Question 4 - A-Level Maths

Edexcel FS1 AS Specimen — Question 4 11 marks

Exam Board	Edexcel
Module	FS1 AS (Further Statistics 1 AS)
Session	Specimen
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Chi-squared goodness of fit
Type	Chi-squared goodness of fit: Poisson
Difficulty	Standard +0.3 This is a standard chi-squared goodness of fit test for a Poisson distribution with the mean given (not estimated). Part (a) requires basic Poisson probability calculations, (b) is routine mean calculation, (c) involves finding expected frequencies, and (d) is a standard chi-squared test procedure. All steps are textbook applications with no novel insight required, making it slightly easier than average for Further Maths statistics.
Spec	5.02i Poisson distribution: random events model 5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities 5.02l Poisson conditions: for modelling 5.02m Poisson: mean = variance = lambda 5.06b Fit prescribed distribution: chi-squared test 5.06c Fit other distributions: discrete and continuous

The discrete random variable \(X\) follows a Poisson distribution with mean 1.4
1. Write down the value of
  1. \(\mathrm { P } ( \mathrm { X } = 1 )\)
  2. \(\mathrm { P } ( \mathrm { X } \leqslant 4 )\)
The manager of a bank recorded the number of mortgages approved each week over a 40 week period.
Number of mortgages approved 0 1 2 3 4 5 6
Frequency 10 16 7 4 2 0 1
Show that the mean number of mortgages approved over the 40 week period is 1.4 The bank manager believes that the Poisson distribution may be a good model for the number of mortgages approved each week. She uses a Poisson distribution with a mean of 1.4 to calculate expected frequencies as follows.
Number of mortgages approved 0 1 2 3 4 5 or more
Expected frequency 9.86 r 9.67 4.51 1.58 s
Find the value of r and the value of s giving your answers to 2 decimal places. The bank manager will test, at the \(5 \%\) level of significance, whether or not the data can be modelled by a Poisson distribution.
Calculate the test statistic and state the conclusion for this test. State clearly the degrees of freedom and the hypotheses used in the test. \section*{Q uestion 4 continued} \section*{Q uestion 4 continued}

Show mark scheme Show mark scheme source

Question 4:

Part (a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
(i) \(\text{P}(X=1) = 0.34523\ldots\) awrt \(\mathbf{0.345}\)	B1	awrt 0.345
(ii) \(\text{P}(X \leqslant 4) = 0.98575\ldots\) awrt \(\mathbf{0.986}\)	B1	awrt 0.986
(2 marks)

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(\frac{(0\times10)+1\times16+2\times7+3\times4+4\times2+(5\times0)+6\times1}{40} = 1.4^*\)	B1* cso	For a fully correct calculation leading to given answer with no errors seen
(1 mark)

Part (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(r = 40 \times 0.34523\ldots\) \(\quad\) \(s = 40 \times (1 - 0.986\ldots)\)	M1	For attempt at \(r\) or \(s\) (may be implied by correct answers)
\(r = \mathbf{13.81}\) \(\quad\) \(s = \mathbf{0.57}\)	A1ft	For both values correct (follow through on answers to part (a))
(2 marks)

Part (d):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(H_0\): The Poisson distribution is a suitable model; \(H_1\): The Poisson distribution is not a suitable model	B1	For both hypotheses correct (lambda should not be defined so correct use of the model)
Cells are combined when expected frequencies \(< 5\); combine the last 3 cells	M1	For understanding the need to combine cells before calculating the test statistic (may be implied)
\(\chi^2 = \sum\frac{(O-E)^2}{E} = \frac{(10-9.86)^2}{9.86} + \ldots + \frac{(7-(4.51+1.58+0.57))^2}{(4.51+1.58+0.57)}\)	M1	For attempt to find test statistic using \(\chi^2 = \sum\frac{(O-E)^2}{E}\)
awrt \(\mathbf{1.1}\)	A1	awrt 1.1
Degrees of freedom \(= 4 - 1 - 1 = 2\)	B1	For realising there are 2 degrees of freedom leading to critical value \(\chi^2_2(0.05) = 5.991\)
Do not reject \(H_0\) since \(1.10 < \chi^2_{2,(0.05)} = 5.991\). The number of mortgages approved each week follows a Poisson distribution	A1	Concluding that a Poisson model is suitable for the number of mortgages approved each week
(6 marks)

