What is S2.1: Single-variable data diagrams?

In a histogram, frequency is proportional to bar area and frequency density is frequency/class width; unequal class widths make bar height alone misleading.

How many marks are S2.1 questions worth?

S2.1 practice drills on this page are worth 2–7 marks each; marks in a real exam vary by question.

Which past exam papers has S2.1 appeared in?

We have not yet indexed a verified past-paper appearance for S2.1 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

S2.1

Interpret diagrams for single-variable data, including understanding that area in a histogram represents frequency; connect to probability distributions.

Draft — not yet indexed

Single-variable data diagrams

Worked answers and methods for S2.1 on Edexcel A-level Maths 9MA0.

Explanation

In a histogram, frequency is proportional to bar area and frequency density is $\frac{\text{frequency}}{\text{class width}}$ ; unequal class widths make bar height alone misleading.
Frequency polygons show class patterns, box plots summarise centre and spread, and cumulative frequency diagrams support estimates of medians, quartiles and counts below a value.
For a continuous probability density histogram, total area is $1$ and the area above an interval is the probability of an observation in that interval.
Read class boundaries and axis scales before calculating.
A common error is to use frequency density as though it were frequency.

In a histogram, bar area represents frequency, so height is frequency density.

Worked example

A cumulative frequency diagram represents $80$ observations. The cumulative frequencies at $x=10$ and $x=20$ are $18$ and $54$ respectively. Estimate the number of observations satisfying $10<x\leq20$ .

1.The cumulative frequency at $20$ counts all $54$ observations up to that value, including the $18$ already counted up to $10$ .
2.Subtract to isolate the interval: $54-18=36$ .

Answer: $36$

Common mistakes

Don't use bar height rather than class width times frequency density to compare histogram frequencies.
Don't read cumulative frequencies directly as class frequencies instead of subtracting the boundary totals.

Exam tip

For cumulative-frequency intervals, subtract the cumulative totals at the two boundaries and respect endpoint inequalities.

Worked practice

Tier 1 · Easy

A histogram class is

12\leq t<17

and has frequency density

3.6

. Find the frequency in this class.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$18$	2
Notes
The class width is $17-12=5$ . Hence frequency $=$ class width $\times$ frequency density $=5\times3.6=18$ .

(2 marks)

Tier 2 · Standard

In a histogram, the class

10\leq x<15

has frequency density

4

, and the class

15\leq x<25

has frequency density

2.5

. Find the frequency in each class and the total frequency represented by these two bars.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Frequencies $20$ and $25$ Total frequency $45$	4
Notes
Histogram frequency equals bar area, so multiply frequency density by class width. The first class has width $5$ and frequency $4(5)=20$ . The second has width $10$ and frequency $2.5(10)=25$ . Their total is $20+25=45$ .

(4 marks)

Tier 3 · Hard

The probability density histogram for a continuous random variable

X

has constant heights

0.12

0\leq x<3

0.08

3\leq x<8

, and

0.12

8\leq x\leq10

. Verify that it defines a probability distribution and find

P(1<X<6)

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Total area $=1$ . $P(1<X<6)=0.48$ .	5
Notes
The total area is $3(0.12)+5(0.08)+2(0.12)=0.36+0.40+0.24=1$ , so the histogram defines a probability distribution. The area from $1$ to $3$ is $2(0.12)=0.24$ , and the area from $3$ to $6$ is $3(0.08)=0.24$ . Therefore $P(1<X<6)=0.24+0.24=0.48$ .

(5 marks)

Tier 1 · Easy

Two bars in a histogram represent equal frequencies. The first class has width

6

and frequency density

4

. The second class has width

9

. Find its frequency density.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$\dfrac{8}{3}$	2
Notes
Equal frequencies give equal bar areas. The first bar has area $6(4)=24$ , so the second density is $24/9=8/3$ .

(2 marks)

Tier 2 · Standard

On a cumulative frequency diagram for

80

observations, the graph segment from

(30,22)

(42,58)

is a straight line. Estimate the median and the

65

th percentile.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	Median $=36$ $65$ th percentile $=40$	4
Notes
The median has cumulative frequency $40$ . This is $18/36=1/2$ of the way from $22$ to $58$ , so its value is $30+(1/2)(42-30)=36$ . The $65$ th percentile has cumulative frequency $0.65(80)=52$ , which is $30/36=5/6$ of the way along the segment, giving $30+(5/6)(12)=40$ .

