| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | June |
| Marks | 14 |
| 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 calculations using frequency density, linear interpolation for median, and mean/standard deviation from grouped frequency tables. All parts follow textbook methods with no problem-solving or novel insight required. The calculations are straightforward but multi-step, making it slightly easier than average. |
| Spec | 2.02b Histogram: area represents frequency2.02g Calculate mean and standard deviation2.02h Recognize outliers2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| Cost (£ \(\boldsymbol { x }\) ) | Frequency (f) | Midpoint (£y) |
| \(20 \leqslant x < 40\) | 12 | 30 |
| \(40 \leqslant x < 45\) | 13 | 42.5 |
| \(45 \leqslant x < 50\) | 25 | 47.5 |
| \(50 \leqslant x < 60\) | 32 | 55 |
| \(60 \leqslant x < 80\) | 8 | 70 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | Width \((w) = 4\) cm | B1 |
| Areas: \(16\) cm² represents 32 offices (o.e.) or their \(h = \frac{6}{\text{their } w}\) (3sf) or \(\frac{8}{3.2} × 0.6\) | M1 | |
| So height \((h) = 1.5\) cm | A1 | For \(h = 1.5\) cm (Correct answer only 2/2) |
| (b) | e.g. \((45) + \frac{20}{25} × 5\) or \((50) - \frac{5}{25} × 5\) (o.e.); \(= (±) 49\) | M1; A1 |
| (c) | \(\sum \frac{f y}{90} = \frac{4420}{90}\) | M1, A1 |
| \(= (±) 49.11\) (or better) (Allow \(\frac{442}{9}\) or \(49\frac{1}{9}\)) | A1 for 49.11 (Allow 49.1 from correct working) (Correct answer only 2/2, 49.1 only M1A0) | |
| (d) | \(\sqrt{\frac{226687.5}{90} - x^2} = \sqrt{106.8487\ldots}\) | M1, A1 |
| \(= 10.3367 = \text{awrt } (±) 10.3\) | ||
| (e) | Mean ≈ median so distribution is symmetric (no skew or very little skew) | B1ft |
| (f) | Symmetric (or little skew) so normal (or Rika's suggestion) may be suitable | B1ft |
| (g) | \(\frac{c - 50}{10} = 0.8416\) | M1, B1 |
| \(c = 58.416 = (±) 58.42\) awrt \(58.4\) | A1 | For awrt 58.4 (accept 3sf here) (Ans only of awrt 58.4 is M1B0A1 but 58.416 or better is 3/3) |
(a) | Width $(w) = 4$ cm | B1 | |
| Areas: $16$ cm² represents 32 offices (o.e.) or their $h = \frac{6}{\text{their } w}$ (3sf) or $\frac{8}{3.2} × 0.6$ | M1 | |
| So height $(h) = 1.5$ cm | A1 | For $h = 1.5$ cm (Correct answer only 2/2) |
(b) | e.g. $(45) + \frac{20}{25} × 5$ or $(50) - \frac{5}{25} × 5$ (o.e.); $= (±) 49$ | M1; A1 | M1 for a correct expression without end point. Allow "$n + 1$" so e.g. $(45) + \frac{20×5}{25}$. A1 for 49 or, if $(n + 1)$ used, allow 49.1 (Correct answer of 49 only 2/2) |
(c) | $\sum \frac{f y}{90} = \frac{4420}{90}$ | M1, A1 | M1 for an attempt at $\frac{\sum f y}{90}$ with at least 3 correct products of $\sum fy$ or $4000<\sum f y ≤5000$ |
| $= (±) 49.11$ (or better) (Allow $\frac{442}{9}$ or $49\frac{1}{9}$) | | A1 for 49.11 (Allow 49.1 from correct working) (Correct answer only 2/2, 49.1 only M1A0) |
(d) | $\sqrt{\frac{226687.5}{90} - x^2} = \sqrt{106.8487\ldots}$ | M1, A1 | M1 for a correct expression including $\sqrt{\ }$, ft their mean. Allow use of $s$. A1 for awrt 10.3 Allow $s =$ awrt 10.4 if clearly used. [NB use of 49.1 gives 10.389 ⟹ A0 (Correct answer of 10.3 with no working is 2/2) |
| $= 10.3367 = \text{awrt } (±) 10.3$ | | |
(e) | Mean ≈ median so distribution is symmetric (no skew or very little skew) | B1ft | For reason and "symmetric" (or other correct) statement [Allow mean $>$ median or $k(\bar{x} - Q_2)$ ($k<0$) so $+$ve skew if compatible with their figures] [If using quartiles we must see $Q_1 = 44.0$ and $Q_3 = 55.5$ used] |
(f) | Symmetric (or little skew) so normal (or Rika's suggestion) may be suitable | B1ft | Suggest normal is or isn't suitable with suitable reason based on (e) or mean and med |
(g) | $\frac{c - 50}{10} = 0.8416$ | M1, B1 | M1 for stand'ing using "$c$", 50 and 10 and setting equal to $± z$ value where $0.84 < z ≤ 0.85$. B1 for using $z = ± 0.8416$ or better (calc gives 0.8416212...) in standard' attempt e.g. $\sqrt{10}$ for 10 |
| $c = 58.416 = (±) 58.42$ awrt $58.4$ | A1 | For awrt 58.4 (accept 3sf here) (Ans only of awrt 58.4 is M1B0A1 but 58.416 or better is 3/3) |2. An estate agent is studying the cost of office space in London. He takes a random sample of 90 offices and calculates the cost, $\pounds x$ per square foot. His results are given in the table below.
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
Cost (£ $\boldsymbol { x }$ ) & Frequency (f) & Midpoint (£y) \\
\hline
$20 \leqslant x < 40$ & 12 & 30 \\
\hline
$40 \leqslant x < 45$ & 13 & 42.5 \\
\hline
$45 \leqslant x < 50$ & 25 & 47.5 \\
\hline
$50 \leqslant x < 60$ & 32 & 55 \\
\hline
$60 \leqslant x < 80$ & 8 & 70 \\
\hline
\end{tabular}
\end{center}
A histogram is drawn for these data and the bar representing $50 \leqslant x < 60$ is 2 cm wide and 8 cm high.
\begin{enumerate}[label=(\alph*)]
\item Calculate the width and height of the bar representing $20 \leqslant x < 40$
\item Use linear interpolation to estimate the median cost.
\item Estimate the mean cost of office space for these data.
\item Estimate the standard deviation for these data.
\item Describe, giving a reason, the skewness.
Rika suggests that the cost of office space in London can be modelled by a normal distribution with mean $\pounds 50$ and standard deviation $\pounds 10$
\item With reference to your answer to part (e), comment on Rika's suggestion.
\item Use Rika's model to estimate the 80th percentile of the cost of office space in London.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 2017 Q2 [14]}}