2022 H2 Maths Paper 2 Q8 — Worked Solution

Part (a)

2 marks

aX - bY \sim N\left( ap - bs,\ a^{2}q^{2} + b^{2}t^{2} \right)

Explanation: since $X$ and $Y$ are independent normal random variables, any linear combination of them is also normal. The mean of $aX - bY$ is $a\, E(X) - b\, E(Y) = ap - bs$ , and since $X$ and $Y$ are independent, $Var(aX - bY) = a^{2}Var(X) + b^{2}Var(Y) = a^{2}q^{2} + b^{2}t^{2}$ .

Part (b)(i)

2 marks

Let $V \sim N(\mu,4)$ , since the standard deviation is $2$ gives variance $2^{2} = 4$ . We are told $P(V > 8) = P(V < 4)$ . By symmetry of the normal curve about its mean $\mu$ , the values $4$ and $8$ must be equidistant from $\mu$ , so $\mu$ is their midpoint:

\mu = \frac{4 + 8}{2} = 6

So $V \sim N(6,4)$ .

$\therefore$ $V \sim N(6,4)$ , sketched above with the shaded region for (b)(ii).

Part (b)(ii)

2 marks

P(V > 10) = 0.0228\ \left( \text{3 s.f.} \right)

Explanation: standardising, with $\mu = 6$ and $\sigma = 2$ ,

Z = \frac{10 - 6}{2} = 2

P(V > 10) = P(Z > 2) = 1 - \Phi(2)

By GC, $\Phi(2) = 0.97725$ , so $1 - \Phi(2) = 0.02275 \approx 0.0228$ (3 s.f.).

To check your answer:

10

is

2

standard deviations above the mean of

6

, so

P(V > 10)

should be small —

0.0228

is consistent with this. ✓

Part (c)

3 marks

Approach

We first use the given relationship between the mean and variance of $W$ to find $p$ , then use this value of $p$ to compute the required probability directly from the two individual terms of the binomial distribution.

$W \sim B(8,p)$ , so $E(W) = 8p$ and $Var(W) = 8p(1 - p)$ . Since $E(W) = 1.2\, Var(W)$ :

8p = 1.2 \times 8p(1 - p)

Dividing both sides by $8p$ (since $p \neq 0$ ),

1 = 1.2(1 - p)

\Rightarrow 1 - p = \frac{1}{1.2} = \frac{5}{6}

\Rightarrow p = \frac{1}{6}

So $W \sim B\left( 8,\frac{1}{6} \right)$ .

P(W < 2) = P(W = 0) + P(W = 1)

= \binom{8}{0}\left( \frac{5}{6} \right)^{8} + \binom{8}{1}\left( \frac{1}{6} \right)\left( \frac{5}{6} \right)^{7}

= \frac{1015625}{1679616}

= 0.605\ \left( \text{3 s.f.} \right)

To check your answer: by GC, cumulative binomial

P(W \leq 1)

with

n = 8,\ p = \frac{1}{6}

gives

0.60468\ldots

, matching

\frac{1015625}{1679616} = 0.60468\ldots

✓

$\therefore$ $p = \frac{1}{6}$ and $P(W < 2) = \frac{1015625}{1679616} = 0.605$ (3 s.f.).

2022 H2 Maths Paper 2, Question 8

Part (a)

Part (b)(i)

Part (b)(ii)

Part (c)

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