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

2017
Views:

398

Pages:

11
Author:

study_genius

AMS 310 Spring 2017 Practice Exam 2A Solutions Fred Rispoli Some Possibly Useful Formulas: P (|X −μ| 0. Mean is , variance is  . 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 1. When a balanced coin is flipped 10,000 times, find the lower bound of the probability that the proportion of heads is obtained will fall between .45 and .55. �=10,000(0.5)=5,000 �=SQRT(10,000(0.5)(0.5)) = 50 �−��=5,000−10(50)=4,500 and �+��=5,000+10(50)=5,500 4,500/10,000=0.45 and 5,500/10,000=0.55 Thus we have �=50, we have �=10. 1 Applying Chebyshev’s Inequality: �(|�−�|<��)≥1− k 2 1 �(|�−5,000|<500)≥1− 100 =0.99 2. When a fair die is flipped 72,000 times, find the lower bound of the probability that the number 5 occurs between 11,5000 and 12,500 times. �=72,000(1/6)=12,000 �=SQRT(72,000(1/6)(5/6) )=100 11,500=�−500 12,500=�+500 ��=500 ⇒ �=500/100=5 �(|�−�|<500)=�(|�−�|<5�)≥1−1/52 =24/25 = 0.96 The probability is at least 0.96. 3. 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: a) Less than 29.75 feet b) Between 29.75 and 29.90 feet 1 The pdf is f(x) = 30.05−29.5 1 = .55 29.75 a) P(X < 29.75) = ∫ f ( x ) dx 29.50 29.90 b) (29.75 < X < 29.90) = ∫ f ( x ) dx 29.75 4. The number of gallons of Gatorade consumed by a football team during a game follows a normal distribution with mean 20. The standard deviation is 3. a) If a game is selected at random, find the probability that the number of gallons consumed will be greater than 23 gallons. b) If a game is selected at random, find the probability that the number of gallons consumed will be between 22 and 25 gallons. c) Find the 90th percentile. a) z = 23−20 3 = 1.0 P(X > 23) = 1 – P(X ≤ 23) = 1 – P(Z ≤ 1) = 1 – (1) = 1 – .8413 = .1587 b) P(22 < X < 25) = P(.67 < Z < 1.67) = (1.67) - (.67) = .9525 - .7486 = .2039 z= 22−20 3 = .67 z= 25−20 3 = 1.67 c) Set F(x) = .95 Use the table in reverse to get Z = 1.645, then solve 1.645 = 5. Let X be a random variable with cdf F(x) = { 0, x <2 x−2 ,2 ≤ x < 4 2 1, x ≥ 4 a) Find the pdf of X. 2 b) Find P( 3 < X < 3) c) Find P( X > 3.5) d) Find the 60th percentile e) Find P(X=3) (a) �(�)=�′(�)={ 1/2, 2≤�≤4 0, otherwise (b) �(2/3<�<3) = �(3)−�(2/3)=1/2−0=1/2 (c) �(� > 3.5) = 1−�(3.5)=1−3/4=1/4 (d) 0.6 = �(�)=(�−2)/2⇒ �−2 = 1.2 ⇒ �=3.2 (e) �(�=3) = 0 x−20 3 to get x= 24.935 6. The random variables X and Y have the following joint probability distribution. X f(x,y) 1 2 3 Y a) Find 2 .05 .05 .05 3 .04 .15 .10 P( X +Y ≤ 5) . b) Find the marginal probability distributions c) Find 1 .1 .2 .15 f 1 (x ) and f 2( y) . P( X <2∨Y =3) . d) Are X and Y independent? Thoroughly explain your answer. e) Find E(X) and Var(X) f) Find the correlation coefficient of X and Y. a) The X,Y combinations that sum to 5 