Question 1 - A-Level Maths

Edexcel S1 2024 October — Question 1

Exam Board	Edexcel
Module	S1 (Statistics 1)
Year	2024
Session	October
Paper	Download PDF ↗
Topic	Measures of Location and Spread
Type	Interpret or analyse given back-to-back stem-and-leaf
Difficulty	Easy -1.2 This is a routine S1 statistics question testing basic skills: reading stem-and-leaf diagrams, finding quartiles using standard position formulas, calculating mean and standard deviation from given sums, and applying a given skewness formula. All techniques are straightforward recall with no problem-solving or novel insight required, making it easier than average A-level questions.
Spec	2.02a Interpret single variable data: tables and diagrams 2.02f Measures of average and spread 2.02g Calculate mean and standard deviation 2.02h Recognize outliers

The back-to-back stem and leaf diagram on page 3 shows information about the running times of 31 Action films and 31 Comedy films.
The running times are given to the nearest minute.
1. Write down the modal running time for these Action films.
Some of the quartiles for these two distributions are shown in the table below.
Action films Comedy films
Lower quartile 121 $a$
Median $b$ 117
Upper quartile 138 $c$
Find the value of $a$, the value of $b$ and the value of $c$
For these Action films find, to one decimal place,
1. the mean running time,
2. the standard deviation of the running times.
  (You may use $\sum x = 4016$ and $\sum x ^ { 2 } = 525056$ where $x$ is the running time, in minutes, of an Action film.) One measure of skewness is found using $$\frac { \text { mean - mode } } { \text { standard deviation } }$$
Evaluate this measure and describe the skewness for the running times of these Action films.
Comment on one difference between the distribution of the running times of these Action films and the distribution of the running times of these Comedy films. State the values of any statistics you have used to support your comment.
Totals Action films Comedy films Totals
(1) 0 9 2235 (5)
(0) 10 356689 (6)
(5) 98642 11 02467999 (8)
(10) 9987654310 12 12466777789 (11)
(8) 87775421 13 1 (1)
(7) 7766431 14 (0)
Key: $0 | 9 | 2$ means 90 minutes for an Action film and 92 minutes for a Comedy film

Show mark scheme Show mark scheme source

Answer	Marks	Guidance
(a) [Mode =] 137	B1	Cao Must be seen in part (a)
(b) $a = 106$	B1	$a = 106$ Must be attached to $a$ (Condone $Q_1$/lower quartile for $a$)
$b = 129$	B1	$b = 129$ Must be attached to $b$ (Condone $Q_2$/median for $b$)
$c = 126$	B1	$c = 126$ Must be attached to $c$ (Condone $Q_3$/upper quartile for $c$)
(c)(i) $\left[\text{mean} = \frac{4016}{31}\right] = 129.5$	B1	awrt 130
(c)(ii) $\left[\text{Standard deviation }=\right] \sqrt{\frac{525056}{31}-\left(\frac{4016}{31}\right)^2}$ or $\sqrt{\frac{30}{31}\left(\frac{525056}{31}-\left(\frac{4016}{31}\right)^2\right)}$	M1	For a correct method (including the square root) to find the standard deviation. Ft their mean. May be implied by awrt 12.4 or awrt 12.9 or awrt 12.6 if sample standard deviation is calculated
$= 12.4$ or $12.6$	A1	awrt 12.4 or awrt 12.6 if sample standard deviation is calculated Correct answer only scores 2/2
(d) $\frac{\text{'129.5'}-\text{'137'}}{\text{'12.4'}} \left[= -0.6\right]$	M1	For substitution of the mode, the mean and the standard deviation into the expression Ft their mode, mean and standard deviation
Negative [skew]	A1ft	A correct interpretation ft their expression. Ignore any reference to correlation
(e) A correct difference of the average or a correct difference of the spread e.g. On average action films run for longer than comedy films as the median is greater $129 > 117$	B1ft	for a correct comment, referring to length, with reference to a correctly named statistic. Must include the correct figures compared. Ft their values. Ignore any reference to skew

Total: 10 marks

**(a)** [Mode =] 137 | B1 | Cao Must be seen in part (a)

**(b)** $a = 106$ | B1 | $a = 106$ Must be attached to $a$ (Condone $Q_1$/lower quartile for $a$)
$b = 129$ | B1 | $b = 129$ Must be attached to $b$ (Condone $Q_2$/median for $b$)
$c = 126$ | B1 | $c = 126$ Must be attached to $c$ (Condone $Q_3$/upper quartile for $c$)

**(c)(i)** $\left[\text{mean} = \frac{4016}{31}\right] = 129.5$ | B1 | awrt 130

**(c)(ii)** $\left[\text{Standard deviation }=\right] \sqrt{\frac{525056}{31}-\left(\frac{4016}{31}\right)^2}$ or $\sqrt{\frac{30}{31}\left(\frac{525056}{31}-\left(\frac{4016}{31}\right)^2\right)}$ | M1 | For a correct method (including the square root) to find the standard deviation. Ft their mean. May be implied by awrt 12.4 or awrt 12.9 or awrt 12.6 if sample standard deviation is calculated
$= 12.4$ or $12.6$ | A1 | awrt 12.4 or awrt 12.6 if sample standard deviation is calculated Correct answer only scores 2/2

