| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Poisson with binomial combination |
| Difficulty | Standard +0.3 This is a standard S2 Poisson question with routine applications: (a) direct Poisson probability calculation, (b) straightforward binomial-Poisson combination, (c) normal approximation requiring reverse lookup and algebraic manipulation. All techniques are textbook exercises with no novel insight required, though part (c) involves slightly more steps than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04d Normal approximation to binomial5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(X \sim \text{Po}(5)\); \(P(X \leq 5) = 0.6160\) | M1 A1 | For writing or using \(P(X \leq 5)\); awrt 0.616 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(X \sim B(4,\ \text{'0.616'})\) | B1ft | For \(X \sim B(4, 0.616)\). Follow through their part (a). May be implied by a correct ft expression for the 2nd M1 |
| \(P(X < 2) = P(X \leq 1)\) | M1 | For writing or using \(P(X \leq 1)\) (may be implied by 2nd M1) |
| \(= 0.384^4 + 4 \times 0.616 \times 0.384^3\) | M1 | For \([^4C_0](1-p)^4 + {^4C_1} \times p \times (1-p)^3\), \(0 < p < 1\) |
| \(= 0.16126...\) | A1 | awrt 0.161; correct answer on its own scores 4 out of 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(X \sim N\!\left(\frac{x}{16},\ \frac{x}{16}\right)\) | B1 | May be implied by values in standardisation |
| \(P(X < 26) = P\!\left(Z < \frac{25.5 - \frac{x}{16}}{\sqrt{\frac{x}{16}}}\right) = 0.5398\) | M1, B1 | For use of continuity correction either 25.5 or 26.5 (Allow 24.5); \(z = \pm 0.1\) allow calculator value if seen \(\pm 0.0999(2986...)\) |
| \(\frac{25.5 - \frac{x}{16}}{\frac{1}{4}\sqrt{x}} = 0.1\) | M1, A1ft | Standardising using 24.5, 25, 25.5, or 26, 26.5 and equate to a \(z\) value. Follow through their mean and variance; a correct equation with compatible signs |
| \(\frac{1}{16}x + \frac{1}{40}\sqrt{x} - 25.5 = 0 \to \sqrt{x} = 20\) (or \(\sqrt{x} = -20.4\)) | M1 | For solving their 3-term equation by factorising, completing the square or use of formula. May be implied by \(-20.4\), otherwise if answer is incorrect working must be shown |
| \((\sqrt{x})^2 = 20^2\) | M1 | For correct squaring of both sides. May be implied by \(416[.16]\) from correct equation. This mark may be scored prior to solving, e.g. \(\left(25.5 - \frac{x}{16}\right)^2 = \left(\frac{1}{40}\sqrt{x}\right)^2\). Do not award if squaring each individual term |
| \(x = 400\) | A1 | \(x = 400\) only. Dependent on all previous marks in (c). SC: Use of \(X \sim N\!\left(\frac{x}{16}, \frac{15x}{256}\right)\) leading to \(x=400\) scores max B0M1B1M1A0M1M1A0 |
# Question 5:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $X \sim \text{Po}(5)$; $P(X \leq 5) = 0.6160$ | M1 A1 | For writing or using $P(X \leq 5)$; awrt 0.616 |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $X \sim B(4,\ \text{'0.616'})$ | B1ft | For $X \sim B(4, 0.616)$. Follow through their part (a). May be implied by a correct ft expression for the 2nd M1 |
| $P(X < 2) = P(X \leq 1)$ | M1 | For writing or using $P(X \leq 1)$ (may be implied by 2nd M1) |
| $= 0.384^4 + 4 \times 0.616 \times 0.384^3$ | M1 | For $[^4C_0](1-p)^4 + {^4C_1} \times p \times (1-p)^3$, $0 < p < 1$ |
| $= 0.16126...$ | A1 | awrt 0.161; correct answer on its own scores 4 out of 4 |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $X \sim N\!\left(\frac{x}{16},\ \frac{x}{16}\right)$ | B1 | May be implied by values in standardisation |
| $P(X < 26) = P\!\left(Z < \frac{25.5 - \frac{x}{16}}{\sqrt{\frac{x}{16}}}\right) = 0.5398$ | M1, B1 | For use of continuity correction either 25.5 or 26.5 (Allow 24.5); $z = \pm 0.1$ allow calculator value if seen $\pm 0.0999(2986...)$ |
| $\frac{25.5 - \frac{x}{16}}{\frac{1}{4}\sqrt{x}} = 0.1$ | M1, A1ft | Standardising using 24.5, 25, 25.5, or 26, 26.5 and equate to a $z$ value. Follow through their mean and variance; a correct equation with compatible signs |
| $\frac{1}{16}x + \frac{1}{40}\sqrt{x} - 25.5 = 0 \to \sqrt{x} = 20$ (or $\sqrt{x} = -20.4$) | M1 | For solving their 3-term equation by factorising, completing the square or use of formula. May be implied by $-20.4$, otherwise if answer is incorrect working must be shown |
| $(\sqrt{x})^2 = 20^2$ | M1 | For correct squaring of both sides. May be implied by $416[.16]$ from correct equation. This mark may be scored prior to solving, e.g. $\left(25.5 - \frac{x}{16}\right)^2 = \left(\frac{1}{40}\sqrt{x}\right)^2$. Do not award if squaring each individual term |
| $x = 400$ | A1 | $x = 400$ only. Dependent on all previous marks in (c). SC: Use of $X \sim N\!\left(\frac{x}{16}, \frac{15x}{256}\right)$ leading to $x=400$ scores max B0M1B1M1A0M1M1A0 |\begin{enumerate}
\item A company produces steel cable.
\end{enumerate}
Defects in the steel cable produced by this company occur at random, at a constant rate of 1 defect per 16 metres.
On one day the company produces a piece of steel cable 80 metres long.\\
(a) Find the probability that there are at most 5 defects in this piece of steel cable.
The company produces a piece of steel cable 80 metres long on each of the next 4 days.\\
(b) Find the probability that fewer than 2 of these 4 pieces of steel cable contain at most 5 defects.
The following week the company produces a piece of steel cable $x$ metres long.\\
Using a normal approximation, the probability that this piece of steel cable has fewer than 26 defects is 0.5398\\
(c) Find the value of $x$
\hfill \mbox{\textit{Edexcel S2 2023 Q5 [14]}}