| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Histogram from continuous grouped data |
| Difficulty | Moderate -0.8 This is a routine S1 statistics question covering standard techniques: drawing a histogram from grouped data, calculating mean and standard deviation from a frequency table, identifying outliers using the mean ± 2SD rule, and comparing distributions. All parts involve direct application of formulae with no problem-solving insight required. The calculations are straightforward and typical of basic A-level statistics coursework. |
| Spec | 2.02b Histogram: area represents frequency2.02g Calculate mean and standard deviation2.02h Recognize outliers |
| Weight | \(30 \leqslant w < 50\) | \(50 \leqslant w < 60\) | \(60 \leqslant w < 70\) | \(70 \leqslant w < 80\) | \(80 \leqslant w < 90\) |
| Frequency | 11 | 10 | 18 | 14 | 7 |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (i) | Weight Frequency Group Width Frequency density |
| Answer | Marks |
|---|---|
| 80 ≤ w < 90 7 10 0.7 | M1 |
| Answer | Marks |
|---|---|
| G1 | For fd’s - at least 3 correct |
| Answer | Marks |
|---|---|
| width of bars | M1 can be also be gained from |
| Answer | Marks | Guidance |
|---|---|---|
| Weight | Frequency | Group Width |
| 30 ≤ w < 50 | 11 | 20 |
| 50 ≤ w < 60 | 10 | 10 |
| 60 ≤ w < 70 | 18 | 10 |
| 70 ≤ w < 80 | 14 | 10 |
| 80 ≤ w < 90 | 7 | 10 |
| Answer | Marks | Guidance |
|---|---|---|
| [5] | height of bars | Height of bars – must be linear |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (ii) | Mean = |
| Answer | Marks |
|---|---|
| 59 | M1 |
| Answer | Marks |
|---|---|
| [4] | For midpoints |
| Answer | Marks |
|---|---|
| (exact answer14.2166…) | For midpoints (at least 3 correct) |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (iii) | x – 2s = 63.4 – (2 14.2) = 35 |
| Answer | Marks |
|---|---|
| upper end | M1 |
| Answer | Marks |
|---|---|
| [3] | For either |
| Answer | Marks |
|---|---|
| both ends | Only follow through numerical |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (iv) | 3624.5 |
| Answer | Marks |
|---|---|
| 49 | B1 |
| Answer | Marks |
|---|---|
| [3] | CAO Ignore units |
| Answer | Marks |
|---|---|
| working) | M1 for 265416 - 50 their |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (v) | Variety A have lower average than Variety B oe |
| Variety A have higher variation than Variety B oe | E1 |
| Answer | Marks |
|---|---|
| [2] | FT their means |
| Answer | Marks |
|---|---|
| FT their sd | Allow ‘on the whole’ or similar |
Question 6:
6 | (i) | Weight Frequency Group Width Frequency density
30 ≤ w < 50 11 20 0.55
50 ≤ w < 60 10 10 1
60 ≤ w < 70 18 10 1.8
70 ≤ w < 80 14 10 1.4
80 ≤ w < 90 7 10 0.7 | M1
A1
G1
G1 | For fd’s - at least 3 correct
Accept any suitable unit for
fd such as eg freq per 10g.
linear scales on both axes
and labels
Vertical scale starting from
zero (not broken - but can
get final mark for heights if
broken)
width of bars | M1 can be also be gained from
freq per 10 – 5.5, 10, 18, 14, 7
(at least 3 correct) or similar.
If fd not explicitly given, M1
A1 can be gained from all
heights correct (within half a
square) on histogram (and M1A0
if at least 3 correct)
Linear scale and label on vertical
axis IN RELATION to first M1
mark ie fd or frequency density
or if relevant freq/10, etc (NOT
eg fd/10).
However allow scale given as
fd×10, or similar.
Accept f/w or f/cw (freq/width or
freq/class width)
Ignore horizontal label
Can also be gained from an
accurate key
G0 if correct label but not fd’s.
Must be drawn at 30, 50 etc
NOT 29.5 or 30.5 etc NO GAPS
ALLOWED
Must have linear scale.
No inequality labels on their own
such as 30 ≤W<50, 50 ≤W<60
etc but allow if 30, 50, 60 etc
occur at the correct boundary
position. See additional notes.
Allow this mark even if not
using fd’s
Weight | Frequency | Group Width | Frequency density
30 ≤ w < 50 | 11 | 20 | 0.55
50 ≤ w < 60 | 10 | 10 | 1
60 ≤ w < 70 | 18 | 10 | 1.8
70 ≤ w < 80 | 14 | 10 | 1.4
80 ≤ w < 90 | 7 | 10 | 0.7
G1
[5] | height of bars | Height of bars – must be linear
vertical scale.
