Suppose X and Y are independent random variables with X - B (n, p) and Y ~ B (m, p). Show that X +Y ~ B (n + m, p).
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A: As per bartleby guidelines we can solve only first question and rest can be reposted
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A: Note: " Since you have asked multiple questions, we will solve the first question for you. If you…
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Q: . Then Var(X+Y+Z) is equal to:
A: Here x , y and z are independent variables So covariance between them are zero
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Q: Suppose we have two random variables X and Y. For a, b > 0, a2+b2=13. Var(aX) = 16 Var(bY) = 36…
A: Suppose we have two random variables X and Y. For a, b > 0, a2+b2=13. Here we use the…
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Q: probability distribution of two-dimensional random variable (X,Y)
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Q: 3. If X and Y are 2 random variables such that Y = ax + b, then the variance of Y is O A. aVar(Y)…
A: Note According to Bartleby guidelines expert solve only one question and rest can be reposted.
Q: There are three random variables X, Y and Z,for which we know the following information E(X)= 6,…
A: Given,V(X)=18 , V(Y)=22 , V(Z)=4Cov(X,Y)=-12 , Cov(X,Z)=5Cov(Y,Z)=8
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A: x1 ~ N(20,25) x2 ~ N(8,8) The distribution of the mean of x1 is x1 ~ N(u,σn) x1 ~ N(20,2516) x1…
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Q: Exercise 8. Suppose X1, X2, , Xn are independent random variables with X; ~ N(Hi, o), 1<i<n. If X =…
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A: Given X is a binomial random variable Number of trials=n=167 Probability of success=p=0.051
Q: If X and Y are two independent random variables, then Ox+y(@) = x (@) 0, (@) %3D
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Q: Show that for two random variables X and Y, 2 2 var(aX+bY) = a¹o²+b²oy² + 2ab Cxx XY where a and b…
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A: Given X is a binomial random variable n=No. of trials=125 p=Probability of success=0.016
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Q: and let Y; = Xí, i = 1, ..., n. %3D Find the distribution of W 2(Y1+Y2+Y3) 3(Y4+Y5)
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- Quiz 2 Suppose that X and Y are jointly discrete random variables with x+y for x = 0, 1, 2 and y = 0, 1, 2, 3 otherwise p(x, y) = 30 (a) Compute and plot px(x) and py(y). (b) Are X and Y independent? (c) Compute and plot Fx(x) and F-(y).Question * Suppose that X and Y are random variables with E(X)=2. E(Y) = 5 and E(X²) = 8, E(Y) = 30 and cov(X + 2Y.-3X+4Y)=-12, then cov(2X +3,-3Y+4) is equal to: Hint: cov(ax + by.cx+dY) = acV(X) + bdv(Y) + (bc + ad)cov(X,Y) O -120 -30 -59 -64 Ohelp please!
- The joint PMF of two random variables X and Y is given by y), Pxr(x, y) = {k(2x + 3 0, Where k is a constant a) What is the conditional PMF of Y given X b) What is the conditional PMF of X given Y x = 1,2; y = 1,2 elsewhereIt's desired to model the random variable X with a shape that rises to a peak near x=6 and whose possible values are integers 1-6. Copy paste the following lines of code into R: x <- 1:10 shape <- x*(12-x) barplot(shape, names. arg=x) This "shape" isn't a valid PMF because the numbers don't sum to 1. Convert the numbers in "shape" to valid probabilities and report P(X-5). Copy/paste all digits from R into your answer here.please solve a,b and c
- The spinner is spun one time. Find the probabilities. P(A) = P(C) = %3D A A E P(BUC)= P(B©) = Sample Space: P(B U A)= P(B U A)C = %3D P(D^)= P(AN C)= Page B.Let X ~ Binomial(3,0.5) and Y ~ Poisson(2) The two random variables are independent. Calculate the quantity, Var(2X-5Y)Construct a random variable X such that its first 2n moments exit and rest of moments don’t exit.