The joint pdf of the random variable X and Yis given in the below matrix. Find the mean value of X and Y: P(X,Y)= X O2.45 and 2.1 O 1.9 and 2.45 O 1.9 and 2.1 O 21 and 2.45 Y 0.05 0.05 0.1 0.05 0.1 0.35 0 0.2 0.1
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- The joint pdf of the random variable X and Yis given in the below matrix. Find the mean value of X and Y: P(X,Y)= X O2.45 and 2.1 O 1.9 and 2.45 O 1.9 and 2.1 O 21 and 2.45 Y 0.05 0.05 0.1 0.05 0.1 0.35 0 0.2 0.1Let X be a discrete random variable with probability mass function given in the following table:The joint pdf of the random variable X and Yis given in the below matrix. Find the mean value of X and Y: P(X,Y)= X O2.45 and 2.1 O 1.9 and 2.45 O 1.9 and 2.1 O 21 and 2.45 Y 0.05 0.05 0.1 0.05 0.1 0.35 0 0.2 0.1
- Consider the following joint random variable pmf for X and Y: x y -1 0 1/4 0 1 1/8 1 1 1/2 1 0 1/8 What is the correlation of X and Y?Assume the following data displays the joint probability for random variables X and Y: X 10 20 30 20 0.1 0.2 0.05 0.05 0.1 0.2 0.2 0.0 0.1 -1 Y O 1 1. Solve for the correlation of X and Y. 2. Solve for the correlation of X and Y conditional on Y being greater than or equal to zero. 3. Are X and Y independent? Are X and Y conditional on Y being greater than or equal to zero, independent.Let Y be a random variable with pdf f(y)=(3/64)y2(4-y), Osys4, zero elsewhere. Match the following. A. 0.31 B. 12.2 C. 16 D. 2.40 E. 0.64 select 1. E(Y)= select 2. V(Y)= select 3. Let X=3Y+5= E(X)= select 4. Let Z-5Y+3= V(Z)= select 5. P(YS2)3
- Consider random variables X and Y with the joint PDF below find pxyhe joint pat of the random variable X and Yis iven in the below matrix. Find the mean value f X and Y: P(X,Y)= X 2.45 and 2.1 1.9 and 2.45 O 1.9 and 2.1 2.1 and 2.45 Y 0.05 0.05 0.1 0.05 0.1 0.35 0 0.2 0.1b. The table below shows the joint probability mass function for the discrete random variables X and Y. Y = 1 Y = 2 X = 0 X = 1 X = 2 0.07 0.23 0.18 0.05 0.29 0.18 Calculate and interpret the covariance between variables X and Y i.e. Cov(X,Y).