OCR MEI S1 — Question 6 8 marks

Exam BoardOCR MEI
ModuleS1 (Statistics 1)
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMeasures of Location and Spread
TypeCalculate statistics from discrete frequency table
DifficultyEasy -1.2 This is a straightforward S1 question requiring standard application of formulas for mean and standard deviation from a frequency table, followed by routine linear transformation of statistics. The calculations are mechanical with no conceptual challenges or problem-solving required beyond recalling the formulas Σfx/Σf and understanding that multiplying data by a constant multiplies both mean and SD by that constant.
Spec2.02f Measures of average and spread2.02g Calculate mean and standard deviation5.02c Linear coding: effects on mean and variance

6 A retail analyst records the numbers of loaves of bread of a particular type bought by a sample of shoppers in a supermarket.
Number of loaves012345
Frequency372311301
  1. Calculate the mean and standard deviation of the numbers of loaves bought per person.
  2. Each loaf costs \(\pounds 1.04\). Calculate the mean and standard deviation of the amount spent on loaves per person.

Question 6:
Part (i)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\text{Mean} = \frac{0\times37+1\times23+2\times11+3\times3+4\times0+5\times1}{75} = \frac{59}{75} = 0.787\)M1, A1
\(S_{xx} = 0^2\times37+1^2\times23+2^2\times11+3^2\times3+4^2\times0+5^2\times1 - \frac{59^2}{75} = 72.59\)M1, M1dep M1 for \(\Sigma fx^2\) s.o.i.; M1 dep for good attempt at \(S_{xx}\); NOTE M1M0 if their \(S_{xx}<0\)
\(s = \sqrt{\frac{72.59}{74}} = 0.99\)A1 CAO
Part (ii)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
New mean \(= 0.787 \times £1.04 = £0.818\) or 81.8 penceB1 ft their mean
New \(s = 0.99 \times £1.04 = £1.03\) or 103 penceB1 ft their \(s\)
B1For correct units *dep* on at least 1 correct (ft)
## Question 6:

### Part (i)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Mean} = \frac{0\times37+1\times23+2\times11+3\times3+4\times0+5\times1}{75} = \frac{59}{75} = 0.787$ | M1, A1 | |
| $S_{xx} = 0^2\times37+1^2\times23+2^2\times11+3^2\times3+4^2\times0+5^2\times1 - \frac{59^2}{75} = 72.59$ | M1, M1dep | M1 for $\Sigma fx^2$ s.o.i.; M1 dep for good attempt at $S_{xx}$; NOTE M1M0 if their $S_{xx}<0$ |
| $s = \sqrt{\frac{72.59}{74}} = 0.99$ | A1 CAO | |

### Part (ii)

| Answer/Working | Marks | Guidance |
|---|---|---|
| New mean $= 0.787 \times £1.04 = £0.818$ or 81.8 pence | B1 | ft their mean |
| New $s = 0.99 \times £1.04 = £1.03$ or 103 pence | B1 | ft their $s$ |
| | B1 | For correct units *dep* on at least 1 correct (ft) |
6 A retail analyst records the numbers of loaves of bread of a particular type bought by a sample of shoppers in a supermarket.

\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
Number of loaves & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
Frequency & 37 & 23 & 11 & 3 & 0 & 1 \\
\hline
\end{tabular}
\end{center}

(i) Calculate the mean and standard deviation of the numbers of loaves bought per person.\\
(ii) Each loaf costs $\pounds 1.04$. Calculate the mean and standard deviation of the amount spent on loaves per person.

\hfill \mbox{\textit{OCR MEI S1  Q6 [8]}}