Combinations of random variables Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Combinations of random variables questions. See exactly how to solve problems on linear-transformation, expectation, variance, scaling-versus-sampling.

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Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
The random variable XX has E(X)=6E(X)=6 and Var⁡(X)=4\operatorname{Var}(X)=4. Find E(3X+5)E\left(3X+5\right).

Worked solution

  1. Quote the rule for the expectation of this combination

    E(3X+5)=3E(X)+5E\left(3X+5\right)=3E(X)+5

    Expectation is linear, so each coefficient simply multiplies its own mean.

  2. Substitute the given means

    E(3X+5)=3×6+5E\left(3X+5\right)=3\times 6+5

    Replace each EE by the number supplied in the question.

  3. State the expectation

    E(3X+5)=23E\left(3X+5\right)=23

    This is the mean of the combination.

Answer
E(3X+5)=23E\left(3X+5\right)=23
Question 2
2 markseasy
The random variable XX has E(X)=2.5E(X)=2.5 and Var⁡(X)=1.5\operatorname{Var}(X)=1.5. Find E(4X−7)E\left(4X-7\right).

Worked solution

  1. Quote the rule for the expectation of this combination

    E(4X−7)=4E(X)−7E\left(4X-7\right)=4E(X)-7

    Expectation is linear, so each coefficient simply multiplies its own mean.

  2. Substitute the given means

    E(4X−7)=4×2.5−7E\left(4X-7\right)=4\times 2.5-7

    Replace each EE by the number supplied in the question.

  3. Work out the arithmetic

    4×2.5−7=34\times 2.5-7=3

    Multiply each mean by its coefficient and add the results.

  4. State the expectation

    E(4X−7)=3E\left(4X-7\right)=3

    This is the mean of the combination.

Answer
E(4X−7)=3E\left(4X-7\right)=3
Question 3
2 markseasy
The random variable XX has E(X)=4E(X)=4 and Var⁡(X)=3\operatorname{Var}(X)=3, and X1X_{1} and X2X_{2} are independent observations of XX. Which of the following correctly compares Var⁡(2X)\operatorname{Var}\left(2X\right) and Var⁡(X1+X2)\operatorname{Var}\left(X_{1}+X_{2}\right)?

Worked solution

  1. Write down the variance of a single observation

    Var⁡(X)=3\operatorname{Var}(X)=3

    Both quantities are built from this one number.

  2. Treat 2X2X as a scaled single observation

    Var⁡(2X)=22×3=12\operatorname{Var}(2X)=2^{2}\times 3=12

    One item is measured and its value is doubled.

  3. Select the correct comparison

    Var⁡(2X)=4Var⁡(X)>2Var⁡(X)=Var⁡(X1+X2)\operatorname{Var}(2X)=4\operatorname{Var}(X)>2\operatorname{Var}(X)=\operatorname{Var}\left(X_{1}+X_{2}\right)

    The two variables have the same mean but different variances.

Answer
Var⁡(2X)=12, Var⁡(X1+X2)=6\operatorname{Var}(2X)=12,\ \operatorname{Var}\left(X_{1}+X_{2}\right)=6
Question 4
2 markseasy
The random variable XX has E(X)=6E(X)=6 and Var⁡(X)=4\operatorname{Var}(X)=4. Find Var⁡(3X+5)\operatorname{Var}\left(3X+5\right).

Worked solution

  1. Quote the rule for the variance of this combination

    Var⁡(3X+5)=(3)2Var⁡(X)\operatorname{Var}\left(3X+5\right)=\left(3\right)^{2}\operatorname{Var}(X)

    Each coefficient is SQUARED, and the constant term contributes nothing.

  2. Substitute the given variances

    Var⁡(3X+5)=(3)2×4\operatorname{Var}\left(3X+5\right)=\left(3\right)^{2}\times 4

    Replace each variance by the number supplied in the question.

  3. Evaluate the squared coefficients

    Var⁡(3X+5)=9×4\operatorname{Var}\left(3X+5\right)=9\times 4

    A negative coefficient becomes positive once it is squared.

  4. State the variance

    Var⁡(3X+5)=36\operatorname{Var}\left(3X+5\right)=36

    This is the variance of the combination.

Answer
Var⁡(3X+5)=36\operatorname{Var}\left(3X+5\right)=36
Question 5
2 markseasy
The random variable XX has E(X)=10E(X)=10 and Var⁡(X)=7\operatorname{Var}(X)=7. Find Var⁡(−2X+9)\operatorname{Var}\left(-2X+9\right).

Worked solution

  1. Quote the rule for the variance of this combination

    Var⁡(−2X+9)=(−2)2Var⁡(X)\operatorname{Var}\left(-2X+9\right)=\left(-2\right)^{2}\operatorname{Var}(X)

    Each coefficient is SQUARED, and the constant term contributes nothing.

  2. Substitute the given variances

    Var⁡(−2X+9)=(−2)2×7\operatorname{Var}\left(-2X+9\right)=\left(-2\right)^{2}\times 7

    Replace each variance by the number supplied in the question.

  3. State the variance

    Var⁡(−2X+9)=28\operatorname{Var}\left(-2X+9\right)=28

    This is the variance of the combination.

Answer
Var⁡(−2X+9)=28\operatorname{Var}\left(-2X+9\right)=28

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