| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Calculate statistics from discrete frequency table |
| Difficulty | Moderate -0.8 This is a straightforward statistics question requiring basic data cleaning awareness and standard formula application for mean and standard deviation from given summations. The calculations are routine (mean = Σx/n, SD formula), and part (c) tests contextual knowledge of the large data set rather than mathematical reasoning. Easier than average A-level content. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02j Clean data: missing data, errors |
| \(x\) | 0 | \(0.1 - 0.5\) | \(0.6 - 1.0\) | \(1.1 - 1.9\) | \(2.0 - 4.0\) | \(4.1 - 6.9\) | \(7.0 - 12.0\) | \(12.1 - 20.9\) | \(21.0 - 32.0\) | \(\operatorname { tr }\) |
| Frequency | 55 | 18 | 18 | 21 | 17 | 9 | 9 | 6 | 2 | 29 |
| Answer | Marks | Guidance |
|---|---|---|
| Need to replace tr with a numerical value | M1 | For recognising that tr must be replaced (oe) with a numerical value. |
| Value of tr is between 0 and 0.05, suggest using e.g. 0.025, 0 or value \(\approx\) 0.05 | A1 (2) | For using a suitable value e.g. 0.025 (or allow 0), i.e. any value in \([0,\ 0.05]\). These give \(\sum x = 390\) (3sf). |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left[\bar{x} = \dfrac{389.3 \sim 390.8}{184}\right] = 2.119\ldots\) awrt \(\mathbf{2.12}\), allow \(\dfrac{195}{92}\) or \(2\tfrac{11}{92}\) | B1 | For awrt 2.12 or allow simplified fraction or mixed number. B0 for \(\frac{390}{184}\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left[\sigma =\right]\sqrt{\dfrac{(\text{awrt})4336}{184} - \text{"}\bar{x}^2\text{"}}\) or allow \(\left[\sigma^2 =\right]\dfrac{(\text{awrt})4336}{184} - \text{"}\bar{x}^2\text{"}\) or awrt 19.1 | M1 | For a correct expression for standard deviation or variance. Allow \(\sum x^2 =\) awrt 4336. Ignore their label \(\sigma\) or \(\sigma^2\). Can ft their mean. |
| \(= 4.367\ldots\) awrt \(\mathbf{4.37}\) | A1 (3) | For awrt 4.37. [Use of \(s\) gives 4.3791… so for correct use seen allow awrt 4.38.] SC Using \(n=155\): allow M1 for \(\left[\sigma =\right]\sqrt{\frac{(\text{awrt})4336}{155} - \text{"}\bar{x}^2\text{"}} = \sqrt{21.64\ldots}\) or 4.65… |
| Answer | Marks | Guidance |
|---|---|---|
| Only covers May–Oct (so not a suitable sample) | B1 | For comment mentioning data is just from May–Oct (so not representative of whole year). Just saying "only 184 days so not representative" is B0, must mention May–Oct. |
| Answer | Marks | Guidance |
|---|---|---|
| e.g. Winter months are missing when we'd expect more rain, so expect estimate in (b)(i) to be an underestimate (oe) | B1 (2) | For comment that missing/winter months expected to have more rain (oe) and "underestimate"(oe). Looking for all 3 ideas: (1) missing data is from winter/different months, (2) suggestion rainfall in those months is probably more, (3) statement about impact — it would be an underestimate or actual mean will be higher. |
# Question 3:
## Part (a)
Need to replace tr with a numerical value | M1 | For recognising that tr must be replaced (oe) with a numerical value.
Value of tr is between 0 and 0.05, suggest using e.g. 0.025, 0 or value $\approx$ 0.05 | A1 **(2)** | For using a suitable value e.g. 0.025 (or allow 0), i.e. any value in $[0,\ 0.05]$. These give $\sum x = 390$ (3sf).
## Part (b)(i)
$\left[\bar{x} = \dfrac{389.3 \sim 390.8}{184}\right] = 2.119\ldots$ awrt $\mathbf{2.12}$, allow $\dfrac{195}{92}$ or $2\tfrac{11}{92}$ | B1 | For awrt 2.12 or allow simplified fraction or mixed number. B0 for $\frac{390}{184}$.
## Part (b)(ii)
$\left[\sigma =\right]\sqrt{\dfrac{(\text{awrt})4336}{184} - \text{"}\bar{x}^2\text{"}}$ or allow $\left[\sigma^2 =\right]\dfrac{(\text{awrt})4336}{184} - \text{"}\bar{x}^2\text{"}$ or awrt 19.1 | M1 | For a correct expression for standard deviation or variance. Allow $\sum x^2 =$ awrt 4336. Ignore their label $\sigma$ or $\sigma^2$. Can ft their mean.
$= 4.367\ldots$ awrt $\mathbf{4.37}$ | A1 **(3)** | For awrt 4.37. [Use of $s$ gives 4.3791… so for correct use seen allow awrt 4.38.] **SC** Using $n=155$: allow M1 for $\left[\sigma =\right]\sqrt{\frac{(\text{awrt})4336}{155} - \text{"}\bar{x}^2\text{"}} = \sqrt{21.64\ldots}$ or 4.65…
## Part (c)(i)
Only covers May–Oct (so not a suitable sample) | B1 | For comment mentioning data is just from May–Oct (so not representative of whole year). Just saying "only 184 days so not representative" is B0, must mention May–Oct.
## Part (c)(ii)
e.g. Winter months are missing when we'd expect more rain, so expect estimate in (b)(i) to be an underestimate (oe) | B1 **(2)** | For comment that missing/winter months expected to have more rain (oe) **and** "underestimate"(oe). Looking for all 3 ideas: (1) missing data is from winter/different months, (2) suggestion rainfall in those months is probably more, (3) statement about impact — it would be an underestimate or actual mean will be higher.
---\begin{enumerate}
\item Ben is studying the Daily Total Rainfall, $x \mathrm {~mm}$, in Leeming for 1987
\end{enumerate}
He used all the data from the large data set and summarised the information in the following table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | c | c | }
\hline
$x$ & 0 & $0.1 - 0.5$ & $0.6 - 1.0$ & $1.1 - 1.9$ & $2.0 - 4.0$ & $4.1 - 6.9$ & $7.0 - 12.0$ & $12.1 - 20.9$ & $21.0 - 32.0$ & $\operatorname { tr }$ \\
\hline
Frequency & 55 & 18 & 18 & 21 & 17 & 9 & 9 & 6 & 2 & 29 \\
\hline
\end{tabular}
\end{center}
(a) Explain how the data will need to be cleaned before Ben can start to calculate statistics such as the mean and standard deviation.
Using all 184 of these values, Ben estimates $\sum x = 390$ and $\sum x ^ { 2 } = 4336$\\
(b) Calculate estimates for\\
(i) the mean Daily Total Rainfall,\\
(ii) the standard deviation of the Daily Total Rainfall.
Ben suggests using the statistic calculated in part (b)(i) to estimate the annual mean Daily Total Rainfall in Leeming for 1987\\
(c) Using your knowledge of the large data set,\\
(i) give a reason why these data would not be suitable,\\
(ii) state, giving a reason, how you would expect the estimate in part (b)(i) to differ from the actual annual mean Daily Total Rainfall in Leeming for 1987
\hfill \mbox{\textit{Edexcel Paper 3 2023 Q3 [7]}}