2016 H2 Mathematics Paper 2 Question 9

The Binomial Distribution
The Normal Distribution

Answers

(a)

7.41{7.41}

(b)

p=0.166 or p=0.599{p=0.166 \textrm{ or } p = 0.599}

(c)

Out of syllabus

Full solutions

(a)

X∼N(15,a2)Z=X−15a∼N(0,1)\begin{gather*} X \sim \mathrm{N}(15, a^2) \\ Z = \frac{X-15}{a} \sim \mathrm{N}(0, 1) \end{gather*}
P(10<X<20)=0.5P(10−15a<Z<20−15a)=0.5P(−5a<Z<5a)=0.5\begin{gather*} \mathrm{P}(10 < X < 20) = 0.5 \\ \mathrm{P}\left(\frac{10-15}{a} < Z < \frac{20-15}{a}\right) = 0.5 \\ \mathrm{P}\left(-\frac{5}{a} < Z < \frac{5}{a}\right) = 0.5 \\ \end{gather*}
5a=0.67449a=7.41 (3 sf)  ■\begin{align*} \frac{5}{a} &= 0.67449 \\ a &= 7.41 \textrm{ (3 sf)} \; \blacksquare \end{align*}

(b)

P(Y=1)+P(Y=2)=0.5(41)p1(1−p)3+(42)p2(1−p)2=0.54p(1−3p2+3p−p3)+6p2(1−2p+p2)=0.54p−12p2+12p3−4p4+6p2−12p3+6p4=0.5\begin{gather*} \mathrm{P}(Y=1) + \mathrm{P}(Y=2) = 0.5 \\ {4 \choose 1}p^1(1-p)^3 + {4 \choose 2}p^2(1-p)^2 = 0.5 \\ 4p(1-3p^2+3p-p^3) + 6p^2(1-2p+p^2) = 0.5 \\ 4p-12p^2+12p^3-4p^4 + 6p^2-12p^3+6p^4 = 0.5 \\ \end{gather*}
2p4−6p2+4p=0.54p4−12p2+8p=1  ■\begin{gather*} 2p^4 - 6 p^2 + 4p = 0.5 \\ 4 p^4 - 12 p^2 + 8 p = 1 \; \blacksquare \end{gather*}
Solving with a GC, since 0≤p≤1,{0 \leq p \leq 1,}
p=0.166   or   p=0.599 (3 sf)  ■p=0.166 \; \textrm{ or } \; p=0.599 \textrm{ (3 sf)} \; \blacksquare

(c)

Out of syllabus