Edexcel S2 — Question 5 12 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypePoisson hypothesis test
DifficultyStandard +0.3 This is a straightforward application of the Poisson distribution with standard procedures: calculating probabilities using tables/calculator for parts (a)-(b), and a routine one-tailed hypothesis test in part (c). While it requires careful setup (scaling the rate parameter correctly) and proper hypothesis test structure, it involves no novel problem-solving or conceptual challenges beyond standard S2 material.
Spec5.02i Poisson distribution: random events model5.02k Calculate Poisson probabilities5.05a Sample mean distribution: central limit theorem

A certain type of steel is produced in a foundry. It has flaws (small bubbles) randomly distributed, and these can be detected by X-ray analysis. On average, there are 0·1 bubbles per cm³, and the number of bubbles per cm³ has a Poisson distribution. In an ingot of 40 cm³, find
  1. the probability that there are less than two bubbles, [3 marks]
  2. the probability that there are more than 3 but less than 10 bubbles. [3 marks]
A new machine is being considered. Its manufacturer claims that it produces fewer bubbles per cm³. In a sample ingot of 60 cm³, there is just one bubble.
  1. Carry out a hypothesis test at the 1% significance level to decide whether the new machine is better. State your hypotheses and conclusion carefully. [6 marks]

AnswerMarks
(a) \(X \sim \text{Po}(4)\), so \(P(X < 2) = 0.0916\)B1 M1 A1
(b) \(P(3 < X < 10) = 0.9919 - 0.4335 = 0.558\)M1 M1 A1
(c) \(H_0\): mean number of bubbles is still \(0.1\) per cm²; \(H_1\): mean \(< 0.1\)B1 B1
Under \(H_0\), no. of bubbles in 60 cm³ is \(\text{Po}(6)\)B1
Then \(P(X \leq 1) = 0.0174\), so do not reject \(H_0\) at 1% levelM1 A1 A1
Total: 12 marks
(a) $X \sim \text{Po}(4)$, so $P(X < 2) = 0.0916$ | B1 M1 A1 |

(b) $P(3 < X < 10) = 0.9919 - 0.4335 = 0.558$ | M1 M1 A1 |

(c) $H_0$: mean number of bubbles is still $0.1$ per cm²; $H_1$: mean $< 0.1$ | B1 B1 |

Under $H_0$, no. of bubbles in 60 cm³ is $\text{Po}(6)$ | B1 |

Then $P(X \leq 1) = 0.0174$, so do not reject $H_0$ at 1% level | M1 A1 A1 |

**Total: 12 marks**

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A certain type of steel is produced in a foundry. It has flaws (small bubbles) randomly distributed, and these can be detected by X-ray analysis. On average, there are 0·1 bubbles per cm³, and the number of bubbles per cm³ has a Poisson distribution.

In an ingot of 40 cm³, find
\begin{enumerate}[label=(\alph*)]
\item the probability that there are less than two bubbles, [3 marks]
\item the probability that there are more than 3 but less than 10 bubbles. [3 marks]
\end{enumerate}

A new machine is being considered. Its manufacturer claims that it produces fewer bubbles per cm³. In a sample ingot of 60 cm³, there is just one bubble.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Carry out a hypothesis test at the 1% significance level to decide whether the new machine is better. State your hypotheses and conclusion carefully. [6 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2  Q5 [12]}}