Question 5 - A-Level Maths

Edexcel S1 2009 January — Question 5 16 marks

Exam Board	Edexcel
Module	S1 (Statistics 1)
Year	2009
Session	January
Marks	16
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Data representation
Type	Calculate using histogram bar dimensions
Difficulty	Standard +0.3 This is a standard S1 histogram question testing routine procedures: calculating frequency density for bar dimensions, linear interpolation for median/IQR, and mean/SD from grouped data. All techniques are textbook exercises requiring careful arithmetic but no problem-solving insight. Slightly easier than average due to being entirely procedural.
Spec	2.02b Histogram: area represents frequency 2.02f Measures of average and spread 2.02g Calculate mean and standard deviation 2.02h Recognize outliers

5. In a shopping survey a random sample of 104 teenagers were asked how many hours, to the nearest hour, they spent shopping in the last month. The results are summarised in the table below.

Number of hours	Mid-point	Frequency
0-5	2.75	20
6-7	6.5	16
8-10	9	18
11-15	13	25
16-25	20.5	15
26-50	38	10

A histogram was drawn and the group ( \(8 - 10\) ) hours was represented by a rectangle that was 1.5 cm wide and 3 cm high.

Calculate the width and height of the rectangle representing the group (16-25) hours.
Use linear interpolation to estimate the median and interquartile range.
Estimate the mean and standard deviation of the number of hours spent shopping.
State, giving a reason, the skewness of these data.
State, giving a reason, which average and measure of dispersion you would recommend to use to summarise these data.

Show mark scheme Show mark scheme source

Question 5:

(a)

8–10 hours: width \(= 10.5 - 7.5 = 3\), represented by 1.5 cm

Answer	Marks	Guidance
16–25 hours: width \(= 25.5 - 15.5 = 10\), represented by 5 cm	B1, M1	B1 for attempting both frequency densities \(\frac{18}{3}(=6)\) and \(\frac{15}{10}\), and \(\frac{15}{10} \times \text{SF}\) where \(\text{SF} \neq 1\)

8–10 hours: height \(= \text{fd} = 18/3 = 6\), represented by 3 cm

Answer	Marks
16–25 hours: height \(= \text{fd} = 15/10 = 1.5\), represented by 0.75 cm	A1 (3)

(b)

Answer	Marks	Guidance
\(Q_2 = 7.5 + \frac{(52-36)}{18} \times 3 = 10.2\)	M1, A1	M1 for identifying correct interval and correct fraction. 1st A1 for 10.2 (using \(n+1\) allow AWRT 10.3)
\(Q_1 = 5.5 + \frac{(26-20)}{16} \times 2 = 6.25\) or \(6.3\), or \(5.5 + \frac{(26.25-20)}{16} \times 2 [=6.3]\)	A1	2nd A1 for correct expression for either \(Q_1\) or \(Q_3\)
\(Q_3 = 10.5 + \frac{(78-54)}{25} \times 5 [=15.3]\) or \(10.5 + \frac{(78.75-54)}{25} \times 5 [=15.45 \approx 15.5]\)	A1, A1ft	3rd A1 for correct expressions for both \(Q_1\) and \(Q_3\); 4th A1ft for IQR, ft their quartiles
\(\text{IQR} = (15.3 - 6.3) = 9\)	(5)

(c)

Answer	Marks	Guidance
\(\sum fx = 1333.5 \Rightarrow \bar{x} = \frac{1333.5}{104}\) AWRT 12.8	M1 A1	1st M1 for attempting \(\sum fx\) and \(\bar{x}\)
\(\sum fx^2 = 27254 \Rightarrow \sigma_x = \sqrt{\frac{27254}{104} - \bar{x}^2} = \sqrt{262.05 - \bar{x}^2}\) AWRT 9.88	M1 A1 (4)	2nd M1 for attempting \(\sum fx^2\) and \(\sigma_x\); \(\sqrt{\phantom{x}}\) is needed for M1. Allow \(s =\) AWRT 9.93

(d)

Answer	Marks	Guidance
\(Q_3 - Q_2 [=5.1] > Q_2 - Q_1 [=3.9]\) or \(Q_2 < \bar{x}\)	B1ft	1st B1ft for suitable test; values need not be seen but statement must be compatible with values used
So data is positively skew	dB1 (2)	2nd dB1 dependent on test showing positive skew and for stating positive skew. If test shows negative skew score 1st B1 but lose 2nd

(e)

