| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Data representation |
| Type | Draw histogram then estimate mean/standard deviation |
| Difficulty | Moderate -0.8 This is a routine S1 statistics question involving standard histogram construction with unequal class widths and straightforward coded mean/standard deviation calculations using given formulas. All steps are mechanical applications of well-practiced techniques with no problem-solving or conceptual challenges beyond basic recall. |
| Spec | 2.02b Histogram: area represents frequency2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| Number of visitors | Number of days |
| 400-459 | 3 |
| 460-479 | 8 |
| 480-499 | 13 |
| 500-519 | 12 |
| 520-539 | 18 |
| 540-559 | 11 |
| 560-599 | 9 |
| 600-699 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) freq. dens. = 0.05, 0.4, 0.65, 0.6, 0.9, 0.55, 0.225, 0.05 | M1 A1 | |
| Histogram drawn correctly | B2 | |
| (b) \(y\) values = 8, −4, −2, 0, 2, 4, 7, 14 | M1 | |
| \(\sum fy = (8 \times 3) + (-4 \times 8) + ... = 131\) | M1 A1 | |
| (c) \(\sum f = 79\); \(\bar{y} = \frac{131}{79} = 1.658\) | M1 | |
| \(x = (10 \times 1.658) + 509.5 = 526.1\) | M1 A1 | |
| std. dev. of \(y = \sqrt{\frac{2044}{79} - 1.658^2} = 4.805\) | M1 | |
| std. dev. of \(x = 10 \times 4.805 = 48.0\) | M1 A1 | (13 marks total) |
(a) freq. dens. = 0.05, 0.4, 0.65, 0.6, 0.9, 0.55, 0.225, 0.05 | M1 A1 |
Histogram drawn correctly | B2 |
(b) $y$ values = 8, −4, −2, 0, 2, 4, 7, 14 | M1 |
$\sum fy = (8 \times 3) + (-4 \times 8) + ... = 131$ | M1 A1 |
(c) $\sum f = 79$; $\bar{y} = \frac{131}{79} = 1.658$ | M1 |
$x = (10 \times 1.658) + 509.5 = 526.1$ | M1 A1 |
std. dev. of $y = \sqrt{\frac{2044}{79} - 1.658^2} = 4.805$ | M1 |
std. dev. of $x = 10 \times 4.805 = 48.0$ | M1 A1 | (13 marks total)6. The number of people visiting a new art gallery each day is recorded over a three-month period and the results are summarised in the table below.
\begin{center}
\begin{tabular}{|l|l|}
\hline
Number of visitors & Number of days \\
\hline
400-459 & 3 \\
\hline
460-479 & 8 \\
\hline
480-499 & 13 \\
\hline
500-519 & 12 \\
\hline
520-539 & 18 \\
\hline
540-559 & 11 \\
\hline
560-599 & 9 \\
\hline
600-699 & 5 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Draw a histogram on graph paper to illustrate these data.
In order to calculate summary statistics for the data it is coded using $y = \frac { x - 509.5 } { 10 }$, where $x$ is the mid-point of each class.
\item Find $\sum$ fy.
You may assume that $\sum f y ^ { 2 } = 2041$.
\item Using these values for $\sum f y$ and $\sum f y ^ { 2 }$, calculate estimates of the mean and standard deviation of the number of visitors per day.\\
(6 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel S1 Q6 [13]}}