| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Session | Specimen |
| Marks | 9 |
| 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 straightforward AS-level statistics question testing standard procedures: linear interpolation for median, using given summations for standard deviation, and applying coding transformations. All techniques are routine recall with no problem-solving insight required. The final part requires only basic proportional reasoning. Easier than average A-level content. |
| Spec | 2.02f Measures of average and spread2.02g Calculate mean and standard deviation |
| Coded time ( \(x\) minutes) | Frequency (f) | Coded time midpoint (y minutes) |
| \(0 \leq x < 5\) | 3 | 2.5 |
| \(5 \leq x < 10\) | 15 | 7.5 |
| \(10 \leq x < 15\) | 2 | 12.5 |
| \(15 \leq x < 25\) | 9 | 20 |
| \(25 \leq x < 35\) | 1 | 30 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(s = vt - \frac{1}{2}at^2\) | M1 | Complete method to give equation in \(v\) only (could involve 2 or more suvat equations then elimination) with usual rules |
| \(19.6 = 4v - \frac{1}{2} \times 9.8 \times 4^2\) | A1 | Correct equation |
| \(v = 24.5\) or \(25 \ (\text{m s}^{-1})\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0 = 14.7^2 - 2 \times 9.8h\) | M1 | Complete method to find \(h\) |
| \(h = 11.0\) or \(11 \ \text{(m)}\) | A1 | 11.0 or 11 (m) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| New value of speed would be lower | B1 | New value of speed will be lower |
## Question 1:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $s = vt - \frac{1}{2}at^2$ | M1 | Complete method to give equation in $v$ only (could involve 2 or more suvat equations then elimination) with usual rules |
| $19.6 = 4v - \frac{1}{2} \times 9.8 \times 4^2$ | A1 | Correct equation |
| $v = 24.5$ or $25 \ (\text{m s}^{-1})$ | A1 | Correct answer |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 = 14.7^2 - 2 \times 9.8h$ | M1 | Complete method to find $h$ |
| $h = 11.0$ or $11 \ \text{(m)}$ | A1 | 11.0 or 11 (m) |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| New value of speed would be lower | B1 | New value of speed will be lower |\begin{enumerate}
\item A company manager is investigating the time taken, $t$ minutes, to complete an aptitude test. The human resources manager produced the table below of coded times, $x$ minutes, for a random sample of 30 applicants.
\end{enumerate}
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
Coded time ( $x$ minutes) & Frequency (f) & Coded time midpoint (y minutes) \\
\hline
$0 \leq x < 5$ & 3 & 2.5 \\
\hline
$5 \leq x < 10$ & 15 & 7.5 \\
\hline
$10 \leq x < 15$ & 2 & 12.5 \\
\hline
$15 \leq x < 25$ & 9 & 20 \\
\hline
$25 \leq x < 35$ & 1 & 30 \\
\hline
\end{tabular}
\end{center}
(You may use $\sum f y = 355$ and $\sum f y ^ { 2 } = 5675$ )\\
(a) Use linear interpolation to estimate the median of the coded times.\\
(b) Estimate the standard deviation of the coded times.
The company manager is told by the human resources manager that he subtracted 15 from each of the times and then divided by 2 , to calculate the coded times.\\
(c) Calculate an estimate for the median and the standard deviation of $t$.\\
(3)
The following year, the company has 25 positions available. The company manager decides not to offer a position to any applicant who takes 35 minutes or more to complete the aptitude test.
The company has 60 applicants.\\
(d) Comment on whether or not the company manager's decision will result in the company being able to fill the 25 positions available from these 60 applicants. Give a reason for your answer.\\
\hfill \mbox{\textit{Edexcel AS Paper 2 Q1 [9]}}