| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Exact binomial then normal approximation (same context, different n) |
| Difficulty | Moderate -0.3 This is a straightforward application of binomial distribution followed by a standard normal approximation with continuity correction. Part (a) requires direct use of binomial probability formulas or calculator functions, while part (b) is a textbook example of binomial-to-normal approximation. The calculations are routine with no conceptual challenges or problem-solving required beyond recognizing when to apply the normal approximation. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.04d Normal approximation to binomial |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X \sim B(15, 0.48)\) | M1 | Writing or using the binomial distribution. Allow for sight of \(B(15, 0.48)\) or in words: binomial with \(n=15\) and \(p=0.48\). May be implied by one correct answer to 3sf or sight of \(P(X \leqslant 4) = 0.07986...\). Allow \(^{15}C_3 \times 0.48^3 \times 0.52^{12}\) as "correct use". Condone \(B(0.48, 15)\) |
| \(P(X=3) = 0.019668...\) awrt \(0.0197\) | A1 | awrt 0.0197 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([P(X \geqslant 5)] = 1 - P(X \leqslant 4) = 0.92013...\) awrt \(0.920\) | A1 | awrt 0.920 (Allow 0.92) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(Y\) is number of hits: \(Y \sim N(120, 62.4)\) OR \(M\) is number of misses: \(M \sim N(130, 62.4)\) | B1 | Setting up a correct Normal model. Allow sight of \(N(120, 62.4)\) or \(N(130, 62.4)\) or \(N\!\left(120, \frac{312}{5}\right)\) or \(N\!\left(130, \frac{312}{5}\right)\). Normal with mean \(= 120/130\) and variance \(= 62.4\) or \(\text{sd} = \sqrt{62.4}\). Condone \(N(120, \sqrt{62.4})\) or \(N(130, \sqrt{62.4})\) or sd \(= 62.4\). Look out for \(\sigma = \frac{\sqrt{1560}}{5}\) or \(\frac{2\sqrt{390}}{5}\) or awrt 7.90 (condone 7.9). May be implied by sight of 0.897 or 0.8854... |
| \(P(X>110) \approx P\!\left(Z > \frac{110.5 - \text{"120"}}{\sqrt{\text{"62.4"}}}\right)\) OR \(P(X>110) \approx P\!\left(Z < \frac{139.5 - \text{"130"}}{\sqrt{\text{"62.4"}}}\right)\) | M1 | Sight of the continuity correction with a normal distribution. Accept 110.5 or 111.5 or 109.5 / 139.5 or 140.5 or 138.5. NB also allow 129.5 or 130.5 or 128.5 / 120.5 or 119.5 or 121.5. No continuity correction gives awrt 0.897 which is M0 unless CC seen |
| \(= 0.88544...\) | A1 | awrt 0.8854 or awrt 0.885 dependent on sight of \(>110.5\) or \(<129.5\) or \(<139.5\) or \(>120.5\). Allow \(\leqslant\) or \(\geqslant\) instead of \(<\) or \(>\). NB \(0.885548...\) from \(B(250, 0.48)\) scores M0A0 |
# Question 1:
## Part (a)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X \sim B(15, 0.48)$ | M1 | Writing or using the binomial distribution. Allow for sight of $B(15, 0.48)$ or in words: binomial with $n=15$ and $p=0.48$. May be implied by one correct answer to 3sf or sight of $P(X \leqslant 4) = 0.07986...$. Allow $^{15}C_3 \times 0.48^3 \times 0.52^{12}$ as "correct use". Condone $B(0.48, 15)$ |
| $P(X=3) = 0.019668...$ awrt $0.0197$ | A1 | awrt 0.0197 |
## Part (a)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[P(X \geqslant 5)] = 1 - P(X \leqslant 4) = 0.92013...$ awrt $0.920$ | A1 | awrt 0.920 (Allow 0.92) |
**(3 marks total for part a)**
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $Y$ is number of hits: $Y \sim N(120, 62.4)$ OR $M$ is number of misses: $M \sim N(130, 62.4)$ | B1 | Setting up a correct Normal model. Allow sight of $N(120, 62.4)$ or $N(130, 62.4)$ or $N\!\left(120, \frac{312}{5}\right)$ or $N\!\left(130, \frac{312}{5}\right)$. Normal with mean $= 120/130$ and variance $= 62.4$ or $\text{sd} = \sqrt{62.4}$. Condone $N(120, \sqrt{62.4})$ or $N(130, \sqrt{62.4})$ or sd $= 62.4$. Look out for $\sigma = \frac{\sqrt{1560}}{5}$ or $\frac{2\sqrt{390}}{5}$ or awrt 7.90 (condone 7.9). May be implied by sight of 0.897 or 0.8854... |
| $P(X>110) \approx P\!\left(Z > \frac{110.5 - \text{"120"}}{\sqrt{\text{"62.4"}}}\right)$ OR $P(X>110) \approx P\!\left(Z < \frac{139.5 - \text{"130"}}{\sqrt{\text{"62.4"}}}\right)$ | M1 | Sight of the continuity correction with a normal distribution. Accept **110.5** or 111.5 or 109.5 / **139.5** or 140.5 or 138.5. NB also allow **129.5** or 130.5 or 128.5 / **120.5** or 119.5 or 121.5. No continuity correction gives awrt 0.897 which is M0 unless CC seen |
| $= 0.88544...$ | A1 | awrt 0.8854 or awrt 0.885 dependent on sight of $>110.5$ or $<129.5$ or $<139.5$ or $>120.5$. Allow $\leqslant$ or $\geqslant$ instead of $<$ or $>$. NB $0.885548...$ from $B(250, 0.48)$ scores M0A0 |
**(3 marks total for part b)**
**(6 marks total)**\begin{enumerate}
\item George throws a ball at a target 15 times.
\end{enumerate}
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.\\
(a) Find\\
(i) $\mathrm { P } ( X = 3 )$\\
(ii) $\mathrm { P } ( X \geqslant 5 )$
George now throws the ball at the target 250 times.\\
(b) Use a normal approximation to calculate the probability that he will hit the target more than 110 times.
\hfill \mbox{\textit{Edexcel Paper 3 2022 Q1 [6]}}