2019 H2 Mathematics Paper 2 Question 11

The Normal Distribution

Answers

(i)

0.970{0.970}

(ii)

0.802{0.802}

(iii)

M=381{M = 381}

(iv)

0.846{0.846}

Full solutions

(i)

Let W{W} and B{B} denote the mass of a randomly chosen white and black ball respectively
W1+W2+…+W4∼N(4×110,4×42)W1+W2+…+W4∼N(440,64)\begin{align*} W_1 + W_2 + \ldots + W_4 &\sim \textrm{N}(4 \times 110, 4 \times 4^2) \\ W_1 + W_2 + \ldots + W_{4} &\sim \textrm{N}( 440, 64 ) \\ \end{align*}
P(W1+…+W4>425)=0.970 (3 sf)  ■\begin{align*} & \textrm{P}(W_1 + \ldots + W_4 > 425) \\ &= 0.970 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(ii)

W+B∼N(110+55,42+22)W+B∼N(165,20)\begin{align*} W + B &\sim \textrm{N}(110 + 55, 4^2 + 2^2 ) \\ W + B &\sim \textrm{N}( 165, 20 ) \\ \end{align*}
P(161<W+B<175)=0.802 (3 sf)  ■\begin{align*} & \textrm{P}(161 < W + B < 175) \\ &= 0.802 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(iii)

Let T=W1+W2+B1+B2+B3{T=W_1+W_2+}\allowbreak{B_1+B_2+B_3}
T∼N(2×110+3×55,2×42+3×22)T∼N(385,44)\begin{align*} T &\sim \textrm{N}(2\times 110 + 3 \times 55, 2 \times 4^2 + 3 \times 2^2) \\ T &\sim \textrm{N}( 385, 44 ) \\ \end{align*}
P(T<M)=0.271M=381 (3 sf)  ■\begin{align*} \textrm{P}(T < M) &= 0.271 \\ M &= 381 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(iv)

Let R{R} denote the mass of a randomly chosen connecting rod.
Let B∗{B^*} and W∗{W^*} denote the mass of the black and white ball respectively after drilling.
Let A=B1∗+…+B4∗+W∗+R1+…+R4{A = B_1^* + \ldots + B_4^*}+W^*+{R_1 + \ldots + R_4}
E(A)=4(0.9)E(B)+0.7E(W)+4E(R)=355\begin{align*} & \textrm{E}(A) \\ &= 4(0.9)\textrm{E}(B) + 0.7\textrm{E}(W) + 4\textrm{E}(R) \\ &= 355 \end{align*}
Var(A)=4(0.9)2Var(B)+0.72Var(W)+4Var(R)=24.04\begin{align*} & \textrm{Var}(A) \\ &= 4(0.9)^2\textrm{Var}(B) + 0.7^2\textrm{Var}(W) + 4\textrm{Var}(R) \\ &= 24.04 \end{align*}
A∼N(355,24.04)A \sim \textrm{N}( 355, 24.04 )
P(A>350)=0.846 (3 sf)  ■\begin{align*} & \textrm{P}(A > 350) \\ &= 0.846 \textrm{ (3 sf)} \; \blacksquare \end{align*}