FT of heights dep on at least 3
heights correct and all must
agree with their fds
If fds not given and at least 3
heights correct then max
M1A0G1G1G0
Allow restart with correct
heights if given fd wrong (for
last three marks only)
6 | (ii) | Mean =
(40 11) (55 10) (65 18) (75 14) (85 7) 3805
60 60
= 63.4 (or 63.42)
(402 11) (552 10) (652 18) (752 14) (852 7)
x2f =
253225
38052
S 253225- 11924.6
xx 60
11924.6
s 202.11 14.2
59 | M1
A1
M1
A1
[4] | For midpoints
Products are 440, 550,
1170, 1050, 595
CAO
(exact answer 63.41666…)
For attempt at S
xx
Should include sum of at
least 3 correct multiples fx2
–Ʃx2/n
At least 1dp required
Use of mean 63.4 leading to
answer of 14.29199.. with
S = 12051.4 gets full
xx
credit.
63.42 leads to 14.2014…
Do not FT their incorrect
mean
(exact answer14.2166…) | For midpoints (at least 3 correct)
No marks for mean or sd unless
using midpoints
Answer must NOT be left as
improper fraction as this is an
estimate
Accept correct answers for mean
and sd from calculator even if eg
wrong Sxx given
Allow M1 for anything which
rounds to 11900
Allow SC1 for RMSD 14.1
(14.0976…) from calculator.
Only penalise once in part (ii) for
over specification, even if mean
and standard deviation both over
specified.
If using (x – x)2 method, B2 if
14.2 or better (14.3 if use of
63.4), otherwise B0
6 | (iii) | x – 2s = 63.4 – (2 14.2) = 35
x + 2s = 63.4 + (2 14.2) = 91.8
So there are probably some outliers at the lower end, but none at the
upper end | M1
A1
E1
[3] | For either
No marks in (iii) unless
using x + 2s or x– 2s
For both (FT)
Must include an element of
doubt and must mention
both ends | Only follow through numerical
values, not variables such as s, so
if a candidate does not find s but
then writes here ‘limit is 63.4 +
2 standard deviation’, do NOT
award M1
Do not penalise for over-
specification
Must have correct limits to get
this mark
6 | (iv) | 3624.5
Mean = = 72.5g (or exact answer 72.49g)
50
3624.52
S = 265416 = 2676
xx
50
2676
s = = 54.61 = 7.39g
49 | B1
M1
A1
[3] | CAO Ignore units
For S
xx
CAO ignore units
Allow 7.4 but NOT 7.3
(unless RMSD with
working) | M1 for 265416 - 50 their
mean2
BUT NOTE M0 if their S < 0
xx
For s2 of 54.6 (or better) allow
M1A0 with or without working.
For RMSD of 7.3 (or better)
allow M1A0 provided working
seen
For RMSD2 of 53.5 (or better)
allow M1A0 provided working
seen
6 | (v) | Variety A have lower average than Variety B oe
Variety A have higher variation than Variety B oe | E1
E1
[2] | FT their means
Do not condone lower
central tendency or lower
mean
FT their sd | Allow ‘on the whole’ or similar
in place of ‘average’.
Allow ‘more spread’ or similar
but not ‘higher range’ or ‘higher
variance’
Condone less consistent.The weights, $w$ grams, of a random sample of 60 carrots of variety A are summarised in the table below.
\begin{tabular}{|c|c|c|c|c|c|}
\hline
Weight & $30 \leqslant w < 50$ & $50 \leqslant w < 60$ & $60 \leqslant w < 70$ & $70 \leqslant w < 80$ & $80 \leqslant w < 90$ \\
\hline
Frequency & 11 & 10 & 18 & 14 & 7 \\
\hline
\end{tabular}
\begin{enumerate}[label=(\roman*)]
\item Draw a histogram to illustrate these data. [5]
\item Calculate estimates of the mean and standard deviation of $w$. [4]
\item Use your answers to part (ii) to investigate whether there are any outliers. [3]
\end{enumerate}
The weights, $x$ grams, of a random sample of 50 carrots of variety B are summarised as follows.
$n = 50$ \quad $\sum x = 3624.5$ \quad $\sum x^2 = 265416$
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{3}
\item Calculate the mean and standard deviation of $x$. [3]
\item Compare the central tendency and variation of the weights of varieties A and B. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 Q6 [17]}}