Further Mathematics · Pearson Edexcel 9FM0 · Option FS1 · Topic 1

Discrete probability distributions

Random variables that take separate values. You find their mean (expected value) and variance, including for functions of the variable.

  • 1 sub-topic
  • Taught in Year 12
  • 19 practice questions
  • 1 animated lesson

What the specification says

Every sub-topic Pearson Edexcel 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. 1.1Discrete probability distributionsY12

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Easy and medium questions from this topic, each with a worked solutions and the mark for every step. All 19 questions in this topic, free, no account needed.

Discrete probability distributions

Easy

E(X²) for a discrete distribution · Year 12 · 1 mark

  1. A discrete random variable has P(X=2)=0.1\mathrm{P}(X = 2) = 0.1, P(X=3)=0.6\mathrm{P}(X = 3) = 0.6, P(X=4)=0.3\mathrm{P}(X = 4) = 0.3. What is E(X2)\mathrm{E}(X^2)?
    (1)

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  1. E(X2)\mathrm{E}(X^2)=∑x2 P(X=x){}= \sum x^2\,\mathrm{P}(X = x). It is not the square of E(X)\mathrm{E}(X).CC: 10.6

Total1

  • MMethod mark: for a correct method, even with a slip.
  • AAccuracy mark: for a correct answer from a correct method.
  • BIndependent mark: for a correct fact or result on its own, with no method needed.
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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

Total7

  • MMethod mark: for a correct method, even with a slip.
  • AAccuracy mark: for a correct answer from a correct method.
  • BIndependent mark: for a correct fact or result on its own, with no method needed.
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