Question 3 - A-Level Maths

Edexcel AS Paper 2 2022 June — Question 3 8 marks

Exam Board	Edexcel
Module	AS Paper 2 (AS Paper 2)
Year	2022
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Data representation
Type	Use linear interpolation for median or quartiles
Difficulty	Moderate -0.3 Part (a) is a standard linear interpolation exercise requiring identification of the median position (128th value) and interpolation within a histogram class - routine AS-level statistics. Parts (b) and (c) involve integration to find k and then solving for the median of a continuous distribution, which adds some challenge but remains a well-practiced technique at AS level. Overall slightly easier than average due to the structured, multi-part nature guiding students through standard procedures.
Spec	2.02b Histogram: area represents frequency 2.02f Measures of average and spread

The histogram summarises the heights of 256 seedlings two weeks after they were planted. \includegraphics[max width=\textwidth, alt={}, center]{08e3b0b0-2155-4b37-83e3-343c317ca10c-06_1242_1810_287_132}
1. Use linear interpolation to estimate the median height of the seedlings.
  (4)
Chris decides to model the frequency density for these 256 seedlings by a curve with equation $$y = k x ( 8 - x ) \quad 0 \leqslant x \leqslant 8$$ where $k$ is a constant.
Find the value of $k$ Using this model,
write down the median height of the seedlings.

Show mark scheme Show mark scheme source

Answer	Marks	Guidance
3(a)	Class	Frequency
$0-1$	15	15
$1-2$	35	50
$2-3.5$	75	125
$3.5-4.5$	55	180
M1	AO 2.1
A1	AO 1.1b
$[Q_2 =] (3.5) + \frac{256 - ''125''}{''55''} \times (4.5 - 3.5)$ or $(4.5) - \frac{''180'' - 256}{''55''} \times 1$	M1	AO 2.1
A1	AO 1.1b
$= 3.5545\ldots$ awrt 3.55		(4)
3(b) Need area under curve to be 256 so $\int_{(0)}^{(8)} k(8-x) \, dx = 256$	M1	AO 3.1a
$k\left[4x^2 - \frac{x^3}{3}\right]_{(0)}^{(8)} = 256$	M1	AO 1.1b
$\{k[4 \times 8^2 - \frac{8}{3}x^2]\} = 256 \Rightarrow$ $k = 3$	A1	AO 1.1b
3(c) [By symmetry median $=$] 4	B1	AO 2.2a

Total: 8 marks

Notes:

(a) 1st M1 for an attempt to form frequency table (at least 1st 4 rows and freq or cum freq seen must have the frequency of 75 correct and can condone one error/omission in 15, 35, 55). Frequencies or cum freq may be seen on bars of the histogram.

1st A1 for identifying class, freq and cum freq (i.e. highlighted values from the table) or diagram with 125, "128", 180, 3.5 & 4.5. May be implied by values in 2nd M1 expression.

2nd M1 for a correct calculation for $Q_2$ (condone error in end point e.g. 3.45 or 3.49 etc). Can fit their "125" (provided $>$ 100) and their "55". Allow use of $(n + 1)$, usually see $128.5 - \ldots$ leading to $3.5636\ldots$ or awrt 3.56.

2nd A1 awrt 3.55 but 3.555 is fine (allow 3.56 if $(n+1)$ being used)...need sight of $\frac{256}{2}$ etc). Correct answer with no incorrect working scores 4/4.

(b) 1st M1 for identifying the need to find the area under the curve by integrating.

2nd M1 for correct integration and $= 256$ (condone missing limits).

A1 for $k = 3$ [May see use of calculator for the integration so score 2nd M1A1 together].

(c) NB The answer to part (c) may be written within the question.

B1 for 4 (Independent of their value of $k$ but must be their "$x$" value). NB when $k = 0.25$ and $x = 4$ gives $y = 4$ so must be clear they intend median $= 4$. The statement in part (c) "$k = 4$" is B0.

