Edexcel FS1 2024 June — Question 2 7 marks

Exam BoardEdexcel
ModuleFS1 (Further Statistics 1)
Year2024
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypePoisson hypothesis test
DifficultyStandard +0.8 This is a standard Further Maths Statistics question covering Poisson probability calculations and hypothesis testing with critical regions. Part (a) involves routine Poisson probability calculations. Parts (b) and (c) require understanding of two-tailed hypothesis tests, finding critical regions from tables, and calculating Type I error probability. The word count scaling (250 vs 100) adds a minor complication but is straightforward. This is typical FS1 material requiring multiple techniques but no novel insight, placing it moderately above average A-level difficulty.
Spec5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.05c Hypothesis test: normal distribution for population mean

  1. The number of errors made by a secretary is modelled by a Poisson distribution with a mean of 2.4 per 100 words.
A 100-word piece of work completed by the secretary is selected at random.
  1. Find the probability that
    1. there are exactly 3 errors,
    2. there are fewer than 2 errors. After a long holiday, a randomly selected piece of work containing 250 words completed by the secretary is examined to see if the rate of errors has changed.
  2. Stating your hypotheses clearly, and using a \(5 \%\) level of significance, find the critical region for a suitable test.
  3. Find P (Type I error) for the test in part (b)

AnswerMarks Guidance
(a)(i)\(0.20901\ldots\) awrt \(\mathbf{0.209}\) B1
(a)(ii)\(0.30844\ldots\) awrt \(\mathbf{0.308}\) B1
(b)\(H_0: \lambda = 2.4\) (or \(\mu = 6\)) \(\quad H_1: \lambda \neq 2.4\) (or \(\mu \neq 6\)) B1
\([E = \text{no. of errors}] E \sim \text{Po}(6)\)M1 3.3
\(P(E \leqslant 1) = 0.0174\) or \(P(E \leqslant 2) = 0.0620\) and \(P(E \leqslant 11) = 0.980\) or \(P(E \geqslant 12) = 0.0201\)M1 3.4
Critical region: \(E \leqslant 1\) or \(E \geqslant 12\)A1 1.1b
(c)\([P(\text{Type I error}) = 0.0174 + 0.0201 =] \mathbf{0.0375}\) (Calc gives: \(0.017351\ldots + 0.0200919\ldots = 0.037443\ldots\)) B1ft
Notes:
- (a) 1st B1 for awrt 0.209; 2nd B1 for awrt 0.308
- (b) B1 for both hypotheses correct in terms of \(\lambda\) or \(\mu\) (allow \(\lambda = 6\) etc)
- (b) 1st M1 for selecting the correct model. Sight or use of Po(6)
- (b) 2nd M1 for use of the correct model with two probs correct to 2.s.f. (accept \(P(E \geq 12) = 0.02\)). Must see attempt at lower and upper limit. Probabilities may be seen in (c).
- (b) A1 for correct critical region (both parts). Allow \(E \leq 1\) and \(E \geq 12\) or \(E \leq 1\), \(E \geq 12\) etc. Writing CR as probability statements is A0.
- NB: Completely correct CR implies M1M1A1
- SC: 1-tailed test
- B0 as hypotheses are incorrect
- M1 for sight or use of Po(6)
- M1 (dep on \(H_1\)) for sight of \(P(E \leq 1) = 0.0174\) or \(P(E \geq 11) = 0.0426\), in line with their \(H_1\)
- A1 for CR: \(E \leq 1\) or CR: \(E \geq 11\), in line with their hypotheses
- (c) B1ft for 0.0375 or 0.0374 or summing their two appropriate probs (ft their CR)
- NB: If candidate uses a 1-tailed test, this mark cannot be gained
(a)(i) | $0.20901\ldots$ awrt $\mathbf{0.209}$ | B1 | 3.4 |

(a)(ii) | $0.30844\ldots$ awrt $\mathbf{0.308}$ | B1 | 1.1b |

(b) | $H_0: \lambda = 2.4$ (or $\mu = 6$) $\quad H_1: \lambda \neq 2.4$ (or $\mu \neq 6$) | B1 | 2.5 |
| $[E = \text{no. of errors}] E \sim \text{Po}(6)$ | M1 | 3.3 |
| $P(E \leqslant 1) = 0.0174$ or $P(E \leqslant 2) = 0.0620$ and $P(E \leqslant 11) = 0.980$ or $P(E \geqslant 12) = 0.0201$ | M1 | 3.4 |
| Critical region: $E \leqslant 1$ or $E \geqslant 12$ | A1 | 1.1b |

(c) | $[P(\text{Type I error}) = 0.0174 + 0.0201 =] \mathbf{0.0375}$ (Calc gives: $0.017351\ldots + 0.0200919\ldots = 0.037443\ldots$) | B1ft | 1.2 |

**Notes:**
- (a) 1st B1 for awrt 0.209; 2nd B1 for awrt 0.308
- (b) B1 for both hypotheses correct in terms of $\lambda$ or $\mu$ (allow $\lambda = 6$ etc)
- (b) 1st M1 for selecting the correct model. Sight or use of Po(6)
- (b) 2nd M1 for use of the correct model with two probs correct to 2.s.f. (accept $P(E \geq 12) = 0.02$). Must see attempt at lower and upper limit. Probabilities may be seen in (c).
- (b) A1 for correct critical region (both parts). Allow $E \leq 1$ and $E \geq 12$ or $E \leq 1$, $E \geq 12$ etc. Writing CR as probability statements is A0.
- **NB: Completely correct CR implies M1M1A1**
- **SC: 1-tailed test**
  - **B0** as hypotheses are incorrect
  - **M1** for sight or use of Po(6)
  - **M1** (dep on $H_1$) for sight of $P(E \leq 1) = 0.0174$ or $P(E \geq 11) = 0.0426$, in line with their $H_1$
  - **A1** for CR: $E \leq 1$ or CR: $E \geq 11$, in line with their hypotheses
- (c) **B1ft** for 0.0375 or 0.0374 or summing their two appropriate probs (ft their CR)
- **NB: If candidate uses a 1-tailed test, this mark cannot be gained**

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\begin{enumerate}
  \item The number of errors made by a secretary is modelled by a Poisson distribution with a mean of 2.4 per 100 words.
\end{enumerate}

A 100-word piece of work completed by the secretary is selected at random.\\
(a) Find the probability that\\
(i) there are exactly 3 errors,\\
(ii) there are fewer than 2 errors.

After a long holiday, a randomly selected piece of work containing 250 words completed by the secretary is examined to see if the rate of errors has changed.\\
(b) Stating your hypotheses clearly, and using a $5 \%$ level of significance, find the critical region for a suitable test.\\
(c) Find P (Type I error) for the test in part (b)

\hfill \mbox{\textit{Edexcel FS1 2024 Q2 [7]}}