2022 H2 Mathematics Paper 2 Question 8

The Binomial Distribution
The Normal Distribution

Answers

(a)

N(ap−bs,a2q2+b2t2){\mathrm{N} \left( ap-bs, a^2q^2+b^2t^2 \right)}
(bi)
(bii)
0.0228.{0.0228.}

(c)

0.605.{0.605.}

Full solutions

(a)

aX−bY∼N(ap−bs,a2q2+b2t2)  ■aX - bY \sim \mathrm{N} \left( ap-bs, a^2q^2+b^2t^2 \right) \; \blacksquare
(bi)
Since P(V>8){\mathrm{P}(V>8)} is equal to P(V<4),{\mathrm{P}(V<4),} the mean of V{V} is 6.{6.}
V∼N(6,22)V \sim \mathrm{N}(6, 2^2)
(bii)
By symmetry, E(V)=6.{\mathrm{E}(V) = 6.}
V∼N(6,22)P(V>10)=0.0228 (3 sf)  ■\begin{align*} V &\sim \mathrm{N} \left( 6, 2^2 \right) \\ \mathrm{P}(V > 10) &= 0.0228 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(c)

E(W)=1.2Var(W)8p=1.2(8)p(1−p)9.6p2−1.6p=0p(9.6p−1.6)=0\begin{gather*} \mathrm{E}(W) = 1.2\mathrm{Var}(W) \\ 8p = 1.2(8)p(1-p) \\ 9.6p^2 -1.6p = 0 \\ p (9.6p -1.6) = 0 \end{gather*}
Since p≠0,{p \neq 0,}
p=1.69.6=16\begin{align*} p &= \frac{1.6}{9.6} \\ &= \frac{1}{6} \end{align*}
P(W<2)=P(W≤1)=0.605(3 sf)  ■\begin{align*} \textrm{P}(W<2) &= \textrm{P}(W \leq 1) \\ &= 0.605 ( 3 \textrm{ sf}) \; \blacksquare \end{align*}

Question Commentary

This question tests standard techniques when handling normal and binomial distributions (mean and variance formulas, normcdf and binomcdf). It is slightly more abstract than usual, so we will need to be confident with our algebraic manipulation skills in addition to the underlying statistical concepts and formulas.