| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Find median and quartiles from stem-and-leaf diagram |
| Difficulty | Easy -1.3 This is a straightforward S1 question testing basic understanding of quartiles from ordered data and the effect of linear transformations on summary statistics. Part (i) requires simple counting to find median and quartiles from 23 values, parts (ii)-(iii) are bookwork recall, and part (iv) applies the standard result that adding a constant to data adds that constant to the mean but leaves standard deviation unchanged. No problem-solving or novel insight required—purely procedural. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers |
| 5 | 5 | 6 | 7 | 8 | 8 | 9 | ||
| 6 | 1 | 2 | 3 | 5 | 6 | 8 | 9 | |
| 7 | 0 | 0 | 2 | 4 | 5 | 6 | 7 | 8 |
| 8 | 0 | |||||||
| 9 | 7 | |||||||
| 9 |
| 0 | 5 | 6 | 7 | 8 | 8 | 9 | |||
| 1 | 1 | 2 | 3 | 5 | 6 | 8 | 9 | ||
| 2 | 0 | 0 | 2 | 4 | 5 | 6 | 7 | 8 | |
| 3 | 0 | ||||||||
| 4 | 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(75 - 59 = 16\) | M1 | Attempt \(6^{th}\) & \(18^{th}\) or 58-60, 74-76 & subtract |
| A1 3 | Must be from \(75 - 59\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Unaffected by outliers or extremes (allow: less affected by outliers); sd can be skewed by one value | B1 1 | NOT: by anomalies or freaks; easier to calculate |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Shows each data item, retains original data; can see how many data items; can find (or easier to read) mode or modal class; can find (or easier to read) frequencies; can find mean | B1 | NOT: shows frequencies; shows results more clearly; B&W does not show frequencies |
| Harder to read median (or Qs or IQR); Doesn't show median; B&W shows median (or Qs or IQR); B&W easier to compare medians | B1 2 | NOT: B&W easier to compare; B&W shows spread or variance or skew; B&W shows highest & lowest. Assume in order: Adv, Disadv, unless told. Allow disadv of B&W for adv of S&L & vice versa. Ignore extras |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(m = 68.1\) NOT by restart | B1 | |
| \(sd = 9.7\) (or same) NOT by restart | B1 2 | Restart mean or mean & sd: 68.1 or 68.087 & 9.7 or 9.73 — B1 only |
## Question 5:
**Part (i):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $75 - 59 = 16$ | M1 | Attempt $6^{th}$ & $18^{th}$ or 58-60, 74-76 & subtract |
| | A1 3 | Must be from $75 - 59$ |
**Part (ii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Unaffected by outliers or extremes (allow: less affected by outliers); sd can be skewed by one value | B1 1 | NOT: by anomalies or freaks; easier to calculate |
**Part (iii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Shows each data item, retains original data; can see how many data items; can find (or easier to read) mode or modal class; can find (or easier to read) frequencies; can find mean | B1 | NOT: shows frequencies; shows results more clearly; B&W does not show frequencies |
| Harder to read median (or Qs or IQR); Doesn't show median; B&W shows median (or Qs or IQR); B&W easier to compare medians | B1 2 | NOT: B&W easier to compare; B&W shows spread or variance or skew; B&W shows highest & lowest. Assume in order: Adv, Disadv, unless told. Allow disadv of B&W for adv of S&L & vice versa. Ignore extras |
**Part (iv):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $m = 68.1$ NOT by restart | B1 | |
| $sd = 9.7$ (or same) NOT by restart | B1 2 | Restart mean or mean & sd: 68.1 or 68.087 & 9.7 or 9.73 — B1 only |
---5 The stem-and-leaf diagram shows the masses, in grams, of 23 plums, measured correct to the nearest gram.
\begin{center}
\begin{tabular}{ l | l l l l l l l l }
5 & 5 & 6 & 7 & 8 & 8 & 9 & & \\
6 & 1 & 2 & 3 & 5 & 6 & 8 & 9 & \\
7 & 0 & 0 & 2 & 4 & 5 & 6 & 7 & 8 \\
8 & 0 & & & & & & & \\
9 & 7 & & & & & & & \\
9 & & & & & & & & \\
\end{tabular}
\end{center}$\quad$ Key $: 6 \mid 2$ means 62
(i) Find the median and interquartile range of these masses.\\
(ii) State one advantage of using the interquartile range rather than the standard deviation as a measure of the variation in these masses.\\
(iii) State one advantage and one disadvantage of using a stem-and-leaf diagram rather than a box-and-whisker plot to represent data.\\
(iv) James wished to calculate the mean and standard deviation of the given data. He first subtracted 5 from each of the digits to the left of the line in the stem-and-leaf diagram, giving the following.
\begin{center}
\begin{tabular}{ l | l l l l l l l l l }
0 & 5 & 6 & 7 & 8 & 8 & 9 & & & \\
1 & 1 & 2 & 3 & 5 & 6 & 8 & 9 & & \\
2 & 0 & 0 & 2 & 4 & 5 & 6 & 7 & 8 & \\
3 & 0 & & & & & & & & \\
4 & 7 & & & & & & & & \\
\end{tabular}
\end{center}
The mean and standard deviation of the data in this diagram are 18.1 and 9.7 respectively, correct to 1 decimal place. Write down the mean and standard deviation of the data in the original diagram.
\hfill \mbox{\textit{OCR S1 2009 Q5 [8]}}