(4 marks)

Tier 3 · Hard

A histogram of

60

observations has frequency densities

3.2

2.0

and

k

in the classes

0\leq x<5

5\leq x<12

and

12\leq x<20

respectively. Find

k

. Hence find the probability that a randomly selected observation from the data is at least

10

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$k=3.75$ $P(X\geq10)=\dfrac{17}{30}$	6
Notes
The first two frequencies are $5(3.2)=16$ and $7(2.0)=14$ . The last class therefore has frequency $60-16-14=30$ , so $k=30/8=3.75$ . From $10$ to $12$ , the histogram area gives frequency $2(2.0)=4$ ; all $30$ observations in the final class also qualify. Hence the probability is $(4+30)/60=17/30$ .

(6 marks)

Tier 2 · Standard

The vertical scale is missing from a histogram. The class

0\leq x<5

has frequency

30

and a bar height of

8

grid units. The class

5\leq x<11

has a bar height of

4

grid units. Find the frequency density represented by one grid unit and the frequency in the second class.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	One grid unit represents frequency density $0.75$ . The frequency in $5\leq x<11$ is $18$ .	4
Notes
The first class has width $5$ , so its frequency density is $30/5=6$ . A height of $8$ grid units therefore represents density $6$ , giving $6/8=0.75$ density units per grid unit. The second bar has density $4(0.75)=3$ and width $6$ , so its area and frequency are $6(3)=18$ .

(4 marks)

Tier 3 · Hard

A histogram has frequency densities

2

5

and

3

in the classes

0\leq x<5

5\leq x<9

and

9\leq x<15

respectively. Estimate the median and the interquartile range, assuming observations are distributed uniformly within each class.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	Estimated median $=7.8$ Estimated lower quartile $=5.4$ and upper quartile $=11$ Estimated interquartile range $=5.6$	6
Notes
The class frequencies are $5(2)=10$ , $4(5)=20$ and $6(3)=18$ , giving total frequency $48$ . The median is the $24$ th value, so interpolation in the second class gives $5+\frac{24-10}{20}(4)=7.8$ . The lower quartile is the $12$ th value, giving $5+\frac{12-10}{20}(4)=5.4$ . The upper quartile is the $36$ th value, in the third class, giving $9+\frac{36-30}{18}(6)=11$ . Hence the estimated IQR is $11-5.4=5.6$ .

(6 marks)

Tier 3 · Hard

Two histograms use the same class boundaries and both vertical axes show frequency density. In the class

10\leq x<15

, sample A has

80

observations in total and frequency density

4

, while sample B has

120

observations in total and frequency density

5

. Calculate the frequency and the percentage of each sample in this class. Hence explain why comparing the two bar heights alone gives the wrong conclusion about which sample has the larger proportion in the class.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	Sample A has frequency $5(4)=20$ , which is $20/80=25\%$ of sample A. Sample B has frequency $5(5)=25$ , which is $25/120=20.8\%$ of sample B to $3$ significant figures. The bar for B is taller and contains more observations, but A has the larger proportion in the class. Frequency-density heights from samples of different total sizes do not directly compare relative frequencies; percentage-frequency density or calculated proportions are needed.	5
Notes
Both class widths are $5$ , so the bar areas give frequencies $20$ and $25$ . Divide each frequency by its own sample total: $20/80=0.25$ and $25/120=0.20833\ldots$ . The different denominators reverse the comparison made from raw bar heights.

(5 marks)

Q10

Tier 3 · Hard

10.

A histogram of

70

observations has frequency densities

3

5

2

and

4

in the classes

0\leq x<4

4\leq x<10

10\leq x<a

and

a\leq x<20

, respectively, where

10<a<20

. Find

a

. Assuming observations are distributed uniformly within each class, estimate the median and find the probability that a randomly selected observation is at least

14

(7)

(Total for Question 10 is 7 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$a=16$ Estimated median $=8.6$ $P(X\geq14)=\dfrac27$	7
Notes
The total frequency is $4(3)+6(5)+(a-10)(2)+(20-a)(4)=102-2a$ . Equating this to $70$ gives $a=16$ . The first two class frequencies are $12$ and $30$ , so the $35$ th value lies in $4\leq x<10$ and is estimated by $4+\frac{35-12}{30}(6)=8.6$ . From $14$ to $16$ the estimated frequency is $2(2)=4$ , and from $16$ to $20$ it is $4(4)=16$ . Hence $P(X\geq14)=(4+16)/70=2/7$ .

(7 marks)

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