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Further Statistics 1: Probability generating functionsEdexcel A-Level Further Maths: Topic test

20 questions, 54 marks

Edexcel A-Level Further Maths

Further Statistics 1: Probability generating functions topic test

Total 54 marks

Name

Class

Date

  1. 1
    The discrete random variable XX takes the values 11, 22 and 33 and has probability generating function GX(t)=kt(1+t)2\mathrm{G}_X(t)=kt(1+t)^2, where kk is a constant.
    (a)
    What is the value of kk?
    [1 mark]
    • A18\frac18
    • B14\frac14
    • C13\frac13
    • D12\frac12
    (b)
    What is P(X=2)\mathrm{P}(X=2)?
    [1 mark]
    • A14\frac14
    • B13\frac13
    • C12\frac12
    • D34\frac34
    (c)
    Use GX(t)\mathrm{G}_X(t) to find E(X)\mathrm{E}(X).
    [2 marks]

    Total for question 1: 4 marks

  2. 2
    A quality inspector tests components one at a time until the first faulty one is found. Each component is faulty with probability 13\frac13, independently of the others. The random variable XX is the number of components tested.
    (a)
    What is P(X=3)\mathrm{P}(X=3)?
    [1 mark]
    • A427\frac{4}{27}
    • B827\frac{8}{27}
    • C29\frac29
    • D127\frac{1}{27}
    (b)
    What is Var(X)\mathrm{Var}(X)?
    [1 mark]
    • A22
    • B33
    • C99
    • D66
    (c)
    Show that the probability generating function of XX is GX(t)=t3−2t\mathrm{G}_X(t)=\dfrac{t}{3-2t}.
    [2 marks]

    Total for question 2: 4 marks

  3. 3
    The independent random variables XX and YY are the numbers of faulty items produced in one hour by two machines. X∼Po(1.5)X\sim\mathrm{Po}(1.5) and Y∼Po(2.5)Y\sim\mathrm{Po}(2.5). The random variable S=X+YS=X+Y.
    (a)
    Prove that the probability generating function of XX is GX(t)=e1.5(t−1)\mathrm{G}_X(t)=\mathrm{e}^{1.5(t-1)}.
    [3 marks]
    (b)
    Find the probability generating function of SS, identify the distribution of SS and hence find P(S=3)\mathrm{P}(S=3).
    [4 marks]

    Total for question 3: 7 marks

  4. 4
    The random variable XX is the number of trials needed to obtain the second success in a sequence of independent trials, each with probability of success 14\frac14. The probability generating function of XX is GX(t)=t2(4−3t)2\mathrm{G}_X(t)=\dfrac{t^2}{(4-3t)^2}.
    (a)
    Use GX(t)\mathrm{G}_X(t) to show that E(X)=8\mathrm{E}(X)=8 and Var(X)=24\mathrm{Var}(X)=24.
    [6 marks]
    (b)
    Two such sequences of trials are carried out independently and YY is the total number of trials needed. Find the probability generating function of YY, state E(Y)\mathrm{E}(Y) and Var(Y)\mathrm{Var}(Y), and use a binomial expansion to find P(Y=5)\mathrm{P}(Y=5).
    [6 marks]

    Total for question 4: 12 marks

  5. 5
    A farmer sows 55 seeds. Each seed germinates with probability 0.30.3, independently of the others. The random variable XX is the number of seeds that germinate.
    (a)
    Which expression is the probability generating function of XX?
    [1 mark]
    • A(0.3+0.7t)5(0.3+0.7t)^5
    • B5(0.7+0.3t)5(0.7+0.3t)
    • C(0.7+0.3t)5(0.7+0.3t)^5
    • D0.7+0.3t50.7+0.3t^5
    (b)
    What is E(X)\mathrm{E}(X)?
    [1 mark]
    • A3.53.5
    • B1.51.5
    • C1.051.05
    • D55
    (c)
    Find the coefficient of t2t^2 in the expansion of GX(t)\mathrm{G}_X(t) and state what it represents.
    [2 marks]

    Total for question 5: 4 marks

  6. 6
    The discrete random variable NN takes the values 0,1,2,…0,1,2,\dots and has probability generating function GN(t)=12−t\mathrm{G}_N(t)=\dfrac{1}{2-t} for ∣t∣<2|t|<2.
    (a)
    What is P(N=2)\mathrm{P}(N=2)?
    [1 mark]
    • A14\frac14
    • B12\frac12
    • C116\frac1{16}
    • D18\frac18
    (b)
    What is E(N)\mathrm{E}(N)?
    [1 mark]
    • A11
    • B22
    • C12\frac12
    • D14\frac14
    (c)
    Find Var(N)\mathrm{Var}(N).
    [2 marks]

    Total for question 6: 4 marks

  7. 7
    The independent random variables XX and YY are the numbers of defective items in two batches. Batch one has 88 items and batch two has 1212 items. Each item in either batch is defective with probability 14\frac14, independently. The random variable S=X+YS=X+Y.
    (a)
    Show that S∼B(20,14)S\sim\mathrm{B}\left(20,\frac14\right) by using probability generating functions.
    [3 marks]
    (b)
    Use GS(t)\mathrm{G}_S(t) to find E(S)\mathrm{E}(S) and Var(S)\mathrm{Var}(S).
    [4 marks]

    Total for question 7: 7 marks

  8. 8
    The number of emails an employee receives in one hour has a Poisson distribution with parameter λ\lambda, so the number NN has probability generating function GN(t)=eλ(t−1)\mathrm{G}_N(t)=\mathrm{e}^{\lambda(t-1)}. Two employees receive emails independently. In one hour the first receives X∼Po(1.2)X\sim\mathrm{Po}(1.2) emails and the second receives Y∼Po(1.8)Y\sim\mathrm{Po}(1.8) emails. The total is T=X+YT=X+Y.
    (a)
    Use differentiation of GN(t)\mathrm{G}_N(t) to prove that E(N)=Var(N)=λ\mathrm{E}(N)=\mathrm{Var}(N)=\lambda.
    [6 marks]
    (b)
    Find the probability generating function of TT and hence find P(T=0)\mathrm{P}(T=0) and P(T≥2)\mathrm{P}(T\geq2).
    [6 marks]

    Total for question 8: 12 marks

End of questions

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).