University:
Stony Brook University
Course:
AMS 310 | Survey of Probability and Statistics
Academic year:

2020
Views:

338

Pages:

9
Author:

study_genius

AMS 310 Spring 2020 Practice Test 2 Solutions Fred Rispoli Part 1 Multiple Choice, True/False and Fill in the Blank. 1. Let X1, X2, … X100 be a random sample from a distribution with pdf 3 x2 f ( x )= 2 + x , 0 ≤ x ≤1 0, otherwise { Find the mean of X. 3 8 None of these a) b) 1 3 c) 17 24 d) 1 2 e) 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 29.50 to 30.05 feet. Set up the correct integral to determine the probability that a pipe produced has length between 29.75 and 29.90 feet 30.05 1 )dx a) ∫ ( 29.50 0.55 None of these 29.90 1 )dx b) ∫ ( 29.75 0.55 29.90 1 ) dx c) ∫ ( 29.75 30.05 29.90 d) 1 )dx ∫ ( 29.50 29.75 3. The random variables X and Y have the following joint probability distribution. Y f(x,y) X Find 1 2 3 4 1 .1 .2 .10 .05 2 .05 .05 .04 .01 3 .04 .08 .10 0.0 4 .01 .04 .06 .07 P( X ≤ 2∨Y =3) . This study source was downloaded by 100000819498514 from CourseHero.com on 11-06-2024 04:33:10 GMT -06:00 https://www.coursehero.com/file/67998408/AMS-310-Practice-Test-2-Spring-2020-Solutionsdocx/ d) e) a) .12/.22 b) .14/.30 c).57 d) 0.22 e) None of these 4. 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 E(X). a) 2.50 b) 2.13 c) 2.60 d) 2.06 e) None of these 5. 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 True or false: Are X and Y independent. a) True b) False 6. Let X be a random variable with cdf This study source was downloaded by 100000819498514 from CourseHero.com on 11-06-2024 04:33:10 GMT -06:00 https://www.coursehero.com/file/67998408/AMS-310-Practice-Test-2-Spring-2020-Solutionsdocx/ F(x) = { 0, x <2 x−2 ,2 ≤ x< 4 2 1, x ≥ 4 Find the 60th percentile for X a) 3.2 b) 3.0 c) 2.8 d) 3.5 e) None of these 7. Fill in the blank: Three random variables X, Y and Z are independent. Each has a binomial distribution with success probability 0.3 and 5 trials. The joint probability distribution function f(x,y,z) is given by ______________________ (5x) 0 .3 .7 x 5− x ( 5y )0.3 .7 y ) ¿ 5− y (5z ) 0.3 .7 z ) ¿ 5− z ) ¿ 8. 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 ____0.31_______ (round off to second decimal place). 9. The joint pdf of X and Y is given as where a is a constant. The integer value of a is given by __a = 1____. 10. Suppose the random variables X and Y have joint pdf as follows: Find the conditional pdf f(y x). 3y a) 3 x 2 , 0 23) = 1 – P(X ≤ 23) = 1 – P(Z ≤ 1) = 1 – (1) = 1 – .8413 = 0.1587 b) P(22 < X < 25) = P(.67 < Z < 1.67) = (1.67) - (.67) = .9525 - .7486 = 0.2039 z= 22−20 3 = 0.67 z= 25−20 3 = 1.67 c) Set F(x) = 0.90 Use the table in reverse to get Z = 1.28, then solve 1.28 = x−20 3 to get x=20+3(1.28) = 23.84 2. Suppose the random variables X and Y have joint pdf as follows: cx+1 x , y ≥ 0∧x + y <1 f(x,y) = 0 otherwise { a) Find the constant c b) Find the marginal pdf f1(x) of X 1 1 ,Y< 2 2 c) Find P(X < ) a) 1 1−x ∫ ∫ ( cx+1 ) dydx 0 0 1 = ∫ [x ( 1−x ) +1−x ]dx 0 This study source was downloaded by 100000819498514 from CourseHero.com on 11-06-2024 04:33:10 GMT -06:00 https://www.coursehero.com/file/67998408/AMS-310-Practice-Test-2-Spring-2020-Solutionsdocx/ =[ = 2 3 2 1 cx c x x ] − + x− 0 2 3 2 c 1 + 6 2 Set this equal to 1 to get c = 3. b) 3 x+1 (¿)dy 1− x = [3xy + y] ∫¿ 1−x 0 = 3x(1-x) + 1 – x = (1-x)(3x + 1), for 0 < x < 1 0 c) 1/ 2 1/ 2 ∫ ∫ (3 x +1) dydx 0 0 3x 2 +1 1/2 3 x2 x = (¿ 2 ) dx = [ ] = + 0 4 2 1 /2 7 16 ∫¿ 0 3. Let X and Y be independent random variables representing the lifetime (in 100 hours) of type A and B lightbulbs, respectively. Both variables have exponential distributions, and the mean of X is 2 and the mean of Y is 3. a) Find the joint pdf f(x,y) of X and Y. b) Find the conditional pdf f2(yx) c) Given the probability that a type A bulb fails at 200 hours, find the probability that a type B bulb fails lasts longer than 300 hours. a) By independence, 𝑓(𝑥,𝑦)=𝑓1(𝑥)𝑓2(𝑦)={(1/6)�−𝑥/2−𝑦/3, 𝑥 ≥ 0, 𝑦 ≥ 0 0, otherwise b) By independence, 𝑓2(𝑦|𝑥)=𝑓2(𝑦)=1/3�−𝑦/3, 𝑦≥0 c) By independence, 𝑃(Y>3|X =2)=𝑃(Y>3)=1−�2(3)= 1 – (1- �−3/3 ) = e−1 = 0.3679 This study source was downloaded by 100000819498514 from CourseHero.com on 11-06-2024 04:33:10 GMT -06:00 https://www.coursehero.com/file/67998408/AMS-310-Practice-Test-2-Spring-2020-Solutionsdocx/ 4. The weight of adult bottlenose dolphins was found to follow a normal distribution with a mean of 550 pounds and a standard deviation of 50 pounds. a) What percentage of bottlenose dolphins weigh from 400 to 600 pounds? b) In a random sample of 9 adult bottlenose dolphins, what is the probability that exactly 5 of them are heavier than 560 pounds? (a) Let X be the weight of an adult bottlenose dolphin. Then 400−550 600−550 P(400< X <600)=P 560 )=P Z > 560−550 =P ( Z >0.2 )=1−Փ ( 0.2 )=1−0.5793=0.4207 50 ) Let Y be the number of dolphins which are heavier than 560 pounds out of the sample of 9. Then Y Bin ( 9, 0.4207 ) . The remainder of the solution can now be done two different ways. The first is to use the binomial distribution formula. P (Y =5 )= 9 (0.4207)5 (0.5793)4 =0.1870 5 () The second way to complete is to use the normal approximation to the binomial and find P(4.5 < Y < 5.5), where  = np = 9(0.4207) = 3.7863, and  = √ 2.1934 √ 9(0.4207)(0.5793) = 1.481. Now use the normal distribution. 5. Let the random variables X and Y have joint pdf as follows: f(x,y) = 3 2 2 ( x +3 y ) , for 0 < x < 1, 0 < y < 1 4 This study source was downloaded by 100000819498514 from CourseHero.com on 11-06-2024 04:33:10 GMT -06:00 https://www.coursehero.com/file/67998408/AMS-310-Practice-Test-2-Spring-2020-Solutionsdocx/ = a) Find the marginal density of y. b) Find the conditional pdf f(x y) c) Find P(X < 1/3  Y = 1/3) a) fy(y) = b) f(x y) = 1+ 9 y 2 4 2 f ( x , y) f y( y) 2 3 ( x +3 y ) 2 1+ 9 y = 2 1 3 (x + ) 3 c) P(X < 1/3  Y = 1/3) = ∫ 2 dx 0 1/ 3 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) = Uniform( a, b): f(x) = i n−i , i = 0, 1,..., n, Mean is np, variance is np(1-p) i λ i! , i = 0, 1, 2,... , Mean is , variance is  −a (ax)( Nn−x ) ( Nn ) , Mean is na N 1 , Mean is (a + b)/2, variance is (b−a)2/12. b−a This study source was downloaded by 100000819498514 from CourseHero.com on 11-06-2024 04:33:10 GMT -06:00 https://www.coursehero.com/file/67998408/AMS-310-Practice-Test-2-Spring-2020-Solutionsdocx/ Normal(µ, σ2): f(x) = Exponential(λ): f(x) = 1 e−( x−μ ) / 2 σ √2 π σ 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 04:33:10 GMT -06:00 https://www.coursehero.com/file/67998408/AMS-310-Practice-Test-2-Spring-2020-Solutionsdocx/ Powered by TCPDF (www.tcpdf.org)