Test 2 Part 1 Solutions Spring 2020

AMS 310 Spring 2020 Test 2 Part 1 Solutions Fred Rispoli Part 1 Multiple Choice (circle the best answer) or Fill in the Blank. 1. The lifetime of a certain brand of car battery has an exponential distribution with a mean of 60 months. Find the probability that a randomly selected battery of this kind lasts 50 to 70 months (round off to the third decimal place). a) 0.000 b) 0.435 c) 0.123 d) 0.311 e) None of these 2. A quality control manager for a company that manufactures aluminum water pipes believes that the product lengths of one of the pipes produced can be modeled by a uniform probability distribution over the interval 28.50 to 31.75 feet. Determine the probability that a pipe produced has length greater than 30 feet a) 0.325 b) 0.538 c) 0.308 3. Let X be a random variable with pdf f(x) = d) 0.462 5x 2 e) None of these 4 , where -1 < x < 1. Find Var(X) (round off to third decimal place). a) 0.714 b) 0.000 c) 0.286 d) 1.000 e) None of these d) 3.7 e) None of these 4. Let X be a random variable with cdf F(x) = { 0, x <2 x−2 ,2 ≤ x< 4 2 1, x ≥ 4 Find the 85th percentile for X. a) 3.2 b) 3.0 c) 2.8 5. Suppose the annual amount of snowfall X (in megatons) accumulated in a lake follows a gamma distribution with  = 10 and  = 2. Find the probability that at most 15 megatons is accumulated over one year. Round off to the fourth decimal place. Answer _________ 0.2236 This study source was downloaded by 100000819498514 from CourseHero.com on 11-06-2024 04:36:47 GMT -06:00 https://www.coursehero.com/file/68042016/AMS-310-Test-2-Part-1-Solutions-Spring-2020docx/ 6. Let the random variables X and Y have joint pdf f(x,y) = 6y, 0 < y < x < 1. Find the marginal pdf of X. a) f(y) = 6y – 6y2, 0 < y < 1 b) f(x) = 3, 0 < x < 1 c) f(y) = 6y, 0 < y < 1 d) f(x) = 3x2, 0 < x < 1 7. The random variables X and Y have the following joint probability distribution. Y f(x,y) 1 2 3 4 X Find 1 .1 .2 .10 .05 2 .05 .05 .04 .01 3 .04 .08 .10 0.0 4 .01 .04 .06 .07 P(Y ≤2∨X=3) . Express your answer as a fraction reduced to lowest terms. a) 6/11 b) 7/15 c) 3/10 d) 7/11 e) None of these 8. The random variables X and Y have the following joint probability distribution. Y f(x,y) 1 2 3 4 X 1 .1 .2 .10 .05 2 .05 .05 .04 .01 3 .04 .08 .10 0.0 4 .01 .04 .06 .07 Find Var(X). a) 2.51 b) 0.61 c) 6.18 d) 2.36 e) None of these Programmed Incorrectly in Blackboard Var(X) = E(X2) – E(X)2 = 6.46 – 2.362 = 0.89 This study source was downloaded by 100000819498514 from CourseHero.com on 11-06-2024 04:36:47 GMT -06:00 https://www.coursehero.com/file/68042016/AMS-310-Test-2-Part-1-Solutions-Spring-2020docx/ 9. In the United States, the mean and standard deviation of adult men’s heights are 70 inches (5 feet 10 inches) and 4 inches, respectively. Suppose the American adult men’s heights have a normal distribution. The probability that a randomly chosen American man is taller than 6 feet (72 inches) is equal to ___________ (round off to fourth decimal place, use the given table). P(X > 72) = P(Z > 0.50) = 1 – 0.6915 =0.3085 Programmed incorrectly in Blackboard 10. Suppose the random variables X and Y have joint pdf as follows: x< 2,1< y <5 {cxy 0<0 otherwise f(x,y) = Find P(1 < X < 2, 2 < Y < 3). Express your answer as a decimal with 5 decimal places. a) 0.15625 b) 0.04167 c) None of these Some Possibly Useful Formulas: Geometric(p): p(i) = (1 − p)i−1 p, i = 1, 2,.... , Mean is 1/p, variance is (1-p)/p2 Binomial(n,p): p(i) = (ni ) p (1− p) Poisson(λ): p(i) = e− λ Hypergeometric: f(x) = i λi i! n−i , i = 0, 1,..., n, Mean is np, variance is np(1-p) , i = 0, 1, 2,... , Mean is , variance is  −a (ax)( Nn−x ) ( Nn ) , Mean is na N This study source was downloaded by 100000819498514 from CourseHero.com on 11-06-2024 04:36:47 GMT -06:00 https://www.coursehero.com/file/68042016/AMS-310-Test-2-Part-1-Solutions-Spring-2020docx/ Uniform( a, b): f(x) = 1 , Mean is (a + b)/2, variance is (b−a)2/12. b−a Normal(µ, σ2): f(x) = 1 e−( x−μ ) / 2 σ √2 π σ Exponential(λ): f(x) = 2 2 1 −x/ e if x > 0. Mean is , variance is 2. β Gamma(n, λ): f(x) = λe−λx(λx)n−1 /(n−1)! , if x ≥ 0. Mean is n/λ, variance is n/λ2. cov(X,Y ) = E(XY )−E(X)E(Y); var(X) = E(X2)−[E(X)]2 ρ = Cov ( X , Y ) σ X σY This study source was downloaded by 100000819498514 from CourseHero.com on 11-06-2024 04:36:47 GMT -06:00 https://www.coursehero.com/file/68042016/AMS-310-Test-2-Part-1-Solutions-Spring-2020docx/ Powered by TCPDF (www.tcpdf.org)