetimes between 1.5 and years is given by P (1.5 < T < ). Since T is a lognormal, the random variable X = ln T is a Gaussian with mean λ and stardard deviation ζ. Then: ( ) ( ) ln.6619 ln P (1.5 < T < ) = P (ln(1.5) < X < ln ) = Φ Φ.5.5 = Φ(.149) Φ( 1.57) =.5477 (1 Φ(1.57)) =.5477 (1.8461) = =.3938.

2 That is, the proportion of laptop batteries with lifetimes between 1.5 and years is approximately 39.38%. (+ points: formula) (ii) The standard deviation of the lognormal is = µ δ =.5 =.5. Then, one standard deviation from the mean gives a value of T = =.5. We have to find ( ) ln P (T >.5) = P (X > ln(.5)) = 1 P (X ln(.5)) = 1 Φ =.5 ( ) Φ = 1 Φ(1.175) = = (+1 point: calculation of T ) (+1 point: complementary) (iii) Six laptop batteries define a Bernouilli sequence of n = 6. Define Y the number of laptop batteries of the sequence of six with a lifetime greater than one standard deviation from the mean. Y is distributed as a Binomial with n = 6 and p = Then (from the table): P (Y 1) = 1 P (Y ) =.69.. (1 points) Customers who purchase a certain make of a car can order an engine in any of three sizes. Of all cars sold, 45% have the smallest engine, 35% have the medium-sized engine, and % have the largest. Of cars with the smallest engine, 1% fail an emissions test within two years of purchase, while this is the case in 1% of those with the medium size, and 15% with the largest engine size. (i) What is the probability that a randomly chosen car will fail an emission test within two years? (ii) Suppose that a record for a failed emissions test is chosen at random. What is the probability that is for a car with a small engine? (i) Define the events: A 1 = a purchased car has a small-sized engine, A = a purchased car has a medium-sized engine, A 3 = a purchased car has a large engine, B = a purchased car fails the emissions test within two years. Page

3 From the problem statement, we have (+1 point: event definition) P (A 1 ) =.45, P (A ) =.35, P (A 3 ) =.. In addition, we are told that (+1.5 points) P (B A 1 ) =.1, P (B A ) =.1, P (B A 3 ) =.15. (+1.5 points) Since {A 1, A, A 3 } are pairwise mutually exclusive events, and collectively exhaustive, we can use the theorem of total probability to find P (B): P (B) = P (B A 1 ) P (A 1 ) + P (B A ) P (A ) + P (B A 3 ) P (A 3 ) = = =.117. (ii) Here we use Bayes rule to find P (A 1 B): (+ points: theorem of total probability) P (A 1 B) = P (B A 1) P (A 1 ) P (B) = P (B A 1) P (A 1 ) P (B) = =.385. (+ points: Bayes theorem) 3. (8 points) A fast-food restaurant operates both a drive-through facility and a walk-in facility. On a randomly selected day, let X and Y, respectively, be the proportions of the time that the drive-through and walk-in facilities are in use, and suppose that the joint density function of these random variables is { f XY (x, y) = 3 (x + y), x < 1, y < 1,, otherwise. (i) Find the marginal density of X. (ii) Find the marginal density of Y. Are the variables X and Y statistically independent? (iii) Find the probability that the drive-through facility is busy less than half of the time. (i) The marginal density of X is given by f X (x) = 1 3 (x + y)dy = 3 (x + 1). Page 3

4 (+ points: formula + integral) (ii) The marginal density of Y is given by f Y (y) = 1 3 (x + y)dx = ( y). Since f XY (x, y) f X (x)f Y (y), the two variables are not independent. (iii) The probability that the drive-through is busy less that half the time is P (X.5) =.5 f X (x)dx =.5 (+ points: formula + integral) 3 (x + 1)dx = ( ) =.417. (+ points: formula + integral) 4. (1 points) The service stations along a highway are located according to a Poisson process in space, with an average of 1 service station in 1 miles. Because of a gas shortage, there is a probability of.3 that a service station would have no gasoline available. Assume that the availabilities of gasoline at different service stations are statistically independent. (i) What is the probability that there is at most 1 service station in the next 15 miles of highway? (ii) What is the probability that none of the next 3 stations has gasoline for sale? (iii) A driver on this highway notices that the fuel gauge in his car reads empty; from experience he knows that he can go another 15 miles. By realizing that service stations with no gas available also follow a Poisson process with a different occurrence rate, find the probability that he will be stranded on the highway without gasoline. (i) The mean occurrence rate of service stations is ν =.1 miles 1. If X 15 is the number of service stations in the next 15 miles, we have P (X 15 1) = P (X 15 = ) + P (X 15 = 1) = (.1 15) e ! (.1 15)1 e.1 15 =.56. 1! (+1 point: occurrence rate) (+ points: Poisson formula) (ii) Service stations without gas are a Bernoulli sequence, with probability p =.3 per station. The probability that none of the next n = 3 gas stations has gas for sale is given by the Page 4

