2021 H2 Maths Paper 2 Q6 — Worked Solution

Solution

7 marks

Approach

we have two unknowns, $p$ and $q$ , but only one equation follows directly from the probabilities summing to $1$ . A second comes from the given variance, which needs both $E(X)$ and $E\left( X^{2} \right)$ . We express $q$ in terms of $p$ from the first equation, then substitute into $Var(X) = E\left( X^{2} \right) - \left\lbrack E(X) \right\rbrack^{2} = 1.61$ to get a quadratic in $p$ alone.

Let $X$ be the score obtained on the spinner, with distribution

$x$	1	2	3	4	5
$P(X = x)$	$0.2$	$0.3$	$p$	$p$	$q$

Since the probabilities sum to $1$ ,

0.2 + 0.3 + p + p + q = 1

\Rightarrow 2p + q = 0.5\quad\quad(1)

The mean is

E(X) = 1(0.2) + 2(0.3) + 3p + 4p + 5q

= 0.8 + 7p + 5q

Using (1), $q = 0.5 - 2p$ , so

E(X) = 0.8 + 7p + 5(0.5 - 2p)

= 3.3 - 3p\quad\quad(2)

Similarly,

E\left( X^{2} \right) = 1^{2}(0.2) + 2^{2}(0.3) + 3^{2}p + 4^{2}p + 5^{2}q

= 1.4 + 25p + 25q

= 1.4 + 25p + 25(0.5 - 2p)

= 13.9 - 25p\quad\quad(3)

Using $Var(X) = E\left( X^{2} \right) - \left\lbrack E(X) \right\rbrack^{2} = 1.61$ and substituting (2) and (3),

13.9 - 25p - (3.3 - 3p)^{2} = 1.61

\Rightarrow 13.9 - 25p - \left( 10.89 - 19.8p + 9p^{2} \right) = 1.61

\Rightarrow 3.01 - 5.2p - 9p^{2} = 1.61

\Rightarrow 9p^{2} + 5.2p - 1.4 = 0

By the quadratic formula (or GC),

p = \frac{- 5.2 \pm \sqrt{{5.2}^{2} + 4(9)(1.4)}}{2(9)}

= \frac{- 5.2 \pm \sqrt{77.44}}{18}

= \frac{- 5.2 \pm 8.8}{18}

This gives $p = 0.2$ or $p = - \frac{7}{9}$ . Since $p$ is a probability, $0 \leq p \leq 1$ , so $p = - \frac{7}{9}\ \left( \text{rej. p} < 0 \right)$ , and $p = 0.2$ .

q = 0.5 - 2(0.2) = 0.1

To check your answer: with

p = 0.2

q = 0.1

, all five probabilities

0.2,0.3,0.2,0.2,0.1

are non-negative and sum to

1

. Also

E\left( X^{2} \right) = 13.9 - 25(0.2) = 8.9

, giving

Var(X) = 8.9 - {2.7}^{2} = 1.61

, matching the given variance. ✓

Substituting into (2),

E(X) = 3.3 - 3(0.2) = 2.7

$\therefore$ the mean score is $2.7$ .

2021 H2 Maths Paper 2, Question 6

Solution

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