| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics A AS (Further Statistics A AS) |
| Year | 2020 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Mean-variance comparison for Poisson validation |
| Difficulty | Moderate -0.3 This is a straightforward application of standard Poisson distribution techniques. Part (a) requires routine calculation of mean and variance from a discrete probability distribution, (b) tests basic understanding that mean ≈ variance indicates Poisson suitability, (c) involves direct use of Poisson probability formulas, and (d) requires solving a simple inequality using Poisson properties. While it's a multi-part question requiring several techniques, each step is standard textbook material with no novel insight required, making it slightly easier than average. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling5.02m Poisson: mean = variance = lambda |
| \(r\) | 0 | 1 | 2 | 3 | 4 | \(> 4\) |
| \(\mathrm { P } ( X = r )\) | 0.30 | 0.38 | 0.19 | 0.08 | 0.05 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | DR |
| Answer | Marks |
|---|---|
| = 1.22 | M1 |
| Answer | Marks |
|---|---|
| [5] | 1.1a |
| Answer | Marks |
|---|---|
| 1.1 | SC if no working shown |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (b) | Because the variance is close to the mean. |
| [1] | 2.2b | For recognising that the values in part |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (c) | P(X = 2) = 0.2169 |
| Answer | Marks |
|---|---|
| = 0.0338 | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | BC |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (d) | P(0) ≤ 0.01 so e−λ ≤ 0.01 |
| Answer | Marks |
|---|---|
| So k = (4.6052/1.2) × 10 = 38.4 minutes | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.1b |
| Answer | Marks |
|---|---|
| 1.1 | BC |
| Answer | Marks |
|---|---|
| A1 for ‘so k = 39’ | OR: e −0.12k ≤0.01 |
Question 1:
1 | (a) | DR
E(X) = 0×0.30 + 1×0.38 + 2×0.19 + 3×0.08 + 4×0.05
=1.2
E(X2) = 0×0.30 + 1×0.38 + 4×0.19 + 9×0.08 + 16×0.05
= 2.66
Var(X) = 2.66 – 1.22
= 1.22 | M1
A1
M1
M1
A1
[5] | 1.1a
1.1
1.1
1.2
1.1 | SC if no working shown
SC1 for E(X) = 1.2
SC2 for Var(X) = 1.22
1 | (b) | Because the variance is close to the mean. | E1
[1] | 2.2b | For recognising that the values in part
(a) are close.
1 | (c) | P(X = 2) = 0.2169
P(X > 3) = 1 – 0.9662
= 0.0338 | B1
M1
A1
[3] | 1.1
3.4
1.1 | BC
BC
1 | (d) | P(0) ≤ 0.01 so e−λ ≤ 0.01
λ ≥ 4.6052
So k = (4.6052/1.2) × 10 = 38.4 minutes | M1
M1
A1
[3] | 3.1b
1.1
1.1 | BC
If trial and improvement method used
award
A1 for P(X = 0) = 0.0104 with k = 38
or for P(X = 0) = 0.00928 with k =39
A1 for ‘so k = 39’ | OR: e −0.12k ≤0.01
0.12k ≥ 4.6052
So k = 38.4 minutes1 The random variable $X$ represents the number of cars arriving at a car wash per 10-minute period. From observations over a number of days, an estimate was made of the probability distribution of $X$. Table 1 shows this estimated probability distribution.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$r$ & 0 & 1 & 2 & 3 & 4 & $> 4$ \\
\hline
$\mathrm { P } ( X = r )$ & 0.30 & 0.38 & 0.19 & 0.08 & 0.05 & 0 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 1}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item In this question you must show detailed reasoning.
Use Table 1 to calculate estimates of each of the following.
\begin{itemize}
\item $\mathrm { E } ( X )$
\item $\operatorname { Var } ( X )$
\item Explain how your answers to part (a) indicate that a Poisson distribution may be a suitable model for $X$.
\end{itemize}
You should now assume that $X$ can be modelled by a Poisson distribution with mean equal to the value which you calculated in part (a).
\item Find each of the following.
\begin{itemize}
\item $\mathrm { P } ( X = 2 )$
\item $\mathrm { P } ( X > 3 )$
\item Given that the probability that there is at least 1 car arriving in a period of $k$ minutes is at least 0.99 , find the least possible value of $k$.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2020 Q1 [12]}}