| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Evaluate claims about large data set |
| Difficulty | Moderate -0.8 This question tests familiarity with the Edexcel large data set and basic calculation of mean/standard deviation from summary statistics. Part (a) requires recall of dataset variables, part (b) is straightforward application of formulas, and part (c) requires knowledge of Perth's climate. All parts are routine with no problem-solving or novel insight required. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Rain[fall] (allow [Mean] Windspeed) | B1 | Allow Rain[fall] or precipitation (e.g. allow Daily Total [or Mean or max] Rainfall etc); allow Mean Windspeed or just "windspeed" BUT not max windspeed or "gust"; if more than one answer, take the last one |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([\bar{x} =]\ 15.2239\ldots =\) awrt \(\mathbf{15.2}\) | B1 | Do not accept fractions or mixed numbers |
| \(\sigma_x = \sqrt{\dfrac{44695.4}{184} - \text{"15.22..."}^2}\) or \(\sqrt{11.1(422\ldots)}\) | M1 | For a correct expression including square root (ft their mean); may be implied by answer of 3.3 or better |
| \(= 3.33800\ldots\) awrt \(\mathbf{3.34}\) | A1 | Allow \(s = 3.3471\ldots\) i.e. awrt 3.35 if correct formula/expression seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Mean is higher than average OR a summer/spring month | M1 | For a reason mentioning that mean or temperature is higher; e.g. it is a warmer/hotter month; or sight of \(19.4 >\) (their) 15.2; only ft their 15.2 if it is less than 19.4; OR suggesting a summer/spring month |
| latest available month is Oct | A1cso | Must choose just October not a range like August–October; Perth is in southern hemisphere/Australia so October is spring |
# Question 3:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Rain[fall] (allow [Mean] Windspeed) | B1 | Allow Rain[fall] or precipitation (e.g. allow Daily Total [or Mean or max] Rainfall etc); allow Mean Windspeed or just "windspeed" BUT not max windspeed or "gust"; if more than one answer, take the last one |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $[\bar{x} =]\ 15.2239\ldots =$ awrt $\mathbf{15.2}$ | B1 | Do not accept fractions or mixed numbers |
| $\sigma_x = \sqrt{\dfrac{44695.4}{184} - \text{"15.22..."}^2}$ or $\sqrt{11.1(422\ldots)}$ | M1 | For a correct expression including square root (ft their mean); may be implied by answer of 3.3 or better |
| $= 3.33800\ldots$ awrt $\mathbf{3.34}$ | A1 | Allow $s = 3.3471\ldots$ i.e. awrt 3.35 if correct formula/expression seen |
**NB:** If answer in (b)(i) $> 19.4$ and an attempt is made in (c) please send to review.
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| Mean is higher than average OR a summer/spring month | M1 | For a reason mentioning that mean or temperature is higher; e.g. it is a warmer/hotter month; or sight of $19.4 >$ (their) 15.2; only ft their 15.2 if it is less than 19.4; OR suggesting a summer/spring month |
| latest available month is **Oct** | A1cso | Must choose just October not a range like August–October; Perth is in southern hemisphere/Australia so October is spring |
**SC:** M1A1 for "October since Perth is in the southern hemisphere/Australia"; M1A0 for Sep or Nov or Dec or Jan or Feb AND "Perth is in the southern hemisphere/Australia"; M0A0 just "Perth is in the southern hemisphere/Australia" without a month
---\begin{enumerate}
\item Ming is studying the large data set for Perth in 2015
\end{enumerate}
He intended to use all the data available to find summary statistics for the Daily Mean Air Temperature, $x { } ^ { \circ } \mathrm { C }$.\\
Unfortunately, Ming selected an incorrect variable on the spreadsheet.\\
This incorrect variable gave a mean of 5.3 and a standard deviation of 12.4\\
(a) Using your knowledge of the large data set, suggest which variable Ming selected.
The correct values for the Daily Mean Air Temperature are summarised as
$$n = 184 \quad \sum x = 2801.2 \quad \sum x ^ { 2 } = 44695.4$$
(b) Calculate the mean and standard deviation for these data.
One of the months from the large data set for Perth in 2015 has
\begin{itemize}
\item mean $\bar { X } = 19.4$
\item standard deviation $\sigma _ { x } = 2.83$\\
for Daily Mean Air Temperature.\\
(c) Suggest, giving a reason, a month these data may have come from.
\end{itemize}
\hfill \mbox{\textit{Edexcel Paper 3 2024 Q3 [6]}}