b) Use the theorem below to express E(W) = E(5Y) in terms of u, where u = E(Y). Let Y be a discrete random variable with probability function p(y), g(Y) be a function of Y, and c be a constant. Then E[cg(Y)] = CE[g(Y)]. E(W) = (| Does this result agree with your answer to part (a)? O Yes O No c) Recalling that the variance is a measure of spread or dispersion, do you expect the variance of W to be larger than, smaller than, or equal to o? = v(Y)? Why? O The variance of W will be smaller than o?, since the spread of values of W has decreased. O The variance of W will be equal to o?, since the spread of values of W has not changed. O The variance of W will be larger than o?, since the spread of values of W has decreased. O The variance of W will be larger than o?, since the spread of values of W has increased. O The variance of W will be smaller than o2, since the spread of values of W has increased. d) Use the definition below and the result in part (b) to show that V(W) = E{[W – E(W)]²} = E[25(Y – µ)²] = 2502; that is, W = 5Y has variance twenty-five times that of Y. If Y is a random variable with mean E(Y) = µ, the variance of a random variable Y is defined to be the expected value of (Y - µ)?. That is, V(Y) = E[(Y – µ)²]. V(W) = E[(W – E(W))?] %3D - = 2502
b) Use the theorem below to express E(W) = E(5Y) in terms of u, where u = E(Y). Let Y be a discrete random variable with probability function p(y), g(Y) be a function of Y, and c be a constant. Then E[cg(Y)] = CE[g(Y)]. E(W) = (| Does this result agree with your answer to part (a)? O Yes O No c) Recalling that the variance is a measure of spread or dispersion, do you expect the variance of W to be larger than, smaller than, or equal to o? = v(Y)? Why? O The variance of W will be smaller than o?, since the spread of values of W has decreased. O The variance of W will be equal to o?, since the spread of values of W has not changed. O The variance of W will be larger than o?, since the spread of values of W has decreased. O The variance of W will be larger than o?, since the spread of values of W has increased. O The variance of W will be smaller than o2, since the spread of values of W has increased. d) Use the definition below and the result in part (b) to show that V(W) = E{[W – E(W)]²} = E[25(Y – µ)²] = 2502; that is, W = 5Y has variance twenty-five times that of Y. If Y is a random variable with mean E(Y) = µ, the variance of a random variable Y is defined to be the expected value of (Y - µ)?. That is, V(Y) = E[(Y – µ)²]. V(W) = E[(W – E(W))?] %3D - = 2502
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
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![(b) Use the theorem below to express E(W) = E(5Y) in terms of u, where u = E(Y).
Let Y be a discrete random variable with probability function p(y), g(Y) be a function of Y, and c be a constant. Then E[cq(Y)] = CE[g(Y)].
E(W) =
Does this result agree with your answer to part (a)?
O Yes
O No
(c) Recalling that the variance is a measure of spread or dispersion, do you expect the variance of W to be larger than, smaller than, or equal to o? = V(Y)? Why?
O The variance of W will be smaller than o2, since the spread of values of W has decreased.
O The variance of W will be equal to o?, since the spread of values of W has not changed.
O The variance of W will be larger than o, since the spread of values of W has decreased.
O The variance of W will be larger than o2, since the spread of values of W has increased.
O The variance of W will be smaller than o?, since the spread of values of W has increased.
(d) Use the definition below and the result in part (b) to show that
V(W) = E{[W - E(W)]²} = E[25(Y – µ)²] = 250²;
that is, W = 5Y has variance twenty-five times that of Y.
If Y is a random variable with mean E(Y) = u, the variance of a random variable Y is defined to be the expected value of (Y - u)2. That is, V(Y) = E[(Y - u)2].
V(W) = E[(W – E(W))²]
=
DECY -
= 2502](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faad9e4ed-ed50-4fe3-96ca-72ed3d364f6f%2F05be3a34-40be-4edf-97ab-f4eca5c794ba%2Fgqymhb9_processed.png&w=3840&q=75)
Transcribed Image Text:(b) Use the theorem below to express E(W) = E(5Y) in terms of u, where u = E(Y).
Let Y be a discrete random variable with probability function p(y), g(Y) be a function of Y, and c be a constant. Then E[cq(Y)] = CE[g(Y)].
E(W) =
Does this result agree with your answer to part (a)?
O Yes
O No
(c) Recalling that the variance is a measure of spread or dispersion, do you expect the variance of W to be larger than, smaller than, or equal to o? = V(Y)? Why?
O The variance of W will be smaller than o2, since the spread of values of W has decreased.
O The variance of W will be equal to o?, since the spread of values of W has not changed.
O The variance of W will be larger than o, since the spread of values of W has decreased.
O The variance of W will be larger than o2, since the spread of values of W has increased.
O The variance of W will be smaller than o?, since the spread of values of W has increased.
(d) Use the definition below and the result in part (b) to show that
V(W) = E{[W - E(W)]²} = E[25(Y – µ)²] = 250²;
that is, W = 5Y has variance twenty-five times that of Y.
If Y is a random variable with mean E(Y) = u, the variance of a random variable Y is defined to be the expected value of (Y - u)2. That is, V(Y) = E[(Y - u)2].
V(W) = E[(W – E(W))²]
=
DECY -
= 2502
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