| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Session | Specimen |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Estimate mean and standard deviation from frequency table |
| Difficulty | Standard +0.3 This is a multi-part statistics question covering grouped data calculations, histogram interpretation, normal distribution modeling, and a one-sample t-test. While it requires multiple techniques, each part is straightforward: (a) linear interpolation within a class, (b) standard grouped data formulas, (c) visual comparison, (d) normal probability calculation, and (e) routine hypothesis test. The question is slightly easier than average because it's methodical and procedural with clear guidance at each step, requiring no novel insight or complex problem-solving. |
| Spec | 2.02b Histogram: area represents frequency2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Lifetime | \(160 \leq x < 165\) | \(165 \leq x < 168\) | \(168 \leq x < 170\) | \(170 \leq x < 172\) | \(172 \leq x < 175\) | \(175 \leq x < 180\) |
| Frequency | 5 | 14 | 20 | 21 | 16 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| 15 | (a) | 16 1 |
| Answer | Marks |
|---|---|
| 80 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 3.1b | |
| 1.1 | for attempt at interpolation | |
| 15 | (b) | E.g. Midpoints |
| Answer | Marks |
|---|---|
| Standard deviation = 3.4 | M1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | evidnence of valid method for |
| Answer | Marks | Guidance |
|---|---|---|
| 15 | (c) | The histogram |
| Answer | Marks |
|---|---|
| so this does support the manager’s belief e | B1 |
| Answer | Marks |
|---|---|
| [2] | m |
| Answer | Marks |
|---|---|
| 3.5a | e |
| Answer | Marks | Guidance |
|---|---|---|
| 15 | (d) | (i) |
| (ii) | p |
| Answer | Marks |
|---|---|
| Answer is very similar to estimate in part (i) | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.4 |
| Answer | Marks |
|---|---|
| 3.5a | oe |
| Answer | Marks | Guidance |
|---|---|---|
| 15 | (e) | Either |
| Answer | Marks |
|---|---|
| −2.246 < -1.645 so significant | M1 |
| Answer | Marks |
|---|---|
| B1 | 3.4 |
| Answer | Marks |
|---|---|
| 1.1 | Must include √8 |
| Answer | Marks | Guidance |
|---|---|---|
| or | 3.4 | n |
| Answer | Marks |
|---|---|
| (cid:169) (cid:185) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| P X (cid:100)207.3 (cid:32)0.01235 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 0.01235< 0.05 so significant | 0.01235< 0.05 so significant | B1 |
| Answer | Marks |
|---|---|
| mean lifetime is less than 210 minutes. | cA1 |
| Answer | Marks |
|---|---|
| [5] | i |
Question 15:
15 | (a) | 16 1
Estimated number (cid:32)4(cid:14) (cid:32)9
3 3
91
3 (cid:32)0.1166... so proportion is approximately 0.117
80 | M1
A1
[2] | 3.1b
1.1 | for attempt at interpolation
15 | (b) | E.g. Midpoints
Mean = 170
Standard deviation = 3.4 | M1
A1
A1
[3] | 1.1
1.1
1.1 | evidnence of valid method for
estimation
BC Mean in the range 169-171
BC SD in the range 3-3.5
15 | (c) | The histogram
e.g. seems to have a rough bell shape
e.g. is symmetrical (around the estimated mean )
e.g. appears to have all data within 3 s.d. of the mean
so this does support the manager’s belief e | B1
c
B1
[2] | m
3.5a
i
3.5a | e
for one reason
for at least two reasons and ‘supports
belief’
15 | (d) | (i)
(ii) | p
P(Lifetime > 174) for N(170, 3.42)
S
0.1197
Answer is very similar to estimate in part (i) | M1
A1
B1
[3] | 3.4
1.1
3.5a | oe
BC FT their mean and standard
deviation
15 | (e) | Either
207.3(cid:16)210
Test statistic = (cid:32)(cid:16)2.246
3.4/ 8
Lower 5% level 1 tailed critical value of z = −1.645
−2.246 < -1.645 so significant | M1
A1
B1 | 3.4
1.1
1.1 | Must include √8
For comparison leading to correct
conclusion
or | 3.4 | n
(cid:167) 3.42 (cid:183)
Under H 0 , X ~ N(cid:168) (cid:168) 210, (cid:184) (cid:184)
8
(cid:169) (cid:185) | M1
(cid:11) (cid:12)
P X (cid:100)207.3 (cid:32)0.01235 | A1 | 1.1 | e
BC
0.01235< 0.05 so significant | 0.01235< 0.05 so significant | B1 | m
1.1
There is sufficient evidence to reject H
0
There is sufficient evidence to conclude that the
mean lifetime is less than 210 minutes. | cA1
E1
[5] | i
2.2b
2.4
3.4A quality control department checks the lifetimes of batteries produced by a company.
The lifetimes, $x$ minutes, for a random sample of 80 'Superstrength' batteries are shown in the table below.
\begin{tabular}{|l|c|c|c|c|c|c|}
\hline
Lifetime & $160 \leq x < 165$ & $165 \leq x < 168$ & $168 \leq x < 170$ & $170 \leq x < 172$ & $172 \leq x < 175$ & $175 \leq x < 180$ \\
\hline
Frequency & 5 & 14 & 20 & 21 & 16 & 4 \\
\hline
\end{tabular}
\begin{enumerate}[label=(\alph*)]
\item Estimate the proportion of these batteries which have a lifetime of at least 174.0 minutes. [2]
\item Use the data in the table to estimate
\begin{itemize}
\item the sample mean,
\item the sample standard deviation.
\end{itemize} [3]
\end{enumerate}
The data in the table on the previous page are represented in the following histogram, Fig 15.
\includegraphics{figure_15}
A quality control manager models the data by a Normal distribution with the mean and standard deviation you calculated in part (b).
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Comment briefly on whether the histogram supports this choice of model. [2]
\item
\begin{enumerate}[label=(\roman*)]
\item Use this model to estimate the probability that a randomly selected battery will have a lifetime of more than 174.0 minutes.
\item Compare your answer with your answer to part (a). [3]
\end{enumerate}
\end{enumerate}
The company also manufactures 'Ultrapower' batteries, which are stated to have a mean lifetime of 210 minutes.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{4}
\item A random sample of 8 Ultrapower batteries is selected. The mean lifetime of these batteries is 207.3 minutes.
Carry out a hypothesis test at the 5% level to investigate whether the mean lifetime is as high as stated. You should use the following hypotheses $\text{H}_0 : \mu = 210$, $\text{H}_1 : \mu < 210$, where $\mu$ represents the population mean for Ultrapower batteries.
You should assume that the population is Normally distributed with standard deviation 3.4. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 Q15 [15]}}