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- If X is a Poisson variable such that P(X = 2) = 9P(X = 4) +90 P(X = 6), find the mean and variance of X.Suppose a random variable X has expected value E[X] = -1 and variance Var(X) = 16. Compute the following quantities. (a) E[2X-3] (b) E[X²] (c) E[(2x - 3)²] (d) Var[2X - 2]If X is a Poisson variable such that P(X =2) = 9P(X= 4) + 90P(X = 6), find the mean and variance of X.
- Suppose a random variable X has expected value E[X] = -4 and variance Var(X) = 16. Compute the following quantities. (a) E[X-1] (b) E[X²] (c) E[(x - 1)²] (d) Var[3X + 2]If X is a random variable with pdf f(x) = 2x − 2 where x = (1, 2), find the variance of Y = 2X - 3.Problem #2: Suppose a random variable X has expected value E[X] = -2 and variance Var(X) = 4. Compute the following quantities. (a) E[X-2] (b) E[X²] (c) E[(x-2)²] (d) Var[X-3]
- Let X and Y be identically distributed with mean 4 and variance 25.(a) If X and Y are uncorrelated, what is the variance of X + Y?(b) If the covariance between X and Y is 8, what is the variance of X + Y? (c) If X and Y are independent, what is E[(X - Y)^2]?6. Find the mean and variance of fx (x) = 2e eA* la.2...) (x) = deNo, this solution is not correct. The variance of an exponentially distributed random variable is the square of its mean, so Var(L) = (1/Y)^2. However, the given solution incorrectly multiplies this by Var(Y) to obtain an incorrect expression for Var(L). The correct approach to find Var(L) involves using the law of total variance. The law of total variance states that Var(L) = E[Var(L|Y)] + Var(E[L|Y]). We already know that Var(L|Y) = (1/Y)^2, so E[Var(L|Y)] = E[(1/Y)^2]. We also know that E[L|Y] = 1/Y, so Var(E[L|Y]) = Var(1/Y). To find E[(1/Y)^2] and Var(1/Y), we need to use the formulas for the expected value and variance of a function of a random variable. These involve integrating the function multiplied by the probability density function of the random variable. Once we correctly compute E[(1/Y)^2] and Var(1/Y), we can plug the correct values into the law of total variance to find the correct value for Var(L).
- The p.d.f. of a random variable X' is as shown in the figure. The pdf is zero for X 5. Calculate (i) the maximum value of p.d.f. (ii) expectation of X, E(X) (iii) variance of X. fx (x) kSuppose that the probability that a patient admitted in a hospital is diagnosed with a certain type of cancer is 0.03. Suppose that on a given day 10 patients are admitted and X denotes the number of patients diagnosed with this type of cancer. The mean and the variance of X are: None of these E(X)=0.5 and V(X)=0.475 E(X)=0.4 and V(X)=0.384 E(X)=0.3 and V(X)=0.291By Laws of Expected Value and Variance, determine E(Z) and V(Z) if Z = 4X + 1. E(Z) = 4E(X) + 1 and V(Z) = 16V(X) E(Z) = 4E(X) + 1 and V(Z) = 4VX) E(Z) = 4E(X) + 4 and V(Z) = 4V(X) E(Z) = 4E(X) + 4 and V(Z) = 16V(X)