Edexcel S2 2009 January — Question 5 9 marks

Exam BoardEdexcel
ModuleS2 (Statistics 2)
Year2009
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicApproximating the Binomial to the Poisson distribution
TypeCalculate multiple probabilities using Poisson approximation
DifficultyModerate -0.3 This is a straightforward application of binomial distribution with clear parameters (n=10, p=0.01) for parts (a) and (b), followed by a standard Poisson approximation (λ=2.5) for part (c). The question explicitly signals the approximation is needed and all calculations are routine with no conceptual challenges beyond recognizing when to use each distribution.
Spec2.04c Calculate binomial probabilities5.02k Calculate Poisson probabilities

A factory produces components of which 1\% are defective. The components are packed in boxes of 10. A box is selected at random.
  1. Find the probability that the box contains exactly one defective component. [2]
  2. Find the probability that there are at least 2 defective components in the box. [3]
  3. Using a suitable approximation, find the probability that a batch of 250 components contains between 1 and 4 (inclusive) defective components. [4]

Part (a)
\(X\) represents the number of defective components.
AnswerMarks Guidance
\(P(X = 1) = (0.99)^9(0.01) \times 10 = 0.0914\)M1A1 (2 marks)
Part (b)
AnswerMarks Guidance
\(P(X \geq 2) = 1 - P(X \leq 1) = 1 - (p)^{10} - (a) = 0.0043\)M1, A1/, A1 (3 marks)
Part (c)
AnswerMarks Guidance
\(X \sim \text{Po}(2.5)\)B1B1
\(P(1 \leq X \leq 4) = P(X \leq 4) - P(X = 0) = 0.8912 - 0.0821 = 0.809\)M1, A1 (4 marks)
Note: Normal distribution used. B1for mean only
Special case for parts a and b:
If they use 0.1 do not treat as misread as it makes it easier.
- (a) M1 A0 if they have 0.3874
- (b) M1 A1ft A0 they will get 0.2639
- (c) Could get B1 B0 M1 A0
For any other values of \(p\) which are in the table:
Do not use misread. Check using the tables. They could get (a) M1 A0 (b) M1 A1ft A0 (c) B1 B0 M1 A0
## Part (a)
$X$ represents the number of defective components.
$P(X = 1) = (0.99)^9(0.01) \times 10 = 0.0914$ | M1A1 | (2 marks)

## Part (b)
$P(X \geq 2) = 1 - P(X \leq 1) = 1 - (p)^{10} - (a) = 0.0043$ | M1, A1/, A1 | (3 marks)

## Part (c)
$X \sim \text{Po}(2.5)$ | B1B1 |
$P(1 \leq X \leq 4) = P(X \leq 4) - P(X = 0) = 0.8912 - 0.0821 = 0.809$ | M1, A1 | (4 marks)

**Note:** Normal distribution used. B1for mean only

**Special case for parts a and b:**
If they use 0.1 do not treat as misread as it makes it easier.
- (a) M1 A0 if they have 0.3874
- (b) M1 A1ft A0 they will get 0.2639
- (c) Could get B1 B0 M1 A0

**For any other values of $p$ which are in the table:**
Do not use misread. Check using the tables. They could get (a) M1 A0 (b) M1 A1ft A0 (c) B1 B0 M1 A0

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A factory produces components of which 1\% are defective. The components are packed in boxes of 10. A box is selected at random.

\begin{enumerate}[label=(\alph*)]
\item Find the probability that the box contains exactly one defective component. [2]
\item Find the probability that there are at least 2 defective components in the box. [3]
\item Using a suitable approximation, find the probability that a batch of 250 components contains between 1 and 4 (inclusive) defective components. [4]
\end{enumerate}

\hfill \mbox{\textit{Edexcel S2 2009 Q5 [9]}}