| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Draw histogram then find median/quartiles from cumulative frequency |
| Difficulty | Moderate -0.3 This is a standard S1 statistics question covering routine histogram construction (frequency density calculations), linear interpolation for median/quartiles, and coding transformations. All techniques are textbook exercises requiring straightforward application of formulas with no novel problem-solving. The multi-part structure and calculation volume make it slightly below average difficulty rather than trivial. |
| Spec | 2.02a Interpret single variable data: tables and diagrams2.02b Histogram: area represents frequency2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02j Clean data: missing data, errors2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Length \(( \boldsymbol { l } )\) | Frequency \(( \boldsymbol { f } )\) |
| \(15 < l \leqslant 20\) | 19 |
| \(20 < l \leqslant 25\) | 35 |
| \(25 < l \leqslant 27\) | 16 |
| \(27 < l \leqslant 30\) | 15 |
| \(30 < l \leqslant 40\) | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Width \(= 2.5\) cm | B1 | 2.5 oe |
| Height \(= \left(\dfrac{35}{15} \times 6\right) \div \text{"2.5"}\) or \(\dfrac{4}{5} \times 7\) or \(6\text{cm}^2 = 15\) (logs) or \(14\text{cm}^2 = 35\) (logs) | M1 | Correct method to relate area to number of logs (implied by "their \(w\)" \(\times\) "their \(h\)" \(= 14\)) |
| \(= 5.6\) cm | A1 | 5.6 oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Median \(= [20] + \dfrac{25}{35} \times 5\), allow \([20] + \dfrac{25.5}{35} \times 5\) | M1 | Correct fraction multiplied by 5. Allow working downwards e.g. \([25] - \dfrac{10}{35} \times 5\) |
| \(= 23.57\ldots\) allow \(23.64\ldots\) awrt 23.6 | A1 | Correct answer from correct working. Allow exact fraction |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(19 + 35 + 8 (= 62^*)\) or \(88 - 3 - 15 - 8 (= 62^*)\) | B1cso* | Correct calculation seen. Allow \(\frac{1}{2}(16)\) for 8. Allow \(\frac{x-54}{70-54} = \frac{26-25}{27-25}\). Minimum working: \(54 + 8\) or \(70 - 8\) or \(\frac{54+70}{2}\). NB: \(26 + x = 88 \Rightarrow x = 62\) is B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{4 logs fit}) = \left(\dfrac{62}{88}\right) \times \left(\dfrac{61}{87}\right) \times \left(\dfrac{60}{86}\right) \times \left(\dfrac{59}{85}\right)\) | M1 | For \(\left(\dfrac{n}{88}\right)\times\left(\dfrac{n-1}{87}\right)\times\left(\dfrac{n-2}{86}\right)\times\left(\dfrac{n-3}{85}\right)\) (allow any \(n < 88\)) |
| \(= 0.23922\ldots\) awrt 0.239 | A1 | With replacement awrt 0.246 scores M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| mean of \(y = \dfrac{924}{88} [= 10.5]\) | M1 | Correct method for mean of \(y\) (implied by 10.5), or equation \(0.5\sum w - 88 \times 255 = 924\) |
| mean of \(w = (\text{"10.5"} + 255) \times 2\) | M1 | Correct method for mean of \(w\), or \(\sum w = 46728\) and \(\dfrac{46728}{88}\) |
| \(= 531\) | A1 | Cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| variance of \(y = \dfrac{12862}{88} - (\text{"10.5"})^2 [= 35.909\ldots]\) | M1 | Correct method for variance of \(y\) (implied by awrt 35.9), or \(0.25\sum w^2 + 255^2 \times 88 - 255\sum w = 12862\) |
| variance of \(w = \text{"35.909"} \times 4\) or \(\text{"35.909"} \div 0.5^2\) | M1 | Correct method for variance of \(w\), or \(\sum w^2 = 24825208\) and \(\dfrac{24825208}{88} - \text{"531"}^2\) |
| \(= 143.636\ldots\) awrt 144 | A1 |
