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The independent random variables X and Y are such that X ~ N(24, 25) and Y ~ N(40, 36). (a) Find (i) E(2X - 3Y), (ii) Var(2X - 3Y). (b) State the distribution that the random variable 2X - 3Y follows.

Mathematics

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Answers

3 parts
a(i)
-72
b

N(-72, 424)

Step-by-step solution

6 steps

Part a(i)· 2 steps

Find E(2X−3Y)E(2X - 3Y).

  1. Use the linearity property of expectation, E(aX+bY)=aE(X)+bE(Y)E(aX + bY) = aE(X) + bE(Y).

    E(2X−3Y)=2E(X)−3E(Y)E(2X - 3Y) = 2E(X) - 3E(Y)
  2. Substitute the given values E(X)=24E(X) = 24 and E(Y)=40E(Y) = 40.

    E(2X−3Y)=2(24)−3(40)E(2X - 3Y) = 2(24) - 3(40) E(2X−3Y)=48−120E(2X - 3Y) = 48 - 120 E(2X−3Y)=−72E(2X - 3Y) = \boxed{-72}

Part a(ii)424

Find Var(2X−3Y)Var(2X - 3Y).

  1. Use the property for the variance of a linear combination of independent random variables, Var(aX+bY)=a2Var(X)+b2Var(Y)Var(aX + bY) = a^2Var(X) + b^2Var(Y).

    Var(2X−3Y)=(2)2Var(X)+(−3)2Var(Y)Var(2X - 3Y) = (2)^2Var(X) + (-3)^2Var(Y)
  2. Substitute the given values Var(X)=25Var(X) = 25 and Var(Y)=36Var(Y) = 36.

    Var(2X−3Y)=4(25)+9(36)Var(2X - 3Y) = 4(25) + 9(36) Var(2X−3Y)=100+324Var(2X - 3Y) = 100 + 324 Var(2X−3Y)=424Var(2X - 3Y) = \boxed{424}

Part bN(-72, 424)

State the distribution that the random variable 2X−3Y2X - 3Y follows.

  1. Since X and Y are independent normal random variables, any linear combination of X and Y is also a normal random variable.

  2. The mean and variance of this new normal distribution are the values calculated in part (a).

    The distribution is N(μ,σ2)N(\mu, \sigma^2), where μ=E(2X−3Y)=−72\mu = E(2X - 3Y) = -72 and σ2=Var(2X−3Y)=424\sigma^2 = Var(2X - 3Y) = 424.
    The random variable 2X−3Y2X - 3Y follows a normal distribution:

    N(−72,424)\boxed{N(-72, 424)}
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