| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Calculate variance from summary statistics |
| Difficulty | Easy -1.2 Part (i) requires basic conceptual understanding of how mean and standard deviation affect the distribution of weights (increase mean or decrease standard deviation). Part (ii) is a routine calculation using standard formulas for sample mean and standard deviation with given summations—straightforward arithmetic with no problem-solving required. This is easier than average A-level content. |
| Spec | 2.02g Calculate mean and standard deviation5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks |
|---|---|
| (i) | The company could increase the mean weight. |
| Answer | Marks |
|---|---|
| deviation. | B1 CAO |
| B1 | 2 |
| (ii) | Sample mean = 11409/25 = 456.36 |
| Answer | Marks |
|---|---|
| 24 | B1 |
| Answer | Marks |
|---|---|
| A1 | 3 |
| TOTAL | 5 |
| Answer | Marks |
|---|---|
| (iii) | 1 1 |
| Answer | Marks |
|---|---|
| = 6.75 weeks | B1 |
| Answer | Marks |
|---|---|
| A1 | Calculation must be |
| Answer | Marks |
|---|---|
| (iii) | Mean = 83.95/8 = 10.49 |
| Answer | Marks |
|---|---|
| fast time, can also receive full credit. | B1 |
| Answer | Marks |
|---|---|
| E1 | Follow through if |
| Answer | Marks |
|---|---|
| (iii) | r 0 1 2 3 4 |
| Answer | Marks |
|---|---|
| P(X > 2.2) = 22k = 0.44 | B1 |
| Answer | Marks |
|---|---|
| B1 | 1 value correct |
Question 2:
--- 2
(i) ---
2
(i) | The company could increase the mean weight.
The company could decrease the standard
deviation. | B1 CAO
B1 | 2
(ii) | Sample mean = 11409/25 = 456.36
114092
S =5206937− =325.76
xx 25
325.76
Sample s.d = =3.68
24 | B1
M1 for S
xx
A1 | 3
TOTAL | 5
3
(i)
(ii)
(iii) | 1 1
P(X = 4) = (4)(5) = (Answer given)
40 2
1
E(X) = (2+12+36+80)
40
So E(X) = 3.25
1
Var (X) = (2+24+108+320) -- 3.2
40
= 11.35 – 10.5625
= 0.7875
6
Expected number of weeks = x45
40
= 6.75 weeks | B1
M1
A1 cao
M1
M1 dep
A1 cao
M1
A1 | Calculation must be
seen
Sum of rp
Sum of r²p
--3.25
Use of np
4
(i)
(ii)
(iii) | Mean = 83.95/8 = 10.49
83.952
881.2119-
8
Variance =
7
= 0.03737
Standard deviation = 0.193
2 standard deviations below mean
= 10.49 – 2(0.193)
= 10.104
but 10.04 < 10.104
so 10.04 is an outlier.
This time is much faster than the others. This may be
the result of wind assistance, faulty timing, false start
and should be discarded.
Opposite conclusion such as this could be a genuinely
fast time, can also receive full credit. | B1
M1
A1
M1
A1
E1
E1 | Follow through if
divisor n has been
used above.
Appreciating need
for investigation
Comment in
context
5
(i)
(ii)
(iii) | r 0 1 2 3 4
P(X = r) 6k 10k 12k 12k 10k
50k = 1 → k = 1/50
E(X) = 110k = 2.2
P(X > 2.2) = 22k = 0.44 | B1
B1
M1
M1
A1
B1 | 1 value correct
all 3 correct
sum of 1
sum of rp
caoA company sells sugar in bags which are labelled as containing 450 grams.
Although the mean weight of sugar in a bag is more than 450 grams, there is concern that too many bags are underweight. The company can adjust the mean or the standard deviation of the weight of sugar in a bag.
\begin{enumerate}[label=(\roman*)]
\item State two adjustments the company could make. [2]
\end{enumerate}
The weights, $x$ grams, of a random sample of 25 bags are now recorded.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Given that $\sum x = 11409$ and $\sum x^2 = 5206937$, calculate the sample mean and sample standard deviation of these weights. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 Q2 [5]}}