Question 10 - A-Level Maths

CAIE FP2 2017 June — Question 10 12 marks

Exam Board	CAIE
Module	FP2 (Further Pure Mathematics 2)
Year	2017
Session	June
Marks	12
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 with straightforward calculations: computing a mean (1 mark), using Poisson probability formula (4 marks), and performing a routine hypothesis test (7 marks). All steps follow textbook procedures with no novel insight required. While it's a Further Maths topic, the execution is mechanical and easier than average A-level questions.
Spec	5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities 5.06b Fit prescribed distribution: chi-squared test

Roberto owns a small hotel and offers accommodation to guests. Over a period of \(100\) nights, the numbers of rooms, \(x\), that are occupied each night at Roberto's hotel and the corresponding frequencies are shown in the following table.

Number of rooms occupied \((x)\)	0	1	2	3	4	5	6	\(\geqslant 7\)
Number of nights	4	9	18	26	20	16	7	0

Show that the mean number of rooms that are occupied each night is \(3.25\). [1]

The following table shows most of the corresponding expected frequencies, correct to \(2\) decimal places, using a Poisson distribution with mean \(3.25\).

Number of rooms occupied \((x)\)	0	1	2	3	4	5	6	\(\geqslant 7\)
Observed frequency	4	9	18	26	20	16	7	0
Expected frequency	3.88	12.60	20.48	22.18	18.02	11.72

Show how the expected value of \(22.18\), for \(x = 3\), is obtained and find the expected values for \(x = 6\) and for \(x \geqslant 7\). [4]
Use a goodness-of-fit test at the \(5\%\) significance level to determine whether the Poisson distribution is a suitable model for the number of rooms occupied each night at Roberto's hotel. [7]

Show mark scheme Show mark scheme source

Question 10:

Answer	Marks	Guidance
10(i)	x = (1/100) Σ x f(x) = 325/100 = 3⋅25 AG	B1
Total:	1
Question	Answer	Marks

Answer	Marks	Guidance
10(ii)	E = 100 λ3 e–λ /3! with λ = 3⋅25
3	M1 A1	State expression for reqd. expected value E (M1 for

3 E or E )

3 6

E = 100 λ6 e–λ /6! = 6⋅35

Answer	Marks	Guidance
6	A1	Find exp. value E

6 E = 100 – Σ 6 E = 4⋅77

Answer	Marks	Guidance
⩾7 0 i	A1	Find exp. value E

⩾7

Answer	Marks
Total:	4

Answer	Marks	Guidance
10(iii)	H : Distribution fits data (AEF)
0	B1	State (at least) null hypothesis in full

O: 13 18 26 20 16 7

i

E: 16⋅48 20⋅48 22⋅18 18⋅02 11⋅72 11⋅12

Answer	Marks	Guidance
i	M1FT	Combine values consistent with all exp. values ⩾ 5

(FT on E and E )

6 ⩾7

Answer	Marks	Guidance
χ2 = 0⋅735 + 0⋅300 + 0⋅658 + 0⋅218 + 1⋅563 + 1⋅527	M1	Find χ 2
= 5⋅00	A1

No. n of cells: 8 7 6 5 4

χ 2: 12⋅59 11⋅07 9⋅488 7⋅815 5⋅991

Answer	Marks	Guidance
n–2, 0.95	B1FT	State or use consistent tabular value χ 2 (to 3 s.f.)

n-2, 0.95

[FT on number, n, of cells used to find χ2]

Accept H if χ 2 < tabular value (AEF)

0

Answer	Marks	Guidance
5⋅00 [± 0⋅1] < 9⋅49 so distn. fits [data]	M1	State or imply valid method for conclusion

Conclusion (requires both values correct)

Answer	Marks	Guidance
or distn. is a suitable model (AEF)	A1
Total:	7
Question	Answer	Marks

Question 10:
--- 10(i) ---
10(i) | x = (1/100) Σ x f(x) = 325/100 = 3⋅25 AG | B1 | Verify given mean (B0 forx = 325/100 = 3⋅25)
Total: | 1
Question | Answer | Marks | Guidance
--- 10(ii) ---
10(ii) | E = 100 λ3 e–λ /3! with λ = 3⋅25
3 | M1 A1 | State expression for reqd. expected value E (M1 for
3
E or E )
3 6
E = 100 λ6 e–λ /6! = 6⋅35
6 | A1 | Find exp. value E
6
E = 100 – Σ 6 E = 4⋅77
⩾7 0 i | A1 | Find exp. value E
⩾7
Total: | 4
--- 10(iii) ---
10(iii) | H : Distribution fits data (AEF)
0 | B1 | State (at least) null hypothesis in full
O: 13 18 26 20 16 7
i
E: 16⋅48 20⋅48 22⋅18 18⋅02 11⋅72 11⋅12
i | M1FT | Combine values consistent with all exp. values ⩾ 5
(FT on E and E )
6 ⩾7
χ2 = 0⋅735 + 0⋅300 + 0⋅658 + 0⋅218 + 1⋅563 + 1⋅527 | M1 | Find χ 2
= 5⋅00 | A1
No. n of cells: 8 7 6 5 4
χ 2: 12⋅59 11⋅07 9⋅488 7⋅815 5⋅991
n–2, 0.95 | B1FT | State or use consistent tabular value χ 2 (to 3 s.f.)
n-2, 0.95
[FT on number, n, of cells used to find χ2]
Accept H if χ 2 < tabular value (AEF)
0
5⋅00 [± 0⋅1] < 9⋅49 so distn. fits [data] | M1 | State or imply valid method for conclusion
Conclusion (requires both values correct)
or distn. is a suitable model (AEF) | A1
Total: | 7
Question | Answer | Marks | Guidance

Show LaTeX source

Roberto owns a small hotel and offers accommodation to guests. Over a period of $100$ nights, the numbers of rooms, $x$, that are occupied each night at Roberto's hotel and the corresponding frequencies are shown in the following table.

\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
Number of rooms occupied $(x)$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & $\geqslant 7$ \\
\hline
Number of nights & 4 & 9 & 18 & 26 & 20 & 16 & 7 & 0 \\
\hline
\end{tabular}

\begin{enumerate}[label=(\roman*)]
\item Show that the mean number of rooms that are occupied each night is $3.25$. [1]
\end{enumerate}

The following table shows most of the corresponding expected frequencies, correct to $2$ decimal places, using a Poisson distribution with mean $3.25$.

\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
Number of rooms occupied $(x)$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & $\geqslant 7$ \\
\hline
Observed frequency & 4 & 9 & 18 & 26 & 20 & 16 & 7 & 0 \\
\hline
Expected frequency & 3.88 & 12.60 & 20.48 & 22.18 & 18.02 & 11.72 & & \\
\hline
\end{tabular}

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Show how the expected value of $22.18$, for $x = 3$, is obtained and find the expected values for $x = 6$ and for $x \geqslant 7$. [4]
\item Use a goodness-of-fit test at the $5\%$ significance level to determine whether the Poisson distribution is a suitable model for the number of rooms occupied each night at Roberto's hotel. [7]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP2 2017 Q10 [12]}}

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