jdale_STATHW2.pdf - Jack Dale STAT 4705 u2013 HW 2 1...

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Unformatted text preview: Jack Dale 6/4/2019 STAT 4705 – HW 2 1. Discrete (D) or Continuous (C) a. X: Number of automobile accidents per year in Virginia i. D – It is a countable statistic b. Y: Length of time to play 18 holes of golf i. C – It could take any time to play that many holes c. M: Amount of milk produced yearly by a particular cow i. C – Volume measured can be a in a fraction d. N: The number of eggs laid each month by a hen i. D – Eggs are a countable statistic e. P: The number of building permits issued each month in a certain city i. D – Permits issued are a countable variable (you either have a permit or you don’t) f. Q: The weight of grain produced per acre i. C – Weight can be less/more and in a fraction Jack Dale 6/4/2019 2. Grafting, the uniting of the stem of one plant with the stem or root of another, is widely used commercially to grow the stem of one variety that produces fine fruit on the root system of another variety with a hardy root system TABLE 1 x (Density) 0 1 2 3 4 5 f(x) 0.7 0.2 0.05 0.03 0.01 ? a. f(5) ( = 0) + ( = 1) + ( = 2) + ( = 3) + ( = 4) + ( = 5) = 1 0.7 + 0.2 + 0.05 + 0.03 + 0.01 + P(X=5) = 1 0.99 + P(X=5) = 1 F(5) = 0.01 b. Table for CDF function F(x) Cumulative Distribution Function (CDF) x (Density) x<0 0 ≤ x < 1 1 ≤ x < 2 2 ≤ x < 3 3 ≤ x < 4 4 ≤ x < 5 5≤x FX(x) 0 0.7 0.9 0.95 0.98 0.99 1 c. Sketch CDF function F(x) d. Use F to find the probability that at most three grafts fail; that at least two grafts fail • P(x≤3) = F(3) → P(x=0) + P(x=1) + P(x=2) + P(x=3) = 0.98 Jack Dale 6/4/2019 o P(x≤3) = 0.98 • P(x≥2) = 1 – P(x<2) = 1 – [F(0) + F(1)] = 1 – [0.7 – 0.2] = 0.1 o P(x≥2) = 0.1 e. Use F to verify that the probability of exactly three failures is 0.03 P(x=3) = P(x≤3) – P(x≤2) = F(3) – F(2) = 0.98 – 0.95 = 0.03 Jack Dale 6/4/2019 3. An investment firm offers its customers municipal bonds that mature after varying numbers of years. Given that the cumulative distribution function of T , the number of years to maturity for a randomly selected bond, is 0 <1 1⁄ 1 ≤ < 3 4 1 () = ⁄2 3 ≤ < 5 3⁄ 5 ≤ < 7 4 {1 ≥7 a. P(T = 5) P(X = x) = F(X) – F(X – 1) P(T ≤ 5) – P(T ≤ 4) = F(5) – F(4) = ¾ - ½ = 1/4 b. P(T > 3) P(X > x) = 1 – F(X ≤ 3) 1 – F(3) = 1 – (1/2) = 1/2 c. P (1.4 < T < 6) P(a < x < b) = F(b) – F(a) P(6) – P(1.4) = (3/4) – (1/4) = 1/2 Jack Dale 6/4/2019 4. The waiting time x, in hours, between successive speeders spotted by a radar unit is a continuous random variable with cumulative distribution function a. Using the cumulative distribution function of X. 12 = 0.2 ℎ → ( < 0.2) = (0.2) 60 (0.2) = 1 − −8(0.2) = . b. Using the probability density function of X. (Hint: Need to find the probability density function first) [()] = [1 − −8 ] = 8 −8 , ≥ 0 () = 0.2 0.2 0.2 −8 −8 (0.2) = ∫ () = 8 ∫ = 8 [− = − −8(0.2) + 1 8 0 0 0 1 − −8(0.2) = . Jack Dale 6/4/2019 5. Determine the value c so that each of the following functions can serve as a probability distribution of the discrete random variable X a. f(x) = c(X2 + 4) for x = 0; 1; 2; 3 3 ∑ () = 1 (0) + (1) + (2) + (3) = 1 =0 [(0 + 4) + (1 + 4) + (4 + 4) + (9 + 4)] = 1 → 30 = 1 = 2 3 b. () = ( ) ( ) for x=0; 1; 2 3− 2 ∑ () = 1 (0) + (1) + (2) = 