| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Goodness-of-fit test for Poisson |
| Difficulty | Challenging +1.2 This is a standard chi-squared goodness of fit test with Poisson distribution requiring calculation of mean from grouped data, computing expected frequencies, pooling cells, and comparing to critical value. While it involves multiple computational steps and understanding of hypothesis testing, it follows a well-established procedure taught in Further Maths statistics with no novel problem-solving required. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| Number sold | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | \(\geq 8\) |
| Number of Saturdays | 7 | 20 | 39 | 16 | 14 | 2 | 1 | 1 | 0 |
| Answer | Marks |
|---|---|
| 8 | Find mean of sample data for use in Poisson distn.: |
| Answer | Marks | Guidance |
|---|---|---|
| χ2 < 11⋅1 so Poisson distn. fits B1 | 9 | [9] |
| Page 9 | Mark Scheme | Syllabus |
| Cambridge International A Level – October/November 2014 | 9231 | 22 |
| Answer | Marks |
|---|---|
| (iv) | Calculate gradient b in y− y =b(x−x) : Sxy = 513 – 73 × 64 = 45⋅8 |
| Answer | Marks |
|---|---|
| State set of possible values of N: N > 16 A1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | [11] | |
| Page 10 | Mark Scheme | Syllabus |
| Cambridge International A Level – October/November 2014 | 9231 | 22 |
Question 8:
8 | Find mean of sample data for use in Poisson distn.:
225
λ = = 2⋅25 B1
100
State (at least) null hypothesis (A.E.F.): H0: Poisson distn. fits data B1
100λre−λ
Find expected values (to 1 d.p.): 10⋅540 23⋅715 26⋅679 20⋅009 11⋅255
r!
(ignore incorrect final value here for M1) 5⋅065 1⋅899 0⋅6105 0⋅2275 M1 A1
Combine last four cells so that exp. value > 5: Oi : . . . 16 14 4
Ei : . . . 20⋅009 11⋅255 7⋅802 *M1
Calculate value of χ2 (to 1 d.p.; A1 dep *M1): χ2 = 1⋅189 + 0⋅582 + 5⋅690 + 0⋅803
+ 0⋅6695 + 1⋅853
= 10⋅8 (allow 10⋅7) M1 A1
State or use consistent tabular value (to 1 d.p.): χ 4, 0.975 2 = 11⋅14 (if cells combined) B1
[χ 7, 0.975 2 = 16⋅01, χ 5, 0.975 2 = 12⋅83]
χ2
Consistent conclusion (A.E.F., on two values):
χ2 < 11⋅1 so Poisson distn. fits B1 | 9 | [9]
Page 9 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – October/November 2014 | 9231 | 22
9 (i)
(ii)
(iii)
(iv) | Calculate gradient b in y− y =b(x−x) : Sxy = 513 – 73 × 64 = 45⋅8
10
732
Sxx = 651 – = 118⋅1
10
S
b = xy = 0⋅388 M1 A1
S
xx
64 73
Find regression line of y on x: y = + 0⋅388 (x – ) M1
10 10
= 0⋅388 x + 3⋅57
(458x+4215)
or A1
1181
642
Find correlation coefficient r: Syy = 462− = 52⋅4
10
S
r = ( xy ) = 0⋅582 M1 A1
S S
xx yy
Find y when x = 10: y = 7⋅45 B1
State valid comment on reliability, e.g.: Not reliable as value of r is small
or reliable since x = 10 is in range
or is near mean B1
Formulate condition for N: Require one-tail rN,1% < r [0⋅582] M1
Identify critical value near r using table: 15 or 16 ( on r) A1
State set of possible values of N: N > 16 A1 | 4
2
2
3 | [11]
Page 10 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – October/November 2014 | 9231 | 22The numbers of a particular type of laptop computer sold by a store on each of 100 consecutive Saturdays are summarised in the following table.
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline
Number sold & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & $\geq 8$ \\
\hline
Number of Saturdays & 7 & 20 & 39 & 16 & 14 & 2 & 1 & 1 & 0 \\
\hline
\end{tabular}
Fit a Poisson distribution to the data and carry out a goodness of fit test at the 2.5% significance level. [9]
\hfill \mbox{\textit{CAIE FP2 2014 Q8 [9]}}