q13- X and Y are two independent random variables. We know V(X) = 26.43 and V(Y) = 23.67, then V(8X + 5Y) is: a. 862.68 b. 2283.27 c. 329.79 d. 16490.91

Previously, you studied linear combinations of independent random variables. What happens if the variables are not independent? A lot of mathematics can be used to prove the following: Let x and y be random variables with means ?x and ?y, variances ?2x and ?2y, and population correlation coefficient ? (the Greek letter rho). Let a and b be any constants and let w = ax + by for the following formula. μw = aμx + bμyσ2w = a2σ2x + b2σ2y + 2abσxσyρ In this formula, r is the population correlation coefficient, theoretically computed using the population of all (x, y) data pairs. The expression ?x?y? is called the covariance of x and y. If x and y are independent, then ? = 0 and the formula for ?2w reduces to the appropriate formula for independent variables. In most real-world applications the population parameters are not known, so we use sample estimates with the understanding that our conclusions are also estimates.Do you have to be rich to invest in bonds and real estate? No, mutual fund…

An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (t) which we write hth, ttt, etc. For each outcome, let N be the random variable counting the number of tails in each outcome. For example, if the outcome is hth, then N (hth) = 1. Suppose that the random variable X is defined in terms of N as follows: X=2N² − 6N-1. The values of X are given in the table below. Outcome thh tth hhh hth ttt htt hht tht Value of X-5 -5 -1 -5 -1 -5 -5 -5 Calculate the probabilities P(X=x) of the probability distribution of X. First, fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row. Value X of X P(X=x) 0 0 00 X

Suppose that X is a N(3,4) random variable and suppose that Y=5X+2. Determine P(Y>18.5).