CAIE S2 2015 November — Question 2 5 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2015
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicApproximating the Poisson to the Normal distribution
TypeFind parameter from given probability
DifficultyChallenging +1.2 This question requires understanding of normal approximation to Poisson (a Further Maths Statistics topic), applying continuity correction, using inverse normal tables, and solving a quadratic equation. While it involves multiple steps and the algebraic manipulation to get a quadratic in √λ is non-trivial, the individual techniques are standard for S2 level. The question guides students toward the solution method, making it moderately challenging but not requiring deep insight.
Spec2.04d Normal approximation to binomial5.02i Poisson distribution: random events model

2 The number of calls received per 5-minute period at a large call centre has a Poisson distribution with mean \(\lambda\), where \(\lambda > 30\). If more than 55 calls are received in a 5 -minute period, the call centre is overloaded. It has been found that the probability of being overloaded during a randomly chosen 5 -minute period is 0.01 . Use the normal approximation to the Poisson distribution to obtain a quadratic equation in \(\sqrt { } \lambda\) and hence find the value of \(\lambda\).

Question 2:
AnswerMarks Guidance
AnswerMark Guidance
\((\Phi^{-1}(0.99) =)\ 2.326\) seenB1 Must be \(\Phi^{-1}\), not \(\Phi\)
\(N(\lambda, \lambda)\) seen or impliedM1
\(\frac{55.5 - \lambda}{\sqrt{\lambda}} = +\text{"2.326"}\)M1 Allow with wrong or no cc & \(\Phi(0.99)\) \((= 0.8389)\); must be "\(z\)" or attempt at \(z\) \((0.99/0.01\) M0\()\)
\(\lambda + \text{"2.326"}\sqrt{\lambda} - 55.5 = 0\)
\(\sqrt{\lambda} = \frac{-\text{"2.326"} \pm \sqrt{\text{"2.326"}^2 + 4 \times 55.5}}{2}\) \((= 6.377\ldots \text{ or } -8.703\ldots)\)M1 For correct method of solving their quadratic in \(\sqrt{\lambda}\) and squaring to find \(\lambda\)
\(\lambda = 40.7\) (3 sf)A1 [5] cao, one answer only; without cc, \(\lambda = 40.2\): lose final A1
# Question 2:

| Answer | Mark | Guidance |
|--------|------|----------|
| $(\Phi^{-1}(0.99) =)\ 2.326$ seen | B1 | Must be $\Phi^{-1}$, not $\Phi$ |
| $N(\lambda, \lambda)$ seen or implied | M1 | |
| $\frac{55.5 - \lambda}{\sqrt{\lambda}} = +\text{"2.326"}$ | M1 | Allow with wrong or no cc & $\Phi(0.99)$ $(= 0.8389)$; must be "$z$" or attempt at $z$ $(0.99/0.01$ M0$)$ |
| $\lambda + \text{"2.326"}\sqrt{\lambda} - 55.5 = 0$ | | |
| $\sqrt{\lambda} = \frac{-\text{"2.326"} \pm \sqrt{\text{"2.326"}^2 + 4 \times 55.5}}{2}$ $(= 6.377\ldots \text{ or } -8.703\ldots)$ | M1 | For correct method of solving their quadratic in $\sqrt{\lambda}$ and squaring to find $\lambda$ |
| $\lambda = 40.7$ (3 sf) | A1 [5] | cao, one answer only; without cc, $\lambda = 40.2$: lose final A1 |

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2 The number of calls received per 5-minute period at a large call centre has a Poisson distribution with mean $\lambda$, where $\lambda > 30$. If more than 55 calls are received in a 5 -minute period, the call centre is overloaded. It has been found that the probability of being overloaded during a randomly chosen 5 -minute period is 0.01 . Use the normal approximation to the Poisson distribution to obtain a quadratic equation in $\sqrt { } \lambda$ and hence find the value of $\lambda$.

\hfill \mbox{\textit{CAIE S2 2015 Q2 [5]}}