| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Calculate statistics from discrete frequency table |
| Difficulty | Easy -1.2 Part (i) is a routine calculation of mean and standard deviation from a discrete frequency table using standard formulas—pure recall and arithmetic. Part (ii) requires conceptual understanding of how adding zero values affects measures of location and spread, but this is a standard textbook discussion point requiring minimal problem-solving. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| Number of eggs | 1 | 2 | 3 | 4 |
| Frequency | 10 | 40 | 15 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Mean} = \frac{1\times10+2\times40+3\times15+4\times5}{70} = \frac{155}{70} = 2.214\) | M1, A1 CAO | For M1 allow sight of at least 3 double pairs from \(1\times10+2\times40+3\times15+4\times5\) with divisor 70; Allow 155/70 or 2.2 or 2.21 or 31/14 |
| \(S_{xx} = 1^2\times10+2^2\times40+3^2\times15+4^2\times5 - \frac{155^2}{70} = 385-343.21 = 41.79\) | M1, M1dep | M1 for \(\Sigma fx^2\) s.o.i.; M1 dep on first M1 for attempt at \(S_{xx}\) |
| \(s = \sqrt{\frac{41.79}{69}} = 0.778\) | A1 CAO | If 0.778 or better seen ignore previous incorrect working (calculator answer); Allow final answer to 2 sig fig; RMSD (divisor \(n\)) gives 0.772 gets M1M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Mean would decrease | B1 | Do not accept increase/decrease on their own – must be linked to mean and SD |
| Standard deviation would increase | B1 | Allow e.g. "It would skew the mean towards zero"; SC1 for justified argument that SD might either increase or decrease according to number with no eggs (\(n\leq496\) increase, \(n\geq497\) decrease) |
## Question 3:
### Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Mean} = \frac{1\times10+2\times40+3\times15+4\times5}{70} = \frac{155}{70} = 2.214$ | M1, A1 CAO | For M1 allow sight of at least 3 double pairs from $1\times10+2\times40+3\times15+4\times5$ with divisor 70; Allow 155/70 or 2.2 or 2.21 or 31/14 |
| $S_{xx} = 1^2\times10+2^2\times40+3^2\times15+4^2\times5 - \frac{155^2}{70} = 385-343.21 = 41.79$ | M1, M1dep | M1 for $\Sigma fx^2$ s.o.i.; M1 dep on first M1 for attempt at $S_{xx}$ |
| $s = \sqrt{\frac{41.79}{69}} = 0.778$ | A1 CAO | If 0.778 or better seen ignore previous incorrect working (calculator answer); Allow final answer to 2 sig fig; RMSD (divisor $n$) gives 0.772 gets M1M1A0 |
### Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Mean would decrease | B1 | Do not accept increase/decrease on their own – must be linked to mean and SD |
| Standard deviation would increase | B1 | Allow e.g. "It would skew the mean towards zero"; SC1 for justified argument that SD might either increase or decrease according to number with no eggs ($n\leq496$ increase, $n\geq497$ decrease) |
---3 The numbers of eggs laid by a sample of 70 female herring gulls are shown in the table.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
Number of eggs & 1 & 2 & 3 & 4 \\
\hline
Frequency & 10 & 40 & 15 & 5 \\
\hline
\end{tabular}
\end{center}
(i) Find the mean and standard deviation of the number of eggs laid per gull.\\
(ii) The sample did not include female herring gulls that laid no eggs. How would the mean and standard deviation change if these gulls were included?
\hfill \mbox{\textit{OCR MEI S1 Q3 [7]}}