5 binomial distribution: P (X = ) = ( 3 ) =.3 3 =.7. (+1 point: Bernoulli sequence) (+ points: binomial formula and numerical value) (iii) Service stations that sell gasoline occur as a Poisson process with occurrence rate ν = ν (1 p) =.1.7 =.7 miles 1. The probability that the driver will be stranded is P (X 15 = ) = (.7 15) e.7 15 =.35.! (+1 point: occurrence rate) (+ points: Poisson formula and numerical value) 5. (1 points) Semiconductor wafers at a fabrication plant are classified according to their diameter into 1 inch and 3 inch wafers. Let X 1 (resp. X ) denote the number of 1 inch (resp. 3 inch) wafers produced in a day. Assume that X 1 (resp. X ) has a mean and standard deviations of µ 1 = 11, 1 = 1 (resp. µ = 9, =.) (i) Assuming that the production of wafers types are independent processes, find the mean and standard deviation of the total number of wafers produced in a day. (Justify your answer.) (ii) Further, assume each of previous variables is distributed as a Gaussian. What is the probability that at least 19 wafers are produced in a day? (iii) How should the standard deviation of the 1 inch wafer production be lowered so that at least 15 wafers are produced in a day with 99% probability? (i) The total production of wafers is given by Y = X 1 + X. This is is a new random variable with mean µ = µ 1 + µ = =, and, since X 1 and X are independent, Y has a standard deviation = 1 + = 1 + = 3.6. Page 5

6 (ii) We are asked about ( ) 19 P (Y 19) = 1 P (Y < 19) = 1 Φ = 1 Φ(.447) = 3.6 = 1 (1 Φ(.447)) Φ(.45) = (iii) Keeping 1 as an unknown, the standard deviation of Y is a function = Using this, we set up the equation: ( ) 5 P (Y 15) = 1 P (Y < 15) = 1 Φ =.99, which implies ( ) ( ) = Φ = 1 Φ. In other words, ( ) 5.99 = Φ.33 = 5 (+ points: set up equation for ) = 5.33 = Then, we have that 1 = = , and thus 1 = (+ points: solve for and 1 using table) 6. (1 points) A new process for producing a type of resin is supposed to have a cycle time of no more than 3.5 hours per batch. Six batches were produced, and their cycle times, in hours, were: Can you conclude that the mean cycle time is 3.5 hours? Set up a one-sided test to evaluate if µ = 3.5 versus µ > 3.5 with a significance value of α =.1. Justify your answer. (i) We set up a one-sided test for the mean cycle of the mean to decide if µ > 3.5 or not. { H : µ = 3.5, H 1 : µ > 3.5. Page 6

7 (ii) Since the true variance of the population is not know, we take X = x 3.5 s/ to be distributed 6 as a t-distribution of n 1 = 5 d.o.f. (iii) The sample mean of the mean cycle times and and its sample variance are x = 1 ( ) = s = 1 5 ( ( ) + ( ) + ( ) + ( ) =.76 (iv) We compute the test statistic assuming that H z = = =.4..76/ ( ) + ( ) ) (v) We find t 1 α,n 1 = t.99,5 = , which defines the region of rejection. (vi) Since t.99,5 = > z =.4, then the null hypothesis is accepted, or in other words, µ = (8 points) Oven thermostats were tested by setting them to 35 o F and measuring the actual temperature of the oven. In a sample of 67 thermostats, the average temperature was 348. o F. If the standard deviation of the population is known to be = 5.1 o F do the following: (i) Find a two-sided 95% confidence interval for the mean oven temperature. (ii) How many thermostats must be sampled so that the 9% confidence interval specifies the mean within ±.8 o F? (i) The sample temperature T of the oven is normally distributed with sample mean t = 348. o F and a known standard deviation = 5.1 o F. We need a two-sided confidence interval with Page 7

8 1 α =.95. Then, 1 α =.975 and α =.5 and we compute ( ) < µ T >.95 = 348. k.975, k ( = ) 5.1, = (347, 349.4). (+1 point: find 1 α/ and α/) (+ points: critical values) (+1 point: confindence interval) (ii) For the two-sided confidence interval with 1 α =.9, then 1 α =.95. The number of samples n that need to be taken to guarantee a 9% confidence interval with width w =.8 is w =.8 = k n = n n = = Therefore, we can take n = 111. (+1 point: find 1 α/ and α/) (+ points: w in terms of n) (+1 point: value of n) Page 8