| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2022 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Measures of Location and Spread |
| Type | Find median and quartiles from stem-and-leaf diagram |
| Difficulty | Moderate -0.8 This is a routine S1 statistics question testing standard procedures: finding quartiles from ordered data, applying the outlier definition, and using coding transformations. Part (a) requires systematic checking but no insight; parts (b)-(d) are direct formula applications. Easier than average A-level due to being purely procedural with no problem-solving required. |
| Spec | 2.01a Population and sample: terminology2.02f Measures of average and spread2.02g Calculate mean and standard deviation2.02h Recognize outliers2.03d Calculate conditional probability: from first principles |
| 10 | 8 | 9 | (2) | |||||||||
| 11 | 0 | 3 | 6 | 6 | 6 | 8 | 8 | 9 | 9 | 9 | 9 | (11) |
| 12 | 4 | 5 | 5 | 5 | 5 | 5 | 5 | 8 | (8) | |||
| 13 | \(a\) | \(b\) | \(c\) | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Scheme | Marks |
| (a) | lower quartile = 116; upper quartile = 125 | B1 |
| "\(125" + 1.5 \times ("125" - "116")\) or "\(125" + 1.5 \times (9)\) | M1 | Use of Q3 + 1.5 × IQR with their values |
| Outlier is greater than 138.5, so \(c = 9*\) | A1*cso (3) | For 138.5 and conclusion \(c = 9\) (do not accept \(c = 139\)) with no errors. Answer is given so working must be shown. |
| (b) | \(\bar{x} = \frac{-96}{24} [= -4]\) | M1 |
| \(\bar{d} = '\bar{x}' + 125\) | M1 | 2nd M1 use of \(\bar{d} = '\bar{x}' + 125\) |
| \(\sum d = 125 \times 24 - 96[= 2904]\) | M1 | 1st M1 for correct expression for \(\sum d\) |
| \(\bar{d} = \frac{2904'}{24}\) | M1 | 2nd M1 use of "\(\sum d" ÷ 24\) must be clear it is their sum |
| \(\bar{d} = 121\) | A1 (3) | A1 121 |
| (c) | \([\sigma_x = \sigma_d] = \sqrt{\frac{1306}{24}}\) | M1 |
| \([\sigma_d] = 7.3767\ldots\) | A1 | A1 awrt 7.38 final answer |
| (2) | ||
| (d) | \([P(D > 118 \mid X < 0)] = \frac{P(118 < D < 125)}{P(D < 125)}\) or \(\frac{P(-7 < X < 0)}{P(X < 0)}\) or \(\frac{5/24}{14/24}\) | M1 |
| \(= \frac{5}{14}\) | A1 (2) | A1 allow awrt 0.357 |
| [10] |
| Part | Scheme | Marks | Guidance |
|------|--------|-------|----------|
| (a) | lower quartile = 116; upper quartile = 125 | B1 | Both values correct. Both values must be seen either in the calculation or separately. They are not implied by the IQR = 9 |
| | "$125" + 1.5 \times ("125" - "116")$ or "$125" + 1.5 \times (9)$ | M1 | Use of Q3 + 1.5 × IQR with their values |
| | Outlier is greater than 138.5, so $c = 9*$ | A1*cso (3) | For 138.5 and conclusion $c = 9$ (do not accept $c = 139$) with no errors. Answer is given so working must be shown. |
| (b) | $\bar{x} = \frac{-96}{24} [= -4]$ | M1 | 1st M1 for correct expression for $\bar{x}$ |
| | $\bar{d} = '\bar{x}' + 125$ | M1 | 2nd M1 use of $\bar{d} = '\bar{x}' + 125$ |
| | $\sum d = 125 \times 24 - 96[= 2904]$ | M1 | 1st M1 for correct expression for $\sum d$ |
| | $\bar{d} = \frac{2904'}{24}$ | M1 | 2nd M1 use of "$\sum d" ÷ 24$ must be clear it is their sum |
| | $\bar{d} = 121$ | A1 (3) | A1 121 |
| (c) | $[\sigma_x = \sigma_d] = \sqrt{\frac{1306}{24}}$ | M1 | M1 correct expression $\sqrt{\frac{1306}{24}}$ |
| | $[\sigma_d] = 7.3767\ldots$ | A1 | A1 awrt 7.38 final answer |
| | | (2) | |
| (d) | $[P(D > 118 \mid X < 0)] = \frac{P(118 < D < 125)}{P(D < 125)}$ or $\frac{P(-7 < X < 0)}{P(X < 0)}$ or $\frac{5/24}{14/24}$ | M1 | M1 correct probability statement (allow a probability of $\frac{k}{14}$ where $0 < k < 14$ to score M1) |
| | $= \frac{5}{14}$ | A1 (2) | A1 allow awrt 0.357 |
| | | [10] | |
---\begin{enumerate}
\item The stem and leaf diagram shows the number of deliveries made by Pat each day for 24 days
\end{enumerate}
\begin{table}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Key: 10 $\mathbf { 8 }$ represents 108 deliveries}
\begin{tabular}{ l | c c c c c c c c c c c c }
10 & 8 & 9 & & & & & & & & & & (2) \\
11 & 0 & 3 & 6 & 6 & 6 & 8 & 8 & 9 & 9 & 9 & 9 & (11) \\
12 & 4 & 5 & 5 & 5 & 5 & 5 & 5 & 8 & & & & (8) \\
13 & $a$ & $b$ & $c$ & & & & & & & & & (3) \\
\end{tabular}
\end{center}
\end{table}
where $a$, $b$ and $c$ are positive integers with $a < b < c$\\
An outlier is defined as any value greater than $1.5 \times$ interquartile range above the upper quartile.
Given that there is only one outlier for these data,\\
(a) show that $c = 9$
The number of deliveries made by Pat each day is represented by $d$\\
The data in the stem and leaf diagram are coded using
$$x = d - 125$$
and the following summary statistics are obtained
$$\sum x = - 96 \quad \text { and } \quad \sum ( x - \bar { x } ) ^ { 2 } = 1306$$
(b) Find the mean number of deliveries.\\
(c) Find the standard deviation of the number of deliveries.
One of these 24 days is selected at random. The random variable $D$ represents the number of deliveries made by Pat on this day.
The random variable $X = D - 125$\\
(d) Find $\mathrm { P } ( D > 118 \mid X < 0 )$
\begin{center}
\end{center}
\hfill \mbox{\textit{Edexcel S1 2022 Q3 [10]}}