**(d)** $\frac{\text{'129.5'}-\text{'137'}}{\text{'12.4'}} \left[= -0.6\right]$ | M1 | For substitution of the mode, the mean and the standard deviation into the expression Ft their mode, mean and standard deviation
Negative [skew] | A1ft | A correct interpretation ft their expression. Ignore any reference to correlation

**(e)** A correct difference of the average or a correct difference of the spread e.g. On average action films run for longer than comedy films as the median is greater $129 > 117$ | B1ft | for a correct comment, referring to length, with reference to a correctly named statistic. Must include the correct figures compared. Ft their values. Ignore any reference to skew

**Total: 10 marks**

Show LaTeX source

\begin{enumerate}
  \item The back-to-back stem and leaf diagram on page 3 shows information about the running times of 31 Action films and 31 Comedy films.\\
The running times are given to the nearest minute.\\
(a) Write down the modal running time for these Action films.
\end{enumerate}

Some of the quartiles for these two distributions are shown in the table below.

\begin{center}
\begin{tabular}{ | c | c | c | }
\hline
 & Action films & Comedy films \\
\hline
Lower quartile & 121 & $a$ \\
\hline
Median & $b$ & 117 \\
\hline
Upper quartile & 138 & $c$ \\
\hline
\end{tabular}
\end{center}

(b) Find the value of $a$, the value of $b$ and the value of $c$\\
(c) For these Action films find, to one decimal place,\\
(i) the mean running time,\\
(ii) the standard deviation of the running times.\\
(You may use $\sum x = 4016$ and $\sum x ^ { 2 } = 525056$ where $x$ is the running time, in minutes, of an Action film.)

One measure of skewness is found using

$$\frac { \text { mean - mode } } { \text { standard deviation } }$$

(d) Evaluate this measure and describe the skewness for the running times of these Action films.\\
(e) Comment on one difference between the distribution of the running times of these Action films and the distribution of the running times of these Comedy films. State the values of any statistics you have used to support your comment.

\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
Totals & \multicolumn{2}{|c|}{Action films} & Comedy films & Totals \\
\hline
(1) & 0 & 9 & 2235 & (5) \\
\hline
(0) &  & 10 & 356689 & (6) \\
\hline
(5) & 98642 & 11 & 02467999 & (8) \\
\hline
(10) & 9987654310 & 12 & 12466777789 & (11) \\
\hline
(8) & 87775421 & 13 & 1 & (1) \\
\hline
(7) & 7766431 & 14 &  & (0) \\
\hline
\end{tabular}
\end{center}

Key: $0 | 9 | 2$ means 90 minutes for an Action film and 92 minutes for a Comedy film

\hfill \mbox{\textit{Edexcel S1 2024 Q1}}

This paper (8 questions)

View full paper

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8

	Action films	Comedy films
Lower quartile	121	\(a\)
Median	\(b\)	117
Upper quartile	138	\(c\)

Totals	Action films		Comedy films	Totals
(1)	0	9	2235	(5)
(0)		10	356689	(6)
(5)	98642	11	02467999	(8)
(10)	9987654310	12	12466777789	(11)
(8)	87775421	13	1	(1)
(7)	7766431	14		(0)

Answer	Marks	Guidance
(a) [Mode =] 137	B1	Cao Must be seen in part (a)
(b) \(a = 106\)	B1	\(a = 106\) Must be attached to \(a\) (Condone \(Q_1\)/lower quartile for \(a\))
\(b = 129\)	B1	\(b = 129\) Must be attached to \(b\) (Condone \(Q_2\)/median for \(b\))
\(c = 126\)	B1	\(c = 126\) Must be attached to \(c\) (Condone \(Q_3\)/upper quartile for \(c\))
(c)(i) \(\left[\text{mean} = \frac{4016}{31}\right] = 129.5\)	B1	awrt 130
(c)(ii) \(\left[\text{Standard deviation }=\right] \sqrt{\frac{525056}{31}-\left(\frac{4016}{31}\right)^2}\) or \(\sqrt{\frac{30}{31}\left(\frac{525056}{31}-\left(\frac{4016}{31}\right)^2\right)}\)	M1	For a correct method (including the square root) to find the standard deviation. Ft their mean. May be implied by awrt 12.4 or awrt 12.9 or awrt 12.6 if sample standard deviation is calculated
\(= 12.4\) or \(12.6\)	A1	awrt 12.4 or awrt 12.6 if sample standard deviation is calculated Correct answer only scores 2/2
(d) \(\frac{\text{'129.5'}-\text{'137'}}{\text{'12.4'}} \left[= -0.6\right]\)	M1	For substitution of the mode, the mean and the standard deviation into the expression Ft their mode, mean and standard deviation
Negative [skew]	A1ft	A correct interpretation ft their expression. Ignore any reference to correlation
(e) A correct difference of the average or a correct difference of the spread e.g. On average action films run for longer than comedy films as the median is greater \(129 > 117\)	B1ft	for a correct comment, referring to length, with reference to a correctly named statistic. Must include the correct figures compared. Ft their values. Ignore any reference to skew