Question

Reserve Problems Chapter 5 Section 6 Problem 4 Three electron emitters produce electron beams with changing kinetic energies that are uniformly distributed in the ranges (5,8), (3,9), and [4,6). Let Y denote the total kinetic energy produced by these electron emitters. (a) Suppose that the three beam energies are independent. Determine the mean and variance of Y. Round your answers to two decimal places (e.g. 98.76). Mean = Variance (b) Suppose that the covariance between any two beam energies is -0,3. Determine the mean and variance of Y. Round your answers to two decimal places (e.9. 98.76). Mean - Variance (c) Compare and comment on the results in parts (a) and (b): The covariance D mean. variance The covariance plisy 2000-2019 John Wiley Soste n ia

Answer 1

solution- Given dode → Let dig Hz, Rz are the ranges. X2 = [3,9] x = [4,6 @. Here we find the mean and variance of y. mean oby:- formula E(X) = 649 → E (xi) = 85 = 13 = 6.5 E (13) = 9+2 = 1226 E (X3) = 6+4 = 10 = 5 Meern E(4) = E(X) + E(X2)+E(X) = 6.5 +6+5 E (4) = 17:50 variance D6 Y fortala (4) = (b-a) 2

var (xi) = (85) = 0.75 war (x2) = (9-3) - 36=3" ver (X) = (5-4)2 = 0.33 variance (4) = var (X1) + vam (X2) + var (x3) 0.75 +3 +0-33. mean = (17.50] variance - 4.08 tere we find the mean and variance oby. Given Comariance between any two beam energies in -0.3. a Mean ey) - E(X) + E(X2) +E(X₂) = 6:5 + 6 + 5 - 17.5 -> variance (4) = fuese (1+ VER CX2) twar (Kg) + Cu( XcXz) + coulxinal - 0.75 + 3+ 0.33 toto-0.3].

1. = (4.08-0.3] - 3.78 ... mean 17.5 Variance 13.78 E Here we find the result ob part (a) and (b). As comparing the variances of both, electron enitters are more efficient as there has covariance betweer any two beam energies.