2010 H2 Mathematics Paper 2 Question 9

The Normal Distribution

Answers

(i)

0.681{0.681}

(ii)

0.234{0.234}

(iii)

0.362{0.362}

Full solutions

(i)

2X∼N(2×180,22×302)2X∼N(360,3600)Y−2X∼N(400−360,602+3600)Y−2X∼N(40,7200)\begin{align*} 2X &\sim \textrm{N}(2\times 180, 2^2 \times 30^2 ) \\ 2 X &\sim \textrm{N}( 360, 3600 ) \\ Y - 2X &\sim \textrm{N}(400 - 360, 60^2 + 3600 ) \\ Y - 2 X &\sim \textrm{N}( 40, 7200 ) \\ \end{align*}
P(Y>2X)=P(Y−2X>0)=0.681 (3 sf)  ■\begin{align*} & \textrm{P}(Y > 2X) \\ &= \textrm{P}(Y-2X > 0) \\ &= 0.681 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(ii)

0.12X∼N(0.12×180,0.122×302)0.12X∼N(21.6,12.96)0.05X∼N(0.05×400,0.052×602)0.05Y∼N(20,9)\begin{align*} 0.12X &\sim \textrm{N}(0.12\times 180, 0.12^2 \times 30^2 ) \\ 0.12 X &\sim \textrm{N}( 21.6, 12.96 ) \\ 0.05X &\sim \textrm{N}(0.05\times 400, 0.05^2 \times 60^2 ) \\ 0.05 Y &\sim \textrm{N}( 20, 9 ) \\ \end{align*}
Let C=0.12X+0.05X{C = 0.12X + 0.05X}
C∼N(21.6+20,12.96+9)C∼N(41.6,21.96)\begin{align*} C &\sim \textrm{N}(21.6+20, 12.96+9 ) \\ C &\sim \textrm{N}( 41.6, 21.96 ) \\ \end{align*}
P(C>45)=0.234 (3 sf)  ■\textrm{P}(C > 45) = 0.234 \textrm{ (3 sf)} \; \blacksquare

(iii)

Let W=0.12X{W = 0.12X}
W1+W2∼N(2×21.6,2×12.96)W1+W2∼N(43.2,25.92)\begin{align*} W_1 + W_2 &\sim \textrm{N}(2 \times 21.6, 2 \times 12.96 ) \\ W_1 + W_2 &\sim \textrm{N}( 43.2, 25.92 ) \\ \end{align*}
P(W1+W2>45)=0.362 (3 sf)  ■\textrm{P}(W_1 + W_2 > 45) = 0.362 \textrm{ (3 sf)} \; \blacksquare