**3(a)** | Class | Frequency | Cum. Frequency |
|---------|-----------|-----------------|
| $0-1$ | 15 | 15 |
| $1-2$ | 35 | 50 |
| $2-3.5$ | 75 | 125 |
| $3.5-4.5$ | 55 | 180 |

| M1 | AO 2.1
| A1 | AO 1.1b

$[Q_2 =] (3.5) + \frac{256 - ''125''}{''55''} \times (4.5 - 3.5)$ or $(4.5) - \frac{''180'' - 256}{''55''} \times 1$ | M1 | AO 2.1
| A1 | AO 1.1b

$= 3.5545\ldots$ awrt **3.55** | | (4)

**3(b)** Need area under curve to be 256 so $\int_{(0)}^{(8)} k(8-x) \, dx = 256$ | M1 | AO 3.1a

$k\left[4x^2 - \frac{x^3}{3}\right]_{(0)}^{(8)} = 256$ | M1 | AO 1.1b

$\{k[4 \times 8^2 - \frac{8}{3}x^2]\} = 256 \Rightarrow$ **$k = 3$** | A1 | AO 1.1b

**3(c)** [By symmetry median $=$] **4** | B1 | AO 2.2a

**Total: 8 marks**

**Notes:**

**(a)** 1st M1 for an attempt to form frequency table (at least 1st 4 rows and freq or cum freq seen must have the frequency of 75 correct and can condone one error/omission in 15, 35, 55). Frequencies or cum freq may be seen on bars of the histogram.

1st A1 for identifying class, freq and cum freq (i.e. highlighted values from the table) or diagram with 125, "128", 180, 3.5 & 4.5. May be implied by values in 2nd M1 expression.

2nd M1 for a correct calculation for $Q_2$ (condone error in end point e.g. 3.45 or 3.49 etc). Can fit their "125" (provided $>$ 100) and their "55". Allow use of $(n + 1)$, usually see $128.5 - \ldots$ leading to $3.5636\ldots$ or awrt 3.56.

2nd A1 awrt 3.55 but 3.555 is fine (allow 3.56 if $(n+1)$ being used)...need sight of $\frac{256}{2}$ etc). Correct answer with no incorrect working scores 4/4.

**(b)** 1st M1 for identifying the need to find the area under the curve by integrating.

2nd M1 for correct integration and $= 256$ (condone missing limits).

A1 for $k = 3$ [May see use of calculator for the integration so score 2nd M1A1 together].

**(c)** **NB** The answer to part (c) may be written within the question.

B1 for 4 (Independent of their value of $k$ but must be their "$x$" value). NB when $k = 0.25$ and $x = 4$ gives $y = 4$ so must be clear they intend median $= 4$. The statement in part (c) "$k = 4$" is B0.

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Show LaTeX source

\begin{enumerate}
  \item The histogram summarises the heights of 256 seedlings two weeks after they were planted.\\
\includegraphics[max width=\textwidth, alt={}, center]{08e3b0b0-2155-4b37-83e3-343c317ca10c-06_1242_1810_287_132}\\
(a) Use linear interpolation to estimate the median height of the seedlings.\\
(4)
\end{enumerate}

Chris decides to model the frequency density for these 256 seedlings by a curve with equation

$$y = k x ( 8 - x ) \quad 0 \leqslant x \leqslant 8$$

where $k$ is a constant.\\
(b) Find the value of $k$

Using this model,\\
(c) write down the median height of the seedlings.

\hfill \mbox{\textit{Edexcel AS Paper 2 2022 Q3 [8]}}

This paper (5 questions)

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Q1 5 Q2 7 Q3 8 Q4 2 Q5 8

Edexcel AS Paper 2 2022 June — Question 3 8 marks

Total: 8 marks

Notes:

(a) 1st M1 for an attempt to form frequency table (at least 1st 4 rows and freq or cum freq seen must have the frequency of 75 correct and can condone one error/omission in 15, 35, 55). Frequencies or cum freq may be seen on bars of the histogram.

1st A1 for identifying class, freq and cum freq (i.e. highlighted values from the table) or diagram with 125, "128", 180, 3.5 & 4.5. May be implied by values in 2nd M1 expression.

2nd M1 for a correct calculation for \(Q_2\) (condone error in end point e.g. 3.45 or 3.49 etc). Can fit their "125" (provided \(>\) 100) and their "55". Allow use of \((n + 1)\), usually see \(128.5 - \ldots\) leading to \(3.5636\ldots\) or awrt 3.56.

2nd A1 awrt 3.55 but 3.555 is fine (allow 3.56 if \((n+1)\) being used)...need sight of \(\frac{256}{2}\) etc). Correct answer with no incorrect working scores 4/4.

(b) 1st M1 for identifying the need to find the area under the curve by integrating.

2nd M1 for correct integration and \(= 256\) (condone missing limits).

A1 for \(k = 3\) [May see use of calculator for the integration so score 2nd M1A1 together].

(c) NB The answer to part (c) may be written within the question.

B1 for 4 (Independent of their value of \(k\) but must be their "\(x\)" value). NB when \(k = 0.25\) and \(x = 4\) gives \(y = 4\) so must be clear they intend median \(= 4\). The statement in part (c) "\(k = 4\)" is B0.

This paper (5 questions)