1 =0 2 3 3 3 2 2 [( ) ( )+( ) ( )+( ) ( )] = 1 0 3−0 1 3−1 2 3−2 [(1 1) + (2 3) + (1 3)] = [(1) + (6) + (3)] = 10 = 1 = Jack Dale 6/4/2019 6. The joint probability distribution of X and Y is given by a. Find the value of c that makes f(x, y) a valid joint probability density function. X=0 1 2 3 TOTAL Y=0 0c 1c 2c 3c 6c 1 1c 2c 3c 4c 10c 2 2c 3c 4c 5c 14c TOTAL 3c 6c 9c 12c 30c 30 = 1 SO = b. Find P(X > 2, Y < 1) X=0 Y=0 0c 1 1c 2 2c 1 1c 2c 3c = c. Find P(X + Y = 4) X=0 Y=0 0c 1 1c 2 2c 2 2c 3c 4c 3 3c 4c 5c 2 2c 3c 4c 3 3c 4c 5c 1 1c 2c 3c = Jack Dale 6/4/2019 7. Two electronic components of a missile system work in harmony for the success of the total system. Let X and Y denote the life in hours of the two components. The joint density of X and Y is a. Give the marginal density functions for both random variables ∞ ∞ () = ∫ (, ) = ∫ −(1+) = ( ) 0 0 ∞ ∞ −(1+) ∞ 1 = ∫ −(1+) (− ) +∫ (1 + ) (1 + ) 1 + 0 0 −(1+) 0 , ≤ () = {( + ) , ∞ ∞ ∞ −(1+) ℎ() = ∫ (, ) = ∫ = ∫ − − 0 0 0 ∞ − ∞ − − − − [ −∞ = ∫ = [− − 0] = = − − [ − ]∞ 0 = − 0 0 1 − − [ ∞ − 1] = − − [0 − 1] = − − , ≤ () = { , b. What is the probability that the lives of both components will exceed 2 hours? ( > 2, > 2) ∞ ∞ ∞ ∞ −(1+) − ∫ ∫ = ∫ {∫ − } = 2 2 2 2 ∞ ∞ − ∞ −∞ −2 − − ∫ {− } = ∫ {− ( − )} = 2 2 2 ∞ ∞ ∞ ∞ −2 −2 −3 − − −3 ∫ {( = )} = ∫ {( )} = ∫ = 3 2 2 2 2 1 −∞ 1 − ( − −6 ) = ( −6 ) = . 3 3 Jack Dale 6/4/2019 8. X and Y have the following joint probability distribution: f(x, y) X 2 Y 1 0.1 3 0.2 5 0.1 a. Marginal distribution of X x 2 g(x) 0.4 (sum of x=2 column values) b. Marginal dist. of Y y 1 h(y) 0.25 3 0.5 5 0.25 4 0.6 4 0.15 0.3 0.15 Elsewhere 0 Elsewhere 0 c. Give the conditional distribution (|) → in x given that y (2,1) 0.1 ( = 2| = 1) = ( = 0.4 )= ( = 1 ) 0.25 (4,1) 0.15 ( = 4| = 1) = ( = 0.6 )= ( = 1 ) 0.25 (2,3) 0.2 ( = 2| = 3) = ( = 0.4 )= ( = 3 ) 0.5 (4,3) 0.3 ( = 4| = 3) = ( = 0.6 )= ( = 3 ) 0.5 (2,5) 0.1 ( = 2| = 5) = ( = 0.4 )= ( = 5 ) 0.25 (4,3) 0.15 ( = 4| = 3) = ( = 0.6 )= ( = 3 ) 0.25 ( = | = , , ) = . ( = | = , , ) = . Jack Dale 6/4/2019 9. The amount of kerosene, in thousands of liters, in a tank at the beginning of any day is a random amount Y from which a random amount X is sold during that day. Suppose that the tank is not resupplied during the day so that x ≤y, and assume that the joint density function of these variables is a. Find the marginal distribution of X. 1 1 () = ∫ () = ∫ 2 = 2|1 = ( − ); < < b. Find the marginal distribution of Y. () = ∫ () = ∫ 2 = 2|0 = ; < < 0 c. Determine if X and Y are independent (, ) = ()ℎ() = () (), ℎ () () = (2 − 2)(2) = 4 − 4 ≠ (, ) NO, THEY ARE NOT INDEPENDENT d. Give the conditional distribution f(x│y) → in x given that y (|) = (, ) = () Jack Dale 6/4/2019 10.Consider the following joint probability density function of the random variables X and Y a. Find the marginal distribution of X and verify that it is a valid density function. 6 2 () = ∫ (, ) = ∫ 6 = | = 3 2 (0 < < 1) 2 0 0 0 1 1 1 3 3 2 ∫ () = ∫ 3 = | = 1 → () 3 0 0 0 b. What is the probability that proportion Y is less than 0.5, given that X is 0.7. (, ) 6 2 (|) = = 2 = 2 , (0 < < 1) () 3 0.5 0.5 ( < 0.5| = ) = ∫ (|) = ∫ 0 0 0.5 2 2 2 0.52 ( 2 ) = 2 | = 2 2 0 0.52 ( < 0.5| = 0.7) = = . 0.72 ...
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