15 The continuous random variable W has probability density function g given by , Osw<4 , 4sw<5 aw+ b 8(W) -{ bw+ a Find the values of a and b if it is given that P(2sW <4)=: Hence, find 51 a. P( W s3), b. value of u if P(W >u)=32 68
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- Suppose that the random variables X,Y, and Z have the joint probability density function f(x,y,z) = 8xyz for 0<x<1, 0<y<1, and 0<z<1. Determine P(X<0.7).9.19 Let X and Y be two continuous random variables, with joint proba- bility density function f(x, y): - 30 -50x²-50y² +80xy for -4 (b) An insurance company provides customers with both auto and home insurance policies. For a particular customer, Χ is the deduction on his or her auto policy and Y is the deduction on the home policy. Possible values of Χ are K100 and K250, and for Y are K0, K100 and K200. The joint probability density function for ( ) ,YΧ is given by the following table: Χ Y K100 K250 K0 0.20 0.05 K100 0.10 0.15 K200 0.20 0.30 iv. If we look only at those insurance customers selecting the lowest auto mobile insurance deduction (K100), what is the probability that a randomly selected\ customer will also select the lowest home deduction (K0). v. Compute the correlation coefficient of Χ and Y
- Let X be a random variable with probability mass function P ( X = 1 ) = 1/2 , P ( X = 2 ) = 1/3 , a n d P ( X = 5 ) = 1/6 . Then E[1/x]=?For any continuous random variables X, Y , Z and any constants a, b, show the following from the definition of the covariance:Suppose that the random variables X and Y have a joint density function given by: f(x,y) = {c(2x+y) for 2≤x≤6 and 0≤y≤5, 0 otherwise P(3 < X < 5, Y >1), P(X < 3), P(X +Y > 5), Find the joint distribution function (cdf),
- Let X be a Gaussian random variable (0,1). Let M = ln(5*X) be a derived random variable. What is E[M]?(a) Let Y1 , Y2 , · · · , Yn be independent and identically distributed random variables. Show if Y1 < Y2 is independent of the event Y1 < Y3.(b) Let Y1 , Y2 , · · · , Y∞ be dependent random variables with following probability density function; Yn+1 =3Yn with probability 1/3Yn+1 =Yn/3 with probability 2/3 for n ∈ [0, ∞) where Y0 = 1. Compute E[Y1] and E[Y2].