Find the value of the constant c in the given probability mass function. 1 if x = -4,-3, –2, 2, 3, 4, p(x) = %3D |0, otherwise O 12 O 10 O 1/12 O 1/10
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- What is the value if the constant c for p(x) to qualify as a probability mass function? p(x)=c(1/4)x-1 if x = 1, 2, 3, 4, 5, ... and p(x) = 0 otherwise.What is the value of the constant c for p(x) to qualify as a probability mass function? ?(?)=?(1/4)x-1 ?? ? = 1, 2, 3, 4, 5, ... and p(x) = 0 otherwise. How would you get c{1-1/4}-1 = 1Find the conditional expectation E(Y/X=0.47) if the joint probability density function of the random variable X and Y isf(x, y) = 1/x, 0 < y ≤ x ≤ 1.
- a. What value of c will make f(x) a valid probability mass function ?b. Compute P (1 < X < 6).Given a discrete RV X that takes on the values {-2, -1, 3, 4} and whose pmf (probability mass function) is: p-2 = 0.1 , p-1 = 0.2 , p3 = 0.6 , p 4 = 0.1 . Calculate the mean μ X of RV X.Let p(x) = cx^2 for the integers 1, 2, and 3 and 0 otherwise. What value must c be in order for p(x) to be a legitimate probability mass function?
- The possible values of a random variable are 1 and 2. If its probability mass function has the form p(x)=kx2, what must the value of k be? A. 1/3 B. 1/4 C. 1/5 D. 1/62. Identify the probability density function, then find the mean and variance without integrating. b. f(x) =1/6 e^−x/6, [0,∞) c. f(x) =1 / 3√2π e^−(x−16)^2/18, (−∞,∞)The General Social Survey asked a sample of adults how many siblings (brothers and sisters) they had (X) and also how many children they had (Y). We show results for those who had no more than 4 children and no more than 4 siblings. Assume that the joint probability mass function is given in the following contingency table: y x 0 1 2 3 4 0 0.03 0.01 0.02 0.01 0.01 1 0.09 0.05 0.08 0.03 0.01 2 0.09 0.05 0.07 0.04 0.02 3 0.06 0.04 0.07 0.04 0.02 4 0.04 0.03 0.04 0.03 0.02 Find ρ(X, Y). (Round the final answer to four decimal places.) ρ(X, Y) =
- The General Social Survey asked a sample of adults how many siblings (brothers and sisters) they had (X) and also how many children they had (Y). We show results for those who had no more than 4 children and no more than 4 siblings. Assume that the joint probability mass function is given in the following contingency table: y x 0 1 2 3 4 0 0.03 0.01 0.02 0.01 0.01 1 0.09 0.05 0.08 0.03 0.01 2 0.09 0.05 0.07 0.03 0.02 3 0.06 0.04 0.07 0.04 0.02 4 0.04 0.04 0.04 0.03 0.02 Find the conditional expectation E(Y|X = 4). (Round the final answer to four decimal places.)The General Social Survey asked a sample of adults how many siblings (brothers and sisters) they had (X) and also how many children they had (Y). We show results for those who had no more than 4 children and no more than 4 siblings. Assume that the joint probability mass function is given in the following contingency table: y x 0 1 2 3 4 0 0.03 0.01 0.02 0.01 0.01 1 0.09 0.05 0.08 0.03 0.01 2 0.09 0.05 0.07 0.03 0.02 3 0.06 0.04 0.07 0.04 0.02 4 0.04 0.04 0.04 0.03 0.02 Find the conditional probability mass function pY|X(y|4). (Round the final answer to four decimal places.) pY|X(0|4) = pY|X(1|4) = pY|X(2|4) = pY|X(3|4) = pY|X(4|4) = Find the conditional probability mass function pX|Y (x|3). (Round the final answer to four decimal places.) pX|Y(0|3) = pX|Y(1|3) = pX|Y(2|3) = pX|Y(3|3) = pX|Y(4|3) = Find the conditional expectation E(X|Y = 3). (Round the final answer to four decimal places.) E(X|Y = 3) =Given that the probability density function of the continuous random variable X is given as follows, what is E (6X + 3X²) equal to? f(x)={ 2(1-x), 0<x<1 0, e.w a)15/18 b)5/2 c)1/18 d)1/3