or less are in red below. X f(x,y) Y 1 2 3 1 .1 .2 .15 2 .05 .05 .05 3 .04 .15 .10 4 .01 .04 .06 P( X +Y ≤ 5) = .1 + .05 + .04 + .01 + .2 + .05 + .15 + .15 + .05 = .80 b) c) X f1(x) 1 .45 2 .15 3 .29 Y f2(y) 1 .2 2 .44 3 .36 4 .11 P( X <2∨Y =3) = .15 d) Not independent f(1,1) = .1 ≠ f1(1) f2(1) = (.45)(.2) = .09 4 .01 .04 .06 e) E(X) = 1(.45) + 2(.15) + 3(.29) + 4(.11) = .45 + .30 + .87 + .44 = 2.06 X2 f1(x) 1 .45 4 .15 9 .29 16 .11 E(X2) = 1(.45) + 4(.15) + 9(.29) + 16(.11) = .45 + .60 + 2.61 + 1.76 = 5.42 Var(X) = E(X2) - E(X)2 = 5.42 – 2.062 = 1.176 X = SQRT(1.176) = 1.084 f) Use ρ = Cov (X , Y ) σ X σY E(Y) = 1(.02) + 2(.44) + 3(.36) = 1.98 Y2 f2(y) 1 .02 4 .44 9 .36 E(Y2) = 1(.02) + 4(.44) + 9(.36) = 5.42=5.02 Var(Y) = E(Y2) - E(Y)2 = 5.02 – 1.982 = 1.100 X = SQRT(1.176) = 1.049 Cov ( X , Y )=E ( XY ) −E ( X ) E (Y ) Thus we need to know E(XY), the procedure is the following: P(XY=1)=P(X=1,Y=1)=0.1 P(XY=2)=P(X=1,Y=2)+ P(X=2,Y=1)=0.2+0.05=0.25 P(XY=3)=P(X=1,Y=3)+ P(X=3,Y=1)=0.04+0.15=0.19 P(XY=4)=P(X=4,Y=1)+ P(X=2,Y=2)=0.01+0.05=0.06 P(XY=6)=P(X=2,Y=3)+ P(X=3,Y=2)=0.05+0.15=0.2 P(XY=8)=P(X=4,Y=2)=0.04 P(XY=9)=P(X=3,Y=3)=0.1 P(XY=12)=P(X=4,Y=3)=0.06 E(XY)=1(0.1)+2(0.25)+3(0.19)+4(0.06)+6(0.2)+8(0.04)+9(0.1)+12(0.06)=4.55 E ( XY )−E ( X ) E (Y ) 4.55−1.98∗2.06 ρ= = =0.3952 σx σ y 1.084∗1.10 7. Suppose the random variables X and Y have joint pdf as follows: 4 2 xy f(x,y) = 7 x + 3 , 0 < x < 1, 0 < y < 3 ( ) a) Find the marginal pdf f1(x) of X, and f2(y) of Y. 3 3 4 xy 12 6 f 1 ( x )=∫ f ( x , y ) dy =∫ x 2+ dy= x 2 + x 3 7 7 0 0 7 ( ) 1 1 f 2 ( y )=∫ f ( x , y ) dx=∫ 0 0 b) Find the cdf of X and Y. 4 3 f 1 ( t ) dt =¿ x 3 + x 2 7 7 x , for 0 ≤ x ≤1 F 1 ( x )=∫ ¿ 0 Thus { 0, x< 0 4 F1 ( x )= x 3 + 3 x 2 , 0 ≤ x ≤1 7 7 1, x >1 f 2 ( t ) dt =¿ 4 1 y + y2 21 21 y F 2 ( y )=∫ ¿ , for 0 ≤ y ≤ 3 0 Thus { 0, y< 0 F2 ( y )= 4 y+ 1 y 2 , 0≤ y ≤3 21 21 1, y> 3 c) Find P(Y < 2) P (Y < 2 )=F2 ( 2 )= d) Find P( 12 21 1 X > , Y <1 ¿ 2 4 2 xy 12 6 4 2 x + dx= x 2 + x= + y 7 3 7 7 21 21 ( ) 1 1 1 17 P X > ,Y <1 =∫ ∫ f ( x , y ) dydx = 2 84 1 0 ( ) 2 e) Find the conditional pdf f2(y x) f (x , y) 6 x 2 +2 y ( | ) f2 y x = = f 1 ( x) 18 x2 +9 x 8. 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, �(�,�)=�1(�)�2(�)={16�−�/2−�/3, � ≥ 0, � ≥ 0 0, otherwise b) By independence, �2(�|�)=�2(�)=1/3�−�/3, �≥0 c) By independence, �(Y>3|X =2)=�(Y>3)=1−�2(3)=�−3/3 = 0.3679 9. Two random variables X, Y and Z are independent. Each has a binomial distribution with success probability .3 and 5 trials. a) Find the joint probability distribution function f(x,y,z). b) Find the probability P(Z > Y > X > 1). (5x) .3 .7 x a) X, Y and Z are all binomial(5,.3), so f(x,y,x) = ¿ 5− x ( 5y ).3 .7 y ) ¿ 5− y (5z ).3 .7 z ) ¿ 5− z ) b) Add the probabilities for (x,y,z) in (2,3,4), (2,3,5), (2,4,5), (3,4,5)