AMS 310 Second Practice Test 2 Solutions

AMS 310 Fall 2023 Fred Rispoli Second Practice Test 2 There are 12 questions here, but the actual test 2 will only have 10 questions. 1. The time (in minutes) that it takes a plumber to fix a faucet leak has an exponential distribution with mean 30. a) Find P(X < 25), P(X > 35), and P(25 < X < 35) b) Find the 40 th percentile. a) P(X < 25) = ∫ 0 25 1 30 e − x 30 dx = 1 - e − 25 30 = 0.565 P(X > 35) = 1 - P(X ≤ 35) = 1 – (1 - e − 35 30 ) = 0.311 P(25 < X < 35) = (1 - 0.3114) - 0.5654 = .124 b) First note that the cdf for an exponential distribution is F(x) = 1 – e − x λ x > 0 Solve .40 = F(x) = 1 – e − x 30 ln(.6) = – x 30 X = -30 ln(.6) = 15.32, note x = 27.49 gives the 60 th percentile not the 40 th 2. Let X be a random variable with pdf f ( x ) = 1 5 , 0 < x < 5 . a) Find the mean of X and variance of X b) Set up the integral that calculates F (1.5). Find the exact value. c) Set up an integral to find P ( 2.5 < X < 4.25 ) . Find the exact value.

a) μ = ∫ 0 5 xf ( x ) dx = 25 10 = 5 2 x E ( ¿¿ 2 )= ∫ 0 5 x 2 f ( x ) dx = x 3 3 x = 5 x = 0 ¿ = 25 3 = 8.33  2 = x E ( ¿¿ 2 )− μ 2 ¿ = 8.33 – ( 5 2 ) 2 = 2.08 b) F(1.5) = ∫ 0 1.5 1 5 dx = 0.3 c) P ( 2.5 < X < 4.25 ) = ∫ 2.5 4.25 1 5 dx = 0.35 3. The number of gallons of Gatorade consumed by the UCONN Women’s Basketball team during a game follows a normal distribution with mean 15. The standard deviation is 2. a) If a game is selected at random, find the probability that the number of gallons consumed will be greater than 18.5 gallons. b) If a game is selected at random, find the probability that the number of gallons consumed will be between 10 and 17.5 gallons. c) Find the values A and B such that P(A < X < B) = .95 a) P(X > 18.5) = P(Z > 1.75) = 1 – P(Z ≤ 1.75) = 1 – 0.9599 = 0.0401 b) P(10 < X < 17.5) = P(-2.5 < Z < 1.25) =  (1.25) – (1 –  (2.5)) = 0.8944 – 0.0054 = 0.889

b) Yes f(x,y) = f 1 (x)f 2 (y), for all pairs x, y c) Var(X) = E(X 2 ) – E(X) 2 = 129 2 (.2) + 130 2 (.7) + 131 2 (.1) – 129.9 2 = 3,328.2 + 11,830 + 1,716.1 – 16,874.01 = .3 Var(Y) = E(Y 2 ) – E(Y) 2 = 15 2 (.6) + 16 2 (.4)– 15.4 2 = 0.24 d) E(XY) = 129(15)(0.12) + 129(16)(0.08) + 130(15)(0.42) + 130(16)(0.28) + 131(15)(0.06) + 131(16)(0.04) = 2000.46 Cov(x,y) = E(XY) – E(X)E(Y) = 2000.46 – 129.9(15.4) = 0.00 We can also get this by 5. Suppose the random variables X and Y have joint pdf as follows: f(x,y) = 12e -4x-3y , y > 0 and x > 0 a) Find the marginal pdf f 1 (x) of X b) Find the conditional pdf f 2 (y  x) c) Find P(1 < Y < 3) a) f 1 (x) = ∫ 0 ∞ f ( x , y ) dy = ∫ 0 ∞ 12 e − 4 x − 3 y dy = 12 e − 4 x ( 1 3 ) = 4 e − 4 x b) f 2 (y  x) = f ( x , y ) f 1 ( x ) = 12 e − 4 x − 3 y 4 e − 4 x = 3 e − 3 y c) f 2 (y) = ∫ 0 ∞ f ( x , y ) dx = ∫ 0 ∞ 12 e − 4 x − 3 y dx = 3 e − 3 y

P(1 < Y < 3) = ∫ 1 3 3 e − 3 y dy = 0.0496

a) We are given αβ = 20 and Var(X) = αβ 2 = 80 So, α = 20 β and hence, 80 = ( 20 β ) β 2 = 20 β Solving gives β = 80/20 = 4 and α = 20/4 = 5 b) f ( x ) = 1 Γ ( α ) β α x α − 1 e − x β = 1 24 ( 4 5 ) x 3 e − x 4 8. A sample of size 10 is taken from a normal distribution and it is found that the sample variance is 5. a) Find the values of χ 2 needed in the formula below. b) Construct the 90% confidence interval to estimate the population variance. Round off to three decimal places. ( ( n − 1 ) s 2 χ α 2 , n − 1 2 , ( n − 1 ) s 2 χ 1 − α 2 , n − 1 2 ) First note that for 90% confidence  = 0.10, α 2 = 0.05 , 1 ─ α 2 = 0.95 and df = 9. a) χ α 2 ,n − 1 2 = χ 0.05 , 9 2 = 16.92 χ 1 − α 2 ,n − 1 2 = χ 0.95 , 9 2 = 3.325 b) The lower bound is ( n − 1 ) s 2 χ α 2 ,n − 1 2 = ( 9 ) 5 16.92 = 2.660 The upper bound is ( n − 1 ) s 2 χ 1 − α 2 , n − 1 2 = ( 9 ) 5 3.325 = 13.534

9. Let X 1 , X 2 , … X 100 be a random sample from a distribution with pdf f ( x ) = 1 5 , 0 < x < 5 . Use the Central Limit Theorem to find the probability of P(2 < ´ X < 3) First note that: μ = ∫ 0 5 xf ( x ) dx = 25 10 = 5 2 = 2.5 x E ( ¿¿ 2 )= ∫ 0 5 x 2 f ( x ) dx = x 3 3 x = 5 x = 0 ¿ = 25 3 = 8.33  2 = x E ( ¿¿ 2 )− μ 2 ¿ = 8.33 – ( 5 2 ) 2 = 2.08 and  = √ 2.08 = 1.44 P ( 2 < ´ X < 3 ) ≈ P ( 2 − 2.5 1.44 / √ 100 < Z < 3 − 2.5 1.44 / √ 100 ) = P ( − 0.5 0.144 < Z < 0.5 0.144 ) 10. Suppose that X has a binomial distribution with n = 250 and the success probability p = 0.6. Find approximations to the following probabilities using a normal approximation to the binomial. Use continuity corrections. a) P(140 ≤ X < 170) b) P(X > 169) First note that μ = np = 250 ( 0.6 ) = 150 ∧ n ( 1 − p ) = 250 ( 0.4 ) = 100 are both large. So we can use the normal approximation to the binomial. σ = √ np ( 1 − p ) = √ 250 ( 0.6 ) ( 0.4 ) = √ 60 = 7.75 (a) P ( 140 ≤ X < 170 ) ≈ P ( 139.5 − 150 7.75 < Z < 169.5 − 150 7.75 ) = P ( − 2.083 < Z < 0.917 ) = Փ ( 2.52 ) − Փ ( − 1.35 ) = 0.99 (b)

t-distribution