Question

The humidity level, X, in Tropicana Field in Tampa at the time of a night game ranges from 40% to 100 %. (.40 to 1.00) The random variable X has cumulative distribution F with the following definition: F(x) 0 for x < 40, F(x)= 5(X-.4)(2-x) for .4 s x 1.0, F(x)=1 for x 2 1.0. a) What is the probability density function for X for .4 s xs 1.0? 1-5(x-4)(2-x) 5(2-x)+5(x-0.4) 5(2-x)-5(x-0.4) Odnorm(x,2,.4 b) What is the probability that X>0.5? c) What is the probability X < 0.5? ? d) What is the probability that X > 0.6? e) What is the probability that X> 0.8? f) What is the probability that X> 0.8 given the X> 0.6 g) Calculate the 60th percentile of X. h) What is the expected value of X? i) What is the expected value of X2? j) What is the variance of X?

Answer 1

- giren CDF of arandom variablex is Ex(x)=10 ; nco-40 501-0.4)(2-31).airs 0.4 enclo 3 nalo a) The pdf of a random variable x is fx (22= d Fx (m) Se on (-04) (2-1)] = [ (0-0-0), (2-4) +(2-132 "" se po (200:10] = {[(-04) x-1 + ( 2-*)x3]

i flan= 5(0.4-1) +5(2-2) 3 -- 5(2-2) + 5(2-0.4) 3 correct choice is b) The probability that x>015 is given by P(x>0.5] = 1- P(x<0.5] =1-FX(0.5) Ein 5(0.5-0.4)(2-0.5) 3 - 580110 15 Cl- = 3 = 1-0.25 = 0.75 iP[x>0.5] = 0.75

Ec P(x20.5] = 1-p (x20.5] =l- 0.75 = 0.25 from b) since x is a continuous random variable. p[x20.5) = P(x>0.5] Ed) P[x>0.6] = 1- P[x 50's] air Ex(0.6) ci5(06-0-4) (2-06) 3 5x0.2 x 1.4 il 51 - 3 = 10:467 = 0.533 ... P(x700] =0.533

e> P[x>0.8] = 1-P[*5018] Ela 560-8-0.4)(2-08) = 15X0.4 x 1.2 = = 1-0.8 0.2 ATP[x20.8] =0.2 /