Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Solution to additional exercises y1 0 y2 5.108 f 1( y1 ) e y1 dy2 y1e y1 , y1 0. f 2( y 2 ) e y1 dy1 e y2 , y 2 0. Then Y1 has a gamma distribution with α = 2 and β = 1, and Y2 has an exponential distribution with β = 1. Thus, E(Y1 + Y2) = 2(1) + 1 = 3. Also, since y1 E(Y1Y2) = y y e 1 y1 2 dy2 dy1 3 , Cov(Y1, Y1) = 3 – 2(1) = 1, 0 0 V(Y1 – Y2) = 2(1)2 + 12 – 2(1) = 1. Since a value of 4 minutes is four three standard deviations above the mean of 1 minute, this is not likely. y2 5.133a f 2 ( y2 ) 6(1 y2 )dy1 6 y2 (1 y2 ), 0 y2 1 . 0 f ( y1 | y2 ) 1/ y2 , 0 y1 y2 1. To find E(Y1 | Y2 y2 ) , note that the conditional distribution of Y1 given Y2 is y uniform on the interval (0, y2). So, E(Y1 | Y2 y2 ) = 2 . 2 y 6.1 6.1 The distribution function of Y is FY ( y ) 2(1 t )dt 2 y y 2 , 0 ≤ y ≤ 1. 0 a. FU1 (u) P(U1 u) P(2Y 1 u) P(Y u1 2 ) FY ( u21 ) 2( u21 ) ( u21 )2 . Thus, fU1 (u ) FU1 (u ) 12u , 1 u 1 . b. FU2 (u) P(U 2 u) P(1 2Y u) P(Y 12u ) FY ( 12u1 ) 1 2( u21 ) ( u21 )2 . Thus, fU 2 (u ) FU2 (u ) u 1 2 , 1 u 1. c. FU3 (u ) P(U 3 u ) P(Y 2 u ) P(Y u ) FY ( u ) 2 u u Thus, fU3 (u) FU3 (u) 1 u 1, 0 u 1. d. E (U1 ) 1 / 3, E (U 2 ) 1 / 3, E (U 3 ) 1 / 6. e. E(2Y 1) 1 / 3, E(1 2Y ) 1 / 3, E(Y 2 ) 1 / 6. 7.42 Let Y denote the sample mean strength of 100 random selected pieces of glass. Thus, the quantity ( Y – 14.5)/.2 has an approximate standard normal distribution. a. P( Y > 14) ≈ P(Z > 2.5) = .0062. b. We have that P(–1.96 < Z < 1.96) = .95. So, denoting the required interval as (a, b) such that P(a < Y < b) = .95, we have that –1.96 = (a – 14)/.2 and 1.96 = (b – 14)/.2. Thus, a = 13.608, b = 14.392. 7.48 a. Although the population is not normally distributed, with n = 35 the sampling distribution of Y will be approximately normal. The probability of interest is P(| Y | 1) P( 1 Y 1) . In order to evaluate this probability, the population standard deviation σ is needed. Since it is unknown, we will estimate its value by using the sample standard deviation s = 12 so that the estimated standard deviation of Y is 12/ 35 = 2.028. Thus, 1 1 P(| Y | 1) P( 1 Y 1) P( 2.028 Z 2.028 ) P( .49 Z .49) = .3758. c. No, the measurements are still only estimates. 7.52 Let Y denote the average resistance for the 25 resistors. With μ = 200 and σ = 10 ohms, a. P(199 ≤ Y ≤ 202) ≈ P(–.5 ≤ Z ≤ 1) = .5328. b. Let X = total resistance of the 25 resistors. Then, P(X ≤ 5100) = P( Y ≤ 204) ≈ P(Z ≤ 2) = .9772.