2009 H2 Mathematics Paper 2 Question 9

The Normal Distribution

Answers

(i)

n=25{n = 25}

(ii)

0.203{0.203}

(iii)

0.0707{0.0707}

(iv)

The thickness of each mechanics textbook and the thickness of each statistics textbook are independent of one another

Full solutions

(i)

M‾∼N(2.5,0.12n)Z=M‾−2.50.1n∼N(0,1)\begin{align*} \overline{M} &\sim N\left( 2.5, \frac{0.1^2}{n} \right) \\ Z = \frac{\overline{M}-2.5}{\frac{0.1}{\sqrt{n}}} &\sim N(0,1) \\ \end{align*}
P(M‾>2.53)=0.0668P(Z>2.53−2.50.1n)=0.0668\begin{align*} \textrm{P}\left( \overline{M} > 2.53 \right) &= 0.0668 \\ \textrm{P}\left( Z > \frac{2.53-2.5}{\frac{0.1}{\sqrt{n}}} \right) &= 0.0668 \\ \end{align*}
0.03n0.1=1.500n=25  ■\begin{align*} \frac{0.03\sqrt{n}}{0.1} &= 1.500 \\ n &= 25 \; \blacksquare \end{align*}

(ii)

Let S{S} denote the thickness of a randomly chosen statistics textbook and T=M1+…+M21+S1+…+S24{T=M_1+ \ldots + M_{21}} \allowbreak {+ S_1+\ldots+S_{24}} denote the total thickness of 21 mechanics textbooks and 24 statistics textbooks
T∼N(21×2.5+24×2.0,21×0.12+24×0.082)T∼N(100.5,0.36360)\begin{align*} T &\sim \textrm{N}(21 \times 2.5 + 24 \times 2.0, 21 \times 0.1^2 + 24 \times 0.08^2 ) \\ T &\sim \textrm{N}( 100.5, 0.36360 ) \\ \end{align*}
P(T<100)=0.203 (3 sf)  ■\begin{align*} & \textrm{P}(T < 100) \\ &= 0.203 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(iii)

Let D=S1+…+S4−3M{D = S_1 + \ldots + S_4 - 3M}
D∼N(4×2.0−3×2.5,4×0.082+32×0.12)D∼N(0.5,0.11560)\begin{align*} D &\sim \textrm{N}(4 \times 2.0 - 3 \times 2.5, 4 \times 0.08^2 + 3^2 \times 0.1^2 ) \\ D &\sim \textrm{N}( 0.5, 0.11560 ) \\ \end{align*}
P(S1+…+S4<3M)=P(D<0)=0.0707 (3 sf)  ■\begin{align*} & \textrm{P}(S_1 + \ldots + S_4 < 3M) \\ &= \textrm{P}(D < 0) \\ &= 0.0707 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(iv)

The thickness of each mechanics textbook and the thickness of each statistics textbook are independent of one another ■{\blacksquare}