## Question 4:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)** $\text{P}(X=1) = 0.34523\ldots$ awrt $\mathbf{0.345}$ | B1 | awrt 0.345 |
| **(ii)** $\text{P}(X \leqslant 4) = 0.98575\ldots$ awrt $\mathbf{0.986}$ | B1 | awrt 0.986 |
| **(2 marks)** | | |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{(0\times10)+1\times16+2\times7+3\times4+4\times2+(5\times0)+6\times1}{40} = 1.4^*$ | B1* cso | For a fully correct calculation leading to given answer with no errors seen |
| **(1 mark)** | | |

### Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = 40 \times 0.34523\ldots$ $\quad$ $s = 40 \times (1 - 0.986\ldots)$ | M1 | For attempt at $r$ or $s$ (may be implied by correct answers) |
| $r = \mathbf{13.81}$ $\quad$ $s = \mathbf{0.57}$ | A1ft | For both values correct (follow through on answers to part (a)) |
| **(2 marks)** | | |

### Part (d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: The Poisson distribution is a suitable model; $H_1$: The Poisson distribution is not a suitable model | B1 | For both hypotheses correct (lambda should not be defined so correct use of the model) |
| Cells are combined when expected frequencies $< 5$; combine the last 3 cells | M1 | For understanding the need to combine cells before calculating the test statistic (may be implied) |
| $\chi^2 = \sum\frac{(O-E)^2}{E} = \frac{(10-9.86)^2}{9.86} + \ldots + \frac{(7-(4.51+1.58+0.57))^2}{(4.51+1.58+0.57)}$ | M1 | For attempt to find test statistic using $\chi^2 = \sum\frac{(O-E)^2}{E}$ |
| awrt $\mathbf{1.1}$ | A1 | awrt 1.1 |
| Degrees of freedom $= 4 - 1 - 1 = 2$ | B1 | For realising there are 2 degrees of freedom leading to critical value $\chi^2_2(0.05) = 5.991$ |
| Do not reject $H_0$ since $1.10 < \chi^2_{2,(0.05)} = 5.991$. The number of mortgages approved each week follows a Poisson distribution | A1 | Concluding that a Poisson model is suitable for the number of mortgages approved each week |
| **(6 marks)** | | |

Show LaTeX source

\begin{enumerate}
  \item The discrete random variable $X$ follows a Poisson distribution with mean 1.4\\
(a) Write down the value of\\
(i) $\mathrm { P } ( \mathrm { X } = 1 )$\\
(ii) $\mathrm { P } ( \mathrm { X } \leqslant 4 )$
\end{enumerate}

The manager of a bank recorded the number of mortgages approved each week over a 40 week period.

\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | }
\hline
Number of mortgages approved & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
Frequency & 10 & 16 & 7 & 4 & 2 & 0 & 1 \\
\hline
\end{tabular}
\end{center}

(b) Show that the mean number of mortgages approved over the 40 week period is 1.4

The bank manager believes that the Poisson distribution may be a good model for the number of mortgages approved each week.

She uses a Poisson distribution with a mean of 1.4 to calculate expected frequencies as follows.

\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
Number of mortgages approved & 0 & 1 & 2 & 3 & 4 & 5 or more \\
\hline
Expected frequency & 9.86 & r & 9.67 & 4.51 & 1.58 & s \\
\hline
\end{tabular}
\end{center}

(c) Find the value of r and the value of s giving your answers to 2 decimal places.

The bank manager will test, at the $5 \%$ level of significance, whether or not the data can be modelled by a Poisson distribution.\\
(d) Calculate the test statistic and state the conclusion for this test. State clearly the degrees of freedom and the hypotheses used in the test.

\section*{Q uestion 4 continued}
\section*{Q uestion 4 continued}

\hfill \mbox{\textit{Edexcel FS1 AS  Q4 [11]}}

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Q1 8 Q2 11 Q3 10 Q4 11

Number of mortgages approved	0	1	2	3	4	5 or more
Expected frequency	9.86	r	9.67	4.51	1.58	s