Question

6. The probability density function of (lifetime of an electronic component in years) X is f, (x)- 4 x exp(-r)U(x) 32 (a) What value of A will make this a valid pdf? (b) What is the probability that it will fail within 6 years, given that normally these units tend to fail within 4 to7 years? (c) What is P[IX-316)? (d) If the unit is known to fail within 6-8 years, what is the probability that it fail within 7 years 7. In a test with 10 Yes/No questions, the probability of being correct is 0.35. If the number of Y's is modeled as discrete random variable X, obtain expressions for the CDF and pdf of this random variable. What are the mean and variance of X? 8. Radar returns (power) are modeled in terms of the following (gamma) density (x)-xe U(x) (a) What is the probability that the received power exceeds 2 units. (b) What is the probability that the received power is less tha 6 units. (c) What is the probability that the received power lies between 2 and 6 units (d) What is the probability that the received power is exactly 5 units. (e) If past observations have shown that the received power is always more than 3 units, what is the probability that the the received power will be less than 6 units.

Answer 1

6a. Since fx(x) is pdf of X then Aexp or, 16A exp(-t)dt = 1: take,-= t or. or, 16A-1 or, A - 6b. P(X < 614 < X < 7) = 32 0 32 0 16 P(X < 6, 4<X<7) P(4X6) P(4 < X s7) P(4 < X s7) = (exp(-4/16) _ exp(-6/16))/(exp(-4/16)-exp(-7/16)) = 0.6873 (c) P(X-3| < 6) = P(-3 < X < 9) = P(0 < X < 9), since fx(x) > 0 for r 0 ア ) dr. =- exp (-t) dt-1-exp(-9/16)-0.4302

(d) P(X < 716-X-8) = POS X-8) P(6<X 8) take = t or, xdx = 1601t 32 = (exp(-6/16) _ exp(-7/16))/(exp(-6/16)-exp(-8/16)) = 0.5156 7. X ~ Binomial (10, 0.35) The pmf of X is fx (x) = -0 otherwise The cdf of X is 10 (0.35)(1 - 0.35)10-*; a - 0,1,2,.., 10 t) = Σ (10) (0.35)"(1-035)10-1 Fx(t) = P(X Mean of X 100.35 3.5 Variance of X = 10 * 0.35 * (1-0.35) = 2.275

8. CDF of X is = F(t) = P(X < t) = | xe-idx (a) P(X > 2)-1-F(2) 3 * erp(-2) 0.4060 (b) P(X < 6) 17*exp(-6) 0.9826 (c) P(2-X-6)-F(6) _ F(2) = 3 * erpl-2)-7 * exp(-6) = 0.3887 (d) P(X = 5) = 0 (since X is continuous random variable) (e) P(X < 6X > 3) P(3 X<6) F(6) 1- F(3) P(X>3) (4 exp(-3) - 7 *erp(-6))/(4 erp(-3))0.9129