2015 H2 Mathematics Paper 2 Question 12

The Normal Distribution

Answers

(i)

0.0127{0.0127}

(ii)

0.0524{0.0524}

(iii)

0.742{0.742}

Full solutions

(i)

Let X{X} and Y{Y} denote the masses in grams of a randomly chosen apple and pear respectively
X∼N(300,202)Y∼N(200,152)\begin{align*} X &\sim \textrm{N}(300, 20^2 ) \\ Y &\sim \textrm{N}(200, 15^2 ) \\ \end{align*}
X1+X2+…+X5∼N(5×300,5×202)X1+X2+…+X5∼N(1500,2000)\begin{align*} X_1 + X_2 + \ldots + X_5 &\sim \textrm{N}(5\times 300, 5 \times 20^2) \\ X_1 + X_2 + \ldots + X_{5} &\sim \textrm{N}( 1500, 2000 ) \end{align*}
P(X1+X2+…+X5>1600)=0.0127 (3 sf)  ■\begin{align*} & \textrm{P}(X_1 + X_2 + \ldots + X_{5} > 1600) \\ &= 0.0127 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(ii)

Y1+Y2+…+Y8∼N(8×200,8×152)Y1+Y2+…+Y8∼N(1600,1800)\begin{align*} Y_1 + Y_2 + \ldots + Y_8 &\sim \textrm{N}(8\times 200, 8 \times 15^2) \\ Y_1 + Y_2 + \ldots + Y_{8} &\sim \textrm{N}( 1600, 1800 ) \end{align*}
Let U=X1+…+X5{U=X_1+\ldots+X_5} and V=Y1+…+Y8{V=Y_1+\ldots+Y_8}
U−V∼N(1500−1600,2000+1800)U−V∼N(−100,3800)\begin{align*} U - V &\sim \textrm{N}(1500-1600, 2000+1800 ) \\ U-V &\sim \textrm{N}( -100, 3800 ) \\ \end{align*}
P(U>V)=P(U−V>0)=0.0524 (3 sf)  ■\begin{align*} & \textrm{P}(U>V) \\ &= \textrm{P}(U-V > 0) \\ &= 0.0524 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(iii)

0.85U∼N(0.85×1500,0.852×2000)0.85U∼N(1275,1445)0.9V∼N(0.9×1600,0.92×1800)0.9V∼N(1440,1458)0.85U+0.9V∼N(1275+1440,1445+1458)0.85U+0.9V∼N(2715,2903)\begin{align*} 0.85 U &\sim \textrm{N}(0.85 \times 1500, 0.85^2 \times 2000 ) \\ 0.85 U &\sim \textrm{N}( 1275, 1445 ) \\ 0.9 V &\sim \textrm{N}(0.9 \times 1600, 0.9^2 \times 1800 ) \\ 0.9 V &\sim \textrm{N}( 1440, 1458 ) \\ 0.85 U + 0.9 V &\sim \textrm{N}(1275+1440, 1445+1458 ) \\ 0.85 U + 0.9 V &\sim \textrm{N}( 2715, 2903 ) \\ \end{align*}
P(0.85U+0.9V<2750)=0.742 (3 sf)  ■\begin{align*} & \textrm{P}(0.85 U + 0.9 V < 2750) \\ & = 0.742 \textrm{ (3 sf)} \; \blacksquare \end{align*}