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    HLPaper 2

    A random variable YYY is normally distributed with mean α\alphaα and standard deviation β\betaβ , such that P(Y<30.31)=0.1180\text{P}(Y < 30.31) = 0.1180P(Y<30.31)=0.1180 and P(Y>42.52)=0.3060\text{P}(Y > 42.52) = 0.3060P(Y>42.52)=0.3060.

    1.

    Find ααα and βββ.

    [6]
    Verified
    Solution

    This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

    P(Y<42.52)=0.6940P(Y < 42.52) = 0.6940P(Y<42.52)=0.6940 (M1)

    either P(Z<30.31−αβ)=0.1180P\left(Z < \frac{{30.31 - \alpha }}{\beta }\right) = 0.1180P(Z<β30.31−α​)=0.1180 or P(Z<42.52−αβ)=0.6940P\left(Z < \frac{{42.52 - \alpha }}{\beta }\right) = 0.6940P(Z<β42.52−α​)=0.6940 (M1)

    30.31−αβ=Φ−1(0.1180)⏟−1.1850…\frac{{30.31 - \alpha }}{\beta } = \underbrace {{\Phi ^{ - 1}}(0.1180)}_{ - 1.1850 \ldots }β30.31−α​=−1.1850…Φ−1(0.1180)​​ (A1)

    42.52−αβ=Φ−1(0.6940)⏟0.5072…\frac{{42.52 - \alpha }}{\beta } = \underbrace {{\Phi ^{ - 1}}(0.6940)}_{0.5072 \ldots }β42.52−α​=0.5072…Φ−1(0.6940)​​ (A1)

    attempting to solve simultaneously (M1)

    α=38.9\alpha = 38.9α=38.9 and β=7.22\beta = 7.22β=7.22 (A1)

    [6 marks]

    2.

    Find P(∣Y−α∣<1.2β){\text{P}}\left( {\left| {Y - \alpha } \right| < 1.2\beta } \right)P(∣Y−α∣<1.2β).

    [2]
    Verified
    Solution

    P(α−1.2β<Y<α+1.2β)P(\alpha - 1.2\beta < Y < \alpha + 1.2\beta)P(α−1.2β<Y<α+1.2β) (or equivalent eg. 2P(α<Y<α+1.2β)2P(\alpha < Y < \alpha + 1.2\beta)2P(α<Y<α+1.2β)) (M1) =0.770= 0.770=0.770 A1

    **Note: **Award (M1)A1 for P(−1.2<Z<1.2)=0.770P(-1.2 < Z < 1.2) = 0.770P(−1.2<Z<1.2)=0.770. [2 marks]

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