Answer	Marks	Guidance
Use median and IQR, since data is skewed or not affected by extreme values or outliers	B1, B1 (2)	1st B1 for choosing median and IQR — must mention both; 2nd B1 for suitable reason. "Use median because data is skewed" scores B0B1 since IQR not mentioned

## Question 5:

**(a)**
8–10 hours: width $= 10.5 - 7.5 = 3$, represented by 1.5 cm

16–25 hours: width $= 25.5 - 15.5 = 10$, represented by 5 cm | B1, M1 | B1 for attempting both frequency densities $\frac{18}{3}(=6)$ and $\frac{15}{10}$, and $\frac{15}{10} \times \text{SF}$ where $\text{SF} \neq 1$

8–10 hours: height $= \text{fd} = 18/3 = 6$, represented by 3 cm

16–25 hours: height $= \text{fd} = 15/10 = 1.5$, represented by 0.75 cm | A1 (3) |

**(b)**
$Q_2 = 7.5 + \frac{(52-36)}{18} \times 3 = 10.2$ | M1, A1 | M1 for identifying correct interval and correct fraction. 1st A1 for 10.2 (using $n+1$ allow AWRT 10.3)

$Q_1 = 5.5 + \frac{(26-20)}{16} \times 2 = 6.25$ or $6.3$, or $5.5 + \frac{(26.25-20)}{16} \times 2 [=6.3]$ | A1 | 2nd A1 for correct expression for either $Q_1$ or $Q_3$

$Q_3 = 10.5 + \frac{(78-54)}{25} \times 5 [=15.3]$ or $10.5 + \frac{(78.75-54)}{25} \times 5 [=15.45 \approx 15.5]$ | A1, A1ft | 3rd A1 for correct expressions for both $Q_1$ and $Q_3$; 4th A1ft for IQR, ft their quartiles

$\text{IQR} = (15.3 - 6.3) = 9$ | (5) |

**(c)**
$\sum fx = 1333.5 \Rightarrow \bar{x} = \frac{1333.5}{104}$ AWRT 12.8 | M1 A1 | 1st M1 for attempting $\sum fx$ and $\bar{x}$

$\sum fx^2 = 27254 \Rightarrow \sigma_x = \sqrt{\frac{27254}{104} - \bar{x}^2} = \sqrt{262.05 - \bar{x}^2}$ AWRT 9.88 | M1 A1 (4) | 2nd M1 for attempting $\sum fx^2$ and $\sigma_x$; $\sqrt{\phantom{x}}$ is needed for M1. Allow $s =$ AWRT 9.93

**(d)**
$Q_3 - Q_2 [=5.1] > Q_2 - Q_1 [=3.9]$ or $Q_2 < \bar{x}$ | B1ft | 1st B1ft for suitable test; values need not be seen but statement must be compatible with values used

So data is positively skew | dB1 (2) | 2nd dB1 dependent on test showing positive skew and for stating positive skew. If test shows negative skew score 1st B1 but lose 2nd

**(e)**
Use median and IQR, since data is skewed or not affected by extreme values or outliers | B1, B1 (2) | 1st B1 for choosing median and IQR — must mention both; 2nd B1 for suitable reason. "Use median because data is skewed" scores B0B1 since IQR not mentioned

Show LaTeX source

5. In a shopping survey a random sample of 104 teenagers were asked how many hours, to the nearest hour, they spent shopping in the last month. The results are summarised in the table below.

\begin{center}
\begin{tabular}{|l|l|l|}
\hline
Number of hours & Mid-point & Frequency \\
\hline
0-5 & 2.75 & 20 \\
\hline
6-7 & 6.5 & 16 \\
\hline
8-10 & 9 & 18 \\
\hline
11-15 & 13 & 25 \\
\hline
16-25 & 20.5 & 15 \\
\hline
26-50 & 38 & 10 \\
\hline
\end{tabular}
\end{center}

A histogram was drawn and the group ( $8 - 10$ ) hours was represented by a rectangle that was 1.5 cm wide and 3 cm high.
\begin{enumerate}[label=(\alph*)]
\item Calculate the width and height of the rectangle representing the group (16-25) hours.
\item Use linear interpolation to estimate the median and interquartile range.
\item Estimate the mean and standard deviation of the number of hours spent shopping.
\item State, giving a reason, the skewness of these data.
\item State, giving a reason, which average and measure of dispersion you would recommend to use to summarise these data.
\end{enumerate}

\hfill \mbox{\textit{Edexcel S1 2009 Q5 [16]}}

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Q1 11 Q2 8 Q3 16 Q4 14 Q5 16 Q6 10