2008 H2 Mathematics Paper 2 Question 11

The Normal Distribution

Answers

(i)

0.0385{0.0385}

(ii)

0.0924{0.0924}

(iii)

E(Y)=110{\mathrm{E}(Y) = 110}
Var(Y)=576{\mathrm{Var}(Y)=576}
a=3{a=3}
b=−40{b=-40}

Full solutions

(i)

X1+X2∼N(2×50,2×82)X1+X2∼N(100,128)\begin{align*} X_1 + X_2 &\sim \textrm{N}(2 \times 50, 2 \times 8^2 ) \\ X_1 + X_2 &\sim \textrm{N}( 100, 128 ) \\ \end{align*}
P(X1+X2>120)=0.0385 (3 sf)  ■\begin{align*} & \textrm{P}(X_1 + X_2 > 120) \\ &= 0.0385 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(ii)

X1−X2∼N(50−50,82+82)X1−X2∼N(0,128)\begin{align*} X_1 - X_2 &\sim \textrm{N}(50-50, 8^2 + 8^2 ) \\ X_1 - X_2 &\sim \textrm{N}( 0, 128 ) \\ \end{align*}
P(X1>X2+15)=P(X1−X2>15)=0.0924 (3 sf)  ■\begin{align*} & \textrm{P}( X_1 > X_2 + 15) \\ &= \textrm{P}(X_1 - X_2 > 15) \\ &= 0.0924 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(iii)

By symmetry,
E(Y)=74+1462=110  ■\begin{align*} \textrm{E}(Y) &= \frac{74+146}{2} \\ &= 110 \; \blacksquare \end{align*}
Let Y∼N(110,σ2)Z=X−110σ∼N(0,1)\begin{gather*} \textrm{Let } Y \sim \textrm{N}(110, \sigma^2) \\ Z = \frac{X-110}{\sigma} \sim \textrm{N}(0, 1) \end{gather*}
P(Y<74)=0.0668P(Z<74−110σ)=0.0668\begin{align*} \textrm{P}(Y < 74) &= 0.0668 \\ \textrm{P}\left(Z < \frac{74-110}{\sigma}\right) &= 0.0668 \\ \end{align*}
120−110σ=−1.5001σ=23.999\begin{align*} \frac{120-110}{\sigma} &= -1.5001 \\ \sigma &= 23.999 \end{align*}
Var(Y)=23.9992=576  ■\begin{align*} \textrm{Var}(Y) &= 23.999^2 \\ &= 576 \; \blacksquare \end{align*}
aX+b=YVar(aX+b)=Var(Y)a2Var(X)=57682a2=576\begin{align*} aX + b &= Y \\ \textrm{Var}(aX+b) &= \textrm{Var}(Y) \\ a^2 \textrm{Var}(X) &= 576 \\ 8^2 a^2 &= 576 \end{align*}
Since a>0,{a > 0,}
a=3  ■a=3 \; \blacksquare
aX+b=YE(aX+b)=E(Y)aE(X)+b=1103(50)+b=110b=−40  ■\begin{align*} aX + b &= Y \\ \textrm{E}(aX+b) &= \textrm{E}(Y) \\ a\textrm{E}(X) + b &= 110 \\ 3(50) + b &= 110 \\ b &= -40 \; \blacksquare \end{align*}