| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Year | 2020 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Distribution |
| Type | Name geometric distribution and parameter |
| Difficulty | Standard +0.3 This is a standard Further Statistics question on geometric distribution and chi-squared goodness-of-fit testing. Part (a) requires recognizing X follows a geometric distribution and estimating p from the sample mean (straightforward). Part (b) involves calculating P(X≥7) using the complement rule. Part (c) is a routine chi-squared test with given calculations. All steps are textbook procedures with no novel insight required, making it slightly easier than average for Further Maths. |
| Spec | 5.02f Geometric distribution: conditions5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.06b Fit prescribed distribution: chi-squared test |
| \(x\) | 1 | 2 | 3 | 4 | 5 | 6 | \(\geqslant 7\) | Total |
| Frequency \(f\) | 20 | 15 | 9 | 13 | 10 | 10 | 23 | 100 |
| \(xf\) | 20 | 30 | 27 | 52 | 50 | 60 | 161 | 400 |
| \(x\) | 1 | 2 | 3 | 4 | 5 | 6 | \(\geqslant 7\) |
| Observed frequency \(O\) | 20 | 15 | 9 | 13 | 10 | 10 | 23 |
| Expected frequency \(E\) | 25 | 18.75 | 14.063 | 10.547 | 7.910 | 5.933 | 17.798 |
| \((O - E)^2/E\) | 1 | 0.75 | 1.823 | 0.571 | 0.552 | 2.789 | 1.520 |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (a) | Geometric |
| Answer | Marks |
|---|---|
| Therefore p = 0.25 | M1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Stated explicitly |
| Answer | Marks |
|---|---|
| Allow even if second M1 not gained | Needs to deduce p in part |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | Probability is 0.756 (= 0.1779785…) | M1 |
| OR: | Or: 0.177978 or 0.177979 or better seen, or 1 – | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Expected frequency = probability×100 = 17.798 | A1 | |
| [2] | 2.1 | 17.798 correctly obtained, with |
| sufficient evidence, www | 100 – Σ(other frequencies): |
| Answer | Marks |
|---|---|
| (c) | H : data consistent with (geometric) distribution |
| Answer | Marks |
|---|---|
| distribution is not a good fit. | B1 |
| Answer | Marks |
|---|---|
| [5] | 1.1 |
| Answer | Marks |
|---|---|
| 2.2b | Both, allow equivalents, but not |
| Answer | Marks |
|---|---|
| Don’t penalise “Geo(0.25)” | E.g. H : X ~ Geo(p) |
| Answer | Marks |
|---|---|
| Exx | α: Reject H . Data is consistent with geometric: M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 7:
7 | (a) | Geometric
Mean = 400 ÷ 100 (= 4) and p = 1/mean
Therefore p = 0.25 | M1
M1
A1
[3] | 1.1
2.4
1.1 | Stated explicitly
Use mean (or P(1) etc) to deduce p
(“Determine”, so justification is
needed for 0.25)
Allow even if second M1 not gained | Needs to deduce p in part
(a), not defer it to (b)
SC Geo(0.2) using
P(1) = 0.2: M1M1A0
(b) | Probability is 0.756 (= 0.1779785…) | M1 | 3.3 | SC Geo(0.2): 0.86 M1A0
OR: | Or: 0.177978 or 0.177979 or better seen, or 1 – | M1 | Allow ± 1 term
[P(1)+ … + P(6)] with evidence, e.g. formula
Expected frequency = probability×100 = 17.798 | A1
[2] | 2.1 | 17.798 correctly obtained, with
sufficient evidence, www | 100 – Σ(other frequencies):
SC B1
(c) | H : data consistent with (geometric) distribution
0
H : not consistent
1
ΣX 2 = 9.005
9.005 < 11.07 (ν = 5)
Do not reject H .
0
Insufficient evidence that a geometric
distribution is not a good fit. | B1
B1
B1
M1ft
A1ft
[5] | 1.1
1.1
1.1
1.1
2.2b | Both, allow equivalents, but not
“evidence that …”.
9.005 or 9.01
Compare their ΣX 2 with 11.07
Correct first conclusion, ft on their
9.005 or on 12.59, needs
like-with-like
Contextualised, not too definite
(needs double negative)
Don’t penalise “Geo(0.25)” | E.g. H : X ~ Geo(p)
0
Allow Geo(0.25)
Allow from comparison with
12.59 but nothing else
Allow addition slip in ΣX 2
SC Geo(0.2): can get full
marks if given data used,
ΣX 2 = 4.54 used gets
B1B1B0M1A1
Exx | α: Reject H . Data is consistent with geometric: M1A0
0
β: Insufficient evidence to reject H . Data is consistent with geometric: M1A1 (BOD)
0
Question | Answer | Marks | AO | GuidanceA biased spinner has five sides, numbered 1 to 5. Elmer spins the spinner repeatedly and counts the number of spins, $X$, up to and including the first time that the number 2 appears. He carries out this experiment 100 times and records the frequency $f$ with which each value of $X$ is obtained. His results are shown in Table 1, together with the values of $xf$.
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline
$x$ & 1 & 2 & 3 & 4 & 5 & 6 & $\geqslant 7$ & Total \\
\hline
Frequency $f$ & 20 & 15 & 9 & 13 & 10 & 10 & 23 & 100 \\
\hline
$xf$ & 20 & 30 & 27 & 52 & 50 & 60 & 161 & 400 \\
\hline
\end{tabular}
Table 1
\begin{enumerate}[label=(\alph*)]
\item State an appropriate distribution with which to model $X$, determining the value(s) of any parameter(s). [3]
\end{enumerate}
Elmer carries out a goodness-of-fit test, at the 5\% level, for the distribution in part (a). Table 2 gives some of his calculations, in which numbers that are not exact have been rounded to 3 decimal places.
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
$x$ & 1 & 2 & 3 & 4 & 5 & 6 & $\geqslant 7$ \\
\hline
Observed frequency $O$ & 20 & 15 & 9 & 13 & 10 & 10 & 23 \\
\hline
Expected frequency $E$ & 25 & 18.75 & 14.063 & 10.547 & 7.910 & 5.933 & 17.798 \\
\hline
$(O - E)^2/E$ & 1 & 0.75 & 1.823 & 0.571 & 0.552 & 2.789 & 1.520 \\
\hline
\end{tabular}
Table 2
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show how the expected frequency corresponding to $x \geqslant 7$ was obtained. [2]
\item Carry out the test. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics 2020 Q7 [10]}}