CAIE S2 2012 June — Question 7 11 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2012
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPoisson distribution
TypeScaled time period sums
DifficultyStandard +0.3 This is a straightforward application of Poisson distribution with standard rate conversions and independence. Part (i) requires basic rate scaling (15/hour → 5/20min), part (ii) combines Poisson with binomial structure (standard S2 technique), and part (iii) adds two independent Poisson variables. All parts follow textbook methods with no novel insight required, making it slightly easier than average.
Spec2.04d 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 probabilities5.02n Sum of Poisson variables: is Poisson

7 At work Jerry receives emails randomly at a constant average rate of 15 emails per hour.
  1. Find the probability that Jerry receives more than 2 emails during a 20 -minute period at work.
  2. Jerry's working day is 8 hours long. Find the probability that Jerry receives fewer than 110 emails per day on each of 2 working days.
  3. At work Jerry also receives texts randomly and independently at a constant average rate of 1 text every 10 minutes. Find the probability that the total number of emails and texts that Jerry receives during a 5 -minute period at work is more than 2 and less than 6 .

(i)
B1 \(\lambda = 5\)
M1 \(1 - e^{-5}\left(1 + 5 + \frac{5^2}{2!}\right)\)
A1 \(= 0.875\)
[3] Any \(\lambda\). Allow one end error; May be implied
(ii)
B1 \(X \sim N(120, 120)\)
M1 \(\frac{109.5-120}{\sqrt{120}} = -0.9585\)
M1 \(1 - \Phi(\text{"0.9585"}) = 1 - 0.8312\)
A1 \(\frac{\text{"0.1688"}}{2} = 0.0285\) to \(0.0286\)
[4] Allow with wrong or no cc or no \(\sqrt{}\)
(iii)
M1 \(\lambda = \frac{15 \times 5 + 0.5}{60} = 1.75\)
A1 \(e^{-1.75}\)
M1 \(\frac{1.75^3}{3!} + \frac{1.75^4}{4!} + \frac{1.75^5}{5!}\)
A1 \(= 0.247\) (3 sfs)
[4] Any \(\lambda\). Allow one end error
**(i)**

**B1** $\lambda = 5$

**M1** $1 - e^{-5}\left(1 + 5 + \frac{5^2}{2!}\right)$

**A1** $= 0.875$

**[3]** Any $\lambda$. Allow one end error; May be implied

**(ii)**

**B1** $X \sim N(120, 120)$

**M1** $\frac{109.5-120}{\sqrt{120}} = -0.9585$

**M1** $1 - \Phi(\text{"0.9585"}) = 1 - 0.8312$

**A1** $\frac{\text{"0.1688"}}{2} = 0.0285$ to $0.0286$

**[4]** Allow with wrong or no cc or no $\sqrt{}$

**(iii)**

**M1** $\lambda = \frac{15 \times 5 + 0.5}{60} = 1.75$

**A1** $e^{-1.75}$

**M1** $\frac{1.75^3}{3!} + \frac{1.75^4}{4!} + \frac{1.75^5}{5!}$

**A1** $= 0.247$ (3 sfs)

**[4]** Any $\lambda$. Allow one end error
7 At work Jerry receives emails randomly at a constant average rate of 15 emails per hour.\\
(i) Find the probability that Jerry receives more than 2 emails during a 20 -minute period at work.\\
(ii) Jerry's working day is 8 hours long. Find the probability that Jerry receives fewer than 110 emails per day on each of 2 working days.\\
(iii) At work Jerry also receives texts randomly and independently at a constant average rate of 1 text every 10 minutes. Find the probability that the total number of emails and texts that Jerry receives during a 5 -minute period at work is more than 2 and less than 6 .

\hfill \mbox{\textit{CAIE S2 2012 Q7 [11]}}