# Question 3:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Width $= 2.5$ cm | B1 | 2.5 oe |
| Height $= \left(\dfrac{35}{15} \times 6\right) \div \text{"2.5"}$ or $\dfrac{4}{5} \times 7$ or $6\text{cm}^2 = 15$ (logs) or $14\text{cm}^2 = 35$ (logs) | M1 | Correct method to relate area to number of logs (implied by "their $w$" $\times$ "their $h$" $= 14$) |
| $= 5.6$ cm | A1 | 5.6 oe |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| Median $= [20] + \dfrac{25}{35} \times 5$, allow $[20] + \dfrac{25.5}{35} \times 5$ | M1 | Correct fraction multiplied by 5. Allow working downwards e.g. $[25] - \dfrac{10}{35} \times 5$ |
| $= 23.57\ldots$ allow $23.64\ldots$ awrt 23.6 | A1 | Correct answer from correct working. Allow exact fraction |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $19 + 35 + 8 (= 62^*)$ **or** $88 - 3 - 15 - 8 (= 62^*)$ | B1cso* | Correct calculation seen. Allow $\frac{1}{2}(16)$ for 8. Allow $\frac{x-54}{70-54} = \frac{26-25}{27-25}$. Minimum working: $54 + 8$ or $70 - 8$ or $\frac{54+70}{2}$. NB: $26 + x = 88 \Rightarrow x = 62$ is B0 |
## Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{4 logs fit}) = \left(\dfrac{62}{88}\right) \times \left(\dfrac{61}{87}\right) \times \left(\dfrac{60}{86}\right) \times \left(\dfrac{59}{85}\right)$ | M1 | For $\left(\dfrac{n}{88}\right)\times\left(\dfrac{n-1}{87}\right)\times\left(\dfrac{n-2}{86}\right)\times\left(\dfrac{n-3}{85}\right)$ (allow any $n < 88$) |
| $= 0.23922\ldots$ awrt 0.239 | A1 | With replacement awrt 0.246 scores M1A0 |
## Part (e)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| mean of $y = \dfrac{924}{88} [= 10.5]$ | M1 | Correct method for mean of $y$ (implied by 10.5), or equation $0.5\sum w - 88 \times 255 = 924$ |
| mean of $w = (\text{"10.5"} + 255) \times 2$ | M1 | Correct method for mean of $w$, or $\sum w = 46728$ and $\dfrac{46728}{88}$ |
| $= 531$ | A1 | Cao |
## Part (e)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| variance of $y = \dfrac{12862}{88} - (\text{"10.5"})^2 [= 35.909\ldots]$ | M1 | Correct method for variance of $y$ (implied by awrt 35.9), or $0.25\sum w^2 + 255^2 \times 88 - 255\sum w = 12862$ |
| variance of $w = \text{"35.909"} \times 4$ or $\text{"35.909"} \div 0.5^2$ | M1 | Correct method for variance of $w$, or $\sum w^2 = 24825208$ and $\dfrac{24825208}{88} - \text{"531"}^2$ |
| $= 143.636\ldots$ awrt 144 | A1 | |
---\begin{enumerate}
\item Gill buys a bag of logs to use in her stove. The lengths, $l \mathrm {~cm}$, of the 88 logs in the bag are summarised in the table below.
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | }
\hline
Length $( \boldsymbol { l } )$ & Frequency $( \boldsymbol { f } )$ \\
\hline
$15 < l \leqslant 20$ & 19 \\
\hline
$20 < l \leqslant 25$ & 35 \\
\hline
$25 < l \leqslant 27$ & 16 \\
\hline
$27 < l \leqslant 30$ & 15 \\
\hline
$30 < l \leqslant 40$ & 3 \\
\hline
\end{tabular}
\end{center}
A histogram is drawn to represent these data.\\
The bar representing logs with length $27 < l \leqslant 30$ has a width of 1.5 cm and a height of 4 cm .\\
(a) Calculate the width and height of the bar representing log lengths of $20 < l \leqslant 25$\\
(b) Use linear interpolation to estimate the median of $l$
The maximum length of log Gill can use in her stove is 26 cm .\\
Gill estimates, using linear interpolation, that $x$ logs from the bag will fit into her stove.\\
(c) Show that $x = 62$
Gill randomly selects 4 logs from the bag.\\
(d) Using $x = 62$, find the probability that all 4 logs will fit into her stove.
The weights, $W$ grams, of the logs in the bag are coded using $y = 0.5 w - 255$ and summarised by
$$n = 88 \quad \sum y = 924 \quad \sum y ^ { 2 } = 12862$$
(e) Calculate\\
(i) the mean of $W$\\
(ii) the variance of $W$
\hfill \mbox{\textit{Edexcel S1 2022 Q3 [14]}}