Further Mathematics · AQA 7367 · Option S · Topic SA

Discrete random variables (DRVs) and expectation

Discrete random variables: probabilities, expectation, variance and the discrete uniform distribution.

  • 7 sub-topics
  • Taught in Year 12
  • 97 practice questions
  • 7 animated lessons
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What the specification says

Every sub-topic AQA can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. SA1Discrete random variables as tables or functionsY12
  2. SA2Probabilities for a DRVY12
  3. SA3Mean, variance, standard deviation, mode and median of a DRVY12
  4. SA4E(X), E(X²) and Var(X)Y12
  5. SA5Expectation and variance of functions of a DRVY12
  6. SA6Discrete uniform distributionY12
  7. SA7Proofs of the mean and variance of the discrete uniform distributionY12

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Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 97 questions in this topic, free, no account needed.

Discrete random variables (DRVs) and expectation

Easy

Variance from E(X) and E(X²) · Year 12 · 1 mark

  1. A discrete random variable has E(X)=2.5\mathrm{E}(X) = 2.5 and E(X2)=7.45\mathrm{E}(X^2) = 7.45. What is Var⁡(X)\operatorname{Var}(X)?
    (1)

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  1. Var⁡(X)\operatorname{Var}(X)=E(X2){}= \mathrm{E}(X^2)−(E(X))2{}- \big(\mathrm{E}(X)\big)^2.A: 1.2

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Discrete probability distributions

Medium

A discrete random variable · Year 12 · 7 marks

The discrete random variable XX has the probability distribution: P(X=1)\mathrm{P}(X = 1)=p, P(X=2){}= p, \ \mathrm{P}(X = 2)=q, P(X=3){}= q, \ \mathrm{P}(X = 3)=0.4, P(X=4){}= 0.4, \ \mathrm{P}(X = 4)=0.1{}= 0.1. Given that E(X)=2.4\mathrm{E}(X) = 2.4,
  1. (a)
    find the values of pp and qq.
    (3)
  2. (b)
    Find Var⁡(X)\operatorname{Var}(X).
    (3)
  3. (c)
    Find Var⁡(1+3X)\operatorname{Var}(1 + 3X).
    (1)
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Part (a)3 marks

  1. The probabilities add to 1.p+q=0.5p + q = 0.5
  2. Use E(X)\mathrm{E}(X)=∑xP(X=x){}= \sum x \mathrm{P}(X = x).p+2q+1.6=2.4p + 2q + 1.6 = 2.4
  3. p=0.2, q=0.3p = 0.2, \ q = 0.3

Part (b)3 marks

  1. ∑x2P(X=x)\sum x^{2} \mathrm{P}(X = x).E(X2)=6.6\mathrm{E}(X^2) = 6.6
  2. Var⁡(X)\operatorname{Var}(X)=6.6−2.42{}= 6.6 - 2.4^2=0.84{}= 0.84

Part (c)1 mark

  1. Var⁡(aX+b)\operatorname{Var}(aX + b)=a2Var⁡(X){}= a^{2} \operatorname{Var}(X).9×0.84=7.569 \times 0.84 = 7.56

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Discrete random variables (DRVs) and expectation

Hard

The discrete uniform distribution · Year 12 · 6 marks

X is equally likely to take each of the values 1, 2, …, 8.
  1. (a)
    Prove that E(X)=n+12\mathrm{E}(X) = \tfrac{n + 1}{2}, where n=8n = 8, and state the value.
    (3)
  2. (b)
    Use the result Var⁡(X)\operatorname{Var}(X)=112(n2−1){}= \tfrac{1}{12}(n^2 - 1) to find the variance of XX.
    (1)
  3. (c)
    Find E(2X−4)\mathrm{E}(2X - 4) and Var⁡(2X−4)\operatorname{Var}(2X - 4).
    (2)

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Part (a)3 marks

  1. E(X)\mathrm{E}(X)=1n∑r=1nr{}= \frac{1}{n}\sum_{r = 1}^{n} r=1n⋅n(n+1)2{}= \frac{1}{n} \cdot \frac{n(n + 1)}{2}
  2. =92= \frac{9}{2}

Part (b)1 mark

  1. 214\frac{21}{4}

Part (c)2 marks

  1. 55
  2. 4×214=214 \times \frac{21}{4} = 21

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