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    A dice manufacturer claims that for a novelty die he produces the probability of scoring thenumbers 1 to 5 are all equal, and the probability of a 6 is two times the probability ofscoring any of the other numbers.To test the manufacture’s claim one of the novelty dice is rolled 350 times and the numbersscored on the die are shown in the table below.A χ2 goodness of fit test is to be used with a 5% significance level.

    Question
    SLPaper 2

    A dice manufacturer claims that for a novelty die he produces the probability of scoring thenumbers 1 to 5 are all equal, and the probability of a 6 is two times the probability ofscoring any of the other numbers.

    To test the manufacture’s claim one of the novelty dice is rolled 350 times and the numbersscored on the die are shown in the table below.

    A χ2 goodness of fit test is to be used with a 5% significance level.

    1.

    Find the probability of scoring a six when rolling the novelty die.

    [3]
    Verified
    Solution

    * This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

    Let the probability of scoring1, … , 5bep ,

    5p+2p=1⇒p=17 (M1)(A1)

    Probability of6=27 A1

    [3 marks]

    2.

    Find the probability of scoring more than 2 sixes when this die is rolled 5 times.

    [4]
    Verified
    Solution

    Let the number of sixes beX

    X~B5, 27 (M1)

    PX>2=PX≥3orPX>2=1-PX≤2 (M1)

    =0.1450.144701… (M1)A1

    [4 marks]

    3.

    Find the expected frequency for each of the numbers if the manufacturer’sclaim is true.

    [2]
    Verified
    Solution

    Expected frequency is350×por350×2p (M1)

    A1

    [2 marks]

    4.

    Write down the null and alternative hypotheses.

    [2]
    Verified
    Solution

    H0: The manufacture’s claim is correct A1
    H1:The manufacturer’s claim is not correct A1

    [2 marks]

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    5.

    State the degrees of freedom for the test.

    [1]
    Verified
    Solution

    Degrees of freedom =5 A1

    [1 mark]

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    6.

    Determine the conclusion of the test, clearly justifying your answer.

    [4]
    Verified
    Solution

    p-value=0.09840.0984037… (M1)A1

    0.0984>0.05 R1

    Hence insufficient evidence to reject the manufacture’s claim. A1

    [4 marks]

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    Related topics

    SL 4.11—Expected, observed, hypotheses, chi squared, gof, t-test

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