June 2022 Paper 3 Q1
1. George throws a ball at a target 15 times.
Each time George throws the ball, the probability of the ball hitting the target is 0.48
The random variable \(X\) represents the number of times George hits the target in 15 throws.
George now throws the ball at the target 250 times.
| Scheme | Marks | AO |
|---|---|---|
| (i) \(X \sim \mathrm{B}(15,\ 0.48)\) | M1 | 3.3 |
| \(\mathrm{P}(X = 3) = 0.019668\ldots\) awrt 0.0197 | A1 | 3.4 |
| (ii) \([\mathrm{P}(X \geqslant 5) = 1 - \mathrm{P}(X \leqslant 4)] = 0.92013\ldots\) awrt 0.920 | A1 | 1.1b |
| (3) |
Notes
M1: Writing or using the binomial distribution in (i) or (ii) Allow for sight of B(15, 0.48) or in words: binomial with \(n = 15\) and \(p = 0.48\) may be implied in (i) or (ii) by one correct answer to 3sf or sight of \(\mathrm{P}(X \leqslant 4) = 0.07986\ldots\) i.e. awrt 0.0799.
Allow for \({}^{15}C_3 \times 0.48^3 \times 0.52^{12}\) as this is “correct use” Condone B(0.48, 15)
(i) A1: awrt 0.0197
(ii) A1: awrt 0.920 (Allow 0.92)
| Scheme | Marks | AO |
|---|---|---|
| \(Y\) is the number of hits: \(Y \sim \mathrm{N}(120,\ 62.4)\) or \(M\) is the number of misses: \(M \sim \mathrm{N}(130,\ 62.4)\) | B1 | 3.3 |
| \(\begin{aligned}&\mathrm{P}(X \gt 110) \approx \mathrm{P}(Y \gt 110.5)\ \left[= \mathrm{P}\left(Z \gt \dfrac{110.5 - \text{``}120\text{''}}{\sqrt{\text{``}62.4\text{''}}}\right)\right]\\[8pt]&\textbf{or }\ \mathrm{P}(X \gt 110) \approx \mathrm{P}(M \lt 139.5)\ \left[= \mathrm{P}\left(Z \lt \dfrac{139.5 - \text{``}130\text{''}}{\sqrt{\text{``}62.4\text{''}}}\right)\right]\end{aligned}\) | M1 | 3.4 |
| \(= 0.88544\ldots\) | A1 | 1.1b |
| (3) | ||
| (6 marks) |
Notes
B1: Setting up a correct Normal model. Allow sight of \(\mathrm{N}(120,\ 62.4)\) or \(\mathrm{N}(130,\ 62.4)\) or \(\mathrm{N}\left(120,\ \dfrac{312}{5}\right)\) or \(\mathrm{N}\left(130,\ \dfrac{312}{5}\right)\) or may be awarded if used correctly in standardisation
or in words: Normal with mean = 120/130 and variance = 62.4 or sd = \(\sqrt{62.4}\) condone \(\mathrm{N}(120,\ \sqrt{62.4})\) or \(\mathrm{N}(130,\ \sqrt{62.4})\) or sd = 62.4
Look out for \(\sigma = \dfrac{\sqrt{1560}}{5}\) or \(\dfrac{2\sqrt{390}}{5}\) or awrt 7.90 (condone 7.9)
This may be implied by sight of 0.897 or 0.8854…
M1: Sight of the continuity correction with a normal distribution
110.5 or 111.5 or 109.5 or 139.5 or 140.5 or 138.5
NB we will also allow 129.5 or 130.5 or 128.5 or 120.5 or 119.5 or 121.5
Continuity correction may be seen in standardisation
NB No continuity correction(CC) gives awrt 0.897 which is M0 unless CC seen
A1: awrt 0.8854 or awrt 0.885 dependent on sight of \(\gt 110.5\) or \(\lt 129.5\) or \(\lt 139.5\) or \(\gt 120.5\)
Allow \(\leqslant\) or \(\geqslant\) instead of \(\lt\) or \(\gt\)
NB 0.885548… from B(250, 0.48) scores M0A0