2016 H2 Maths Paper 2 Q9 — Worked Solution

Part (a)

2 marks

Since $X \sim N\left( 15,\ a^{2} \right)$ , the distribution is symmetric about $\mu = 15$ , and $10$ and $20$ are symmetric about $15$ (each $5$ away from the mean). Since $P(10 < X < 20) = 0.5$ , the two remaining tails together have probability $0.5$ , split equally by symmetry:

P(X < 10) = P(X > 20) = \frac{1 - 0.5}{2} = 0.25

so

P(X < 20) = 1 - 0.25 = 0.75.

Standardising, $Z = \frac{X - 15}{a} \sim N(0,1)$ , and

P\left( Z < \frac{20 - 15}{a} \right) = 0.75

\frac{5}{a} = 0.674\ \text{(by GC, invNorm}(0.75)\text{)}

a = \frac{5}{0.6744897\ldots}

= 7.41\ \text{(3 s.f.)}

$\therefore a = 7.41$ .

Part (b)

4 marks

Approach

$P(Y = 1) + P(Y = 2)$ expands to a polynomial in $p$ ; we expand and collect it to reach the given quartic, then solve the quartic by GC and reject any root outside $0 < p < 1$ (since $p$ is a probability).

Let $Y \sim B(4,p)$ . Then

P(Y = 1) = \binom{4}{1}p(1 - p)^{3} = 4p(1 - p)^{3}

P(Y = 2) = \binom{4}{2}p^{2}(1 - p)^{2} = 6p^{2}(1 - p)^{2}.

Given $P(Y = 1) + P(Y = 2) = 0.5$ ,

4p(1 - p)^{3} + 6p^{2}(1 - p)^{2}

= 4p\left( 1 - 3p + 3p^{2} - p^{3} \right) + 6p^{2}\left( 1 - 2p + p^{2} \right)

= 4p - 12p^{2} + 12p^{3} - 4p^{4} + 6p^{2} - 12p^{3} + 6p^{4}

= 2p^{4} - 6p^{2} + 4p

= 0.5.

Multiplying both sides by $2$ ,

4p^{4} - 12p^{2} + 8p = 1\ \ \text{(shown)}

Rearranging and solving $4p^{4} - 12p^{2} + 8p - 1 = 0$ by GC,

p = 0.166,\ 0.599,\ 1.25\ \text{(rej. as }p > 1\text{)},\ - 2.01\ \text{(rej. as }p < 0\text{)}

all values by GC, the last two rejected since $0 < p < 1$ .

$\therefore\ p = 0.166$ or $p = 0.599$ (3 s.f.).

Part (c)

4 marks

Approach

each question has probability $\frac{1}{3}$ of being guessed correctly, and George answers $100$ questions independently, so the number correct is $B\left( 100,\ \frac{1}{3} \right)$ . With $np = 33.3 > 5$ and $n(1 - p) = 66.7 > 5$ , we approximate this by a normal distribution, applying a continuity correction since we are approximating a discrete distribution by a continuous one.

Let $W$ be the number of questions George guesses correctly, out of $100$ . Then

W \sim B\left( 100,\ \frac{1}{3} \right).

Since $np = 100 \times \frac{1}{3} = 33.3 > 5$ and $n(1 - p) = 100 \times \frac{2}{3} = 66.7 > 5$ , we use the approximation

W \sim N(33.3,\ 22.2)\ \text{approximately,}

where $E(W) = 33.3$ and $Var(W) = 100 \times \frac{1}{3} \times \frac{2}{3} = 22.2$ (to 3 s.f.).

Applying a continuity correction (since $W$ is discrete),

P(W \geq 30)

\approx P(W > 29.5)

= 0.792\ \text{(3 s.f., by GC).}

To check your answer: the exact binomial value is

P(W \geq 30) = 0.791

(3 s.f., by GC), very close to the normal approximation

0.792

. ✓

$\therefore\ P\left( \text{George guesses at least }30\text{ questions right} \right) = 0.792$ .

2016 H2 Maths Paper 2, Question 9

Part (a)

Part (b)

Part (c)

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