| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | October |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Estimate mean and standard deviation from frequency table |
| Difficulty | Moderate -0.8 This is a routine S1 statistics question testing standard procedures: histogram bar calculations using frequency density, linear interpolation for median, mean/SD formulas from given summations, skewness comparison, and basic probability. All techniques are direct applications of textbook methods with no problem-solving insight required, making it easier than average but not trivial due to the multiple parts. |
| 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 deviation2.03a Mutually exclusive and independent events |
| Time (t hours) | Frequency (f) | Time midpoint (m) |
| \(0 \leqslant t < 1\) | 10 | 0.5 |
| \(1 \leqslant t < 2\) | 18 | 1.5 |
| \(2 \leqslant t < 4\) | 15 | 3 |
| \(4 \leqslant t < 6\) | 12 | 5 |
| \(6 \leqslant t < 12\) | 5 | 9 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Width \(= 3\) cm. 1cm² represents 2 cars or \(0.5\)cm² represents 1 car or their \(h \times w = 6\) or area \(= 6\) | B1, M1 | B1 3 only. M1 may be implied by correct height |
| Height \(= \frac{6}{3} = 2\) cm | A1 | |
| (3) | ||
| (b) Median \(= (2) + \frac{30 - 28}{15} \times 2\) or \((2) + \frac{30.5 - 28}{15} \times 2\) (o.e.) | M1 | For any correct equation leading to correct fraction as part of \(m = ...\) or \((m - |
| \(= 2.266...\) (or \(2.33...\)) | A1 | |
| (2) | ||
| (c) \([\bar{t}] = \frac{182}{60} = 3.03...\) | B1 | Awrt 3.03 (allow exact fraction e.g. \(\frac{91}{30}\)) |
| \([\sigma_t] = \sqrt{\frac{883}{60} - \bar{t}^2} = \sqrt{5.5155...} = 2.3485...\) (s \(= 2.3683...\)) | M1, A1 | M1: A correct expression. A1: awrt 2.35 or 2.37 |
| (3) | ||
| (d) Mean > median. Positive skew | B1ft, dB1 (2) | 1st B1: ft their mean and median (Allow "larger frequencies at the start of table"). Do not allow comparison of quartiles unless correct values are seen (2sf comparisons). \(Q_1 = 1.28\) or \(\frac{23}{18}\) [\(n+1=1.29\)]. \(Q_3 = 4.33\) or \(\frac{13}{3}\) [\(n+1=4.42\)] e.g. 2.1 > 0.99 or 2.1 > 1.0. 2nd dB1 dependent on previous B1 being awarded. |
| (e) [75 mins \(= 1.25\) hours] | M1 | 1st M1 for a correct expression for no. of cars longer than 75 mins or shorter than 75 mins |
| \((> 75\) mins\() = 5 + 12 + 15 + \frac{3}{4} \times 18 = 45.5\) or \((< 75) = 10 + \frac{1}{4} \times 18\) or \(28 - \frac{3}{4} \times 18\) | M1 | |
| \(P(T > 1.25) = \frac{45.5}{60}\) or e.g. \(1 - \frac{14.5}{60}\) | M1 | |
| 0.7583... | A1 | awrt 0.758 allow \(\frac{91}{120}\) (o.e.) |
| (3) | ||
| Total 13 |
**(a)** Width $= 3$ cm. 1cm² represents 2 cars or $0.5$cm² represents 1 car or their $h \times w = 6$ or area $= 6$ | B1, M1 | B1 3 only. M1 may be implied by correct height
| Height $= \frac{6}{3} = 2$ cm | A1 |
| | (3) |
**(b)** Median $= (2) + \frac{30 - 28}{15} \times 2$ or $(2) + \frac{30.5 - 28}{15} \times 2$ (o.e.) | M1 | For any correct equation leading to correct fraction as part of $m = ...$ or $(m - |2|) = ...$. Ignore incorrect end point and watch out for "working down"
| $= 2.266...$ (or $2.33...$) | A1 |
| | (2) |
**(c)** $[\bar{t}] = \frac{182}{60} = 3.03...$ | B1 | Awrt 3.03 (allow exact fraction e.g. $\frac{91}{30}$)
| $[\sigma_t] = \sqrt{\frac{883}{60} - \bar{t}^2} = \sqrt{5.5155...} = 2.3485...$ (s $= 2.3683...$) | M1, A1 | M1: A correct expression. A1: awrt 2.35 or 2.37
| | (3) |
**(d)** Mean > median. Positive skew | B1ft, dB1 (2) | 1st B1: ft their mean and median (Allow "larger frequencies at the start of table"). Do not allow comparison of quartiles unless correct values are seen (2sf comparisons). $Q_1 = 1.28$ or $\frac{23}{18}$ [$n+1=1.29$]. $Q_3 = 4.33$ or $\frac{13}{3}$ [$n+1=4.42$] e.g. 2.1 > 0.99 or 2.1 > 1.0. 2nd dB1 dependent on previous B1 being awarded.
**(e)** [75 mins $= 1.25$ hours] | M1 | 1st M1 for a correct expression for no. of cars longer than 75 mins or shorter than 75 mins
| $(> 75$ mins$) = 5 + 12 + 15 + \frac{3}{4} \times 18 = 45.5$ or $(< 75) = 10 + \frac{1}{4} \times 18$ or $28 - \frac{3}{4} \times 18$ | M1 |
| $P(T > 1.25) = \frac{45.5}{60}$ or e.g. $1 - \frac{14.5}{60}$ | M1 |
| 0.7583... | A1 | awrt 0.758 allow $\frac{91}{120}$ (o.e.)
| | (3) |
| | **Total 13** |
---3. The parking times, $t$ hours, for cars in a car park are summarised below.
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
Time (t hours) & Frequency (f) & Time midpoint (m) \\
\hline
$0 \leqslant t < 1$ & 10 & 0.5 \\
\hline
$1 \leqslant t < 2$ & 18 & 1.5 \\
\hline
$2 \leqslant t < 4$ & 15 & 3 \\
\hline
$4 \leqslant t < 6$ & 12 & 5 \\
\hline
$6 \leqslant t < 12$ & 5 & 9 \\
\hline
\end{tabular}
\end{center}
$$\text { (You may use } \sum \mathrm { fm } = 182 \text { and } \sum \mathrm { fm } ^ { 2 } = 883 \text { ) }$$
A histogram is drawn to represent these data.\\
The bar representing the time $1 \leqslant t < 2$ has a width of 1.5 cm and a height of 6 cm .
\begin{enumerate}[label=(\alph*)]
\item Calculate the width and the height of the bar representing the time $4 \leqslant t < 6$
\item Use linear interpolation to estimate the median parking time for the cars in the car park.
\item Estimate the mean and the standard deviation of the parking time for the cars in the car park.
\item Describe, giving a reason, the skewness of the data.
One of these cars is selected at random.
\item Estimate the probability that this car is parked for more than 75 minutes.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2018 Q3 [13]}}