Answer on paper first, then reveal the mark scheme one mark at a time.
1Grade 6–7HigherCalculator4 marks
The table shows information about the times, $t$ minutes, that 85 customers spent in a café.
Time ($t$ minutes)
Frequency
Frequency density
$0 \lt t \le 10$
15
$10 \lt t \le 15$
20
$15 \lt t \le 20$
24
$20 \lt t \le 30$
18
$30 \lt t \le 50$
8
(a) Complete the frequency density column. [2]
(b) Work out an estimate for the number of customers who spent more than 25 minutes in the café. [2]
(a) frequency ÷ class width for at least one class, e.g. $15 \div 10$M1
(a) 1.5, 4, 4.8, 1.8, 0.4A1
(b) $\frac{1}{2} \times 18$ (or $1.8 \times 5$) for the part of the $20 \lt t \le 30$ class above 25M1
(b) 17A1
Answer: (a) 1.5, 4, 4.8, 1.8, 0.4 (b) 17
Examiner tip: Class widths are not all the same – 10, 5, 5, 10, 20 – so divide each frequency by its own width.
2Grade 6–7HigherCalculator6 marks
The histogram shows information about the heights, $h$ cm, of some students.
(a) How many students have a height between 150 cm and 160 cm? [2]
(b) Work out the total number of students. [2]
(c) Work out an estimate for the number of students taller than 165 cm. [2]
(a) $4 \times 5$ or $5.6 \times 5$ (frequency = frequency density × class width)M1
(a) 48A1
(b) at least three of 12, 20, 28, 24, 16 foundM1
(b) 100A1
(c) $2.4 \times 5$ (= 12) for $165 \lt h \le 170$, or $\frac{1}{2} \times 24$M1
(c) 28A1
Answer: (a) 48 (b) 100 (c) 28
Examiner tip: In a histogram the AREA of a bar is the frequency – multiply the height by the width; do not just read the height.
3Grade 8–9HigherCalculator5 marks
The histogram shows information about the times, $t$ minutes, that some people waited for a train. The frequency density scale is not shown.
30 people waited between 15 and 20 minutes.
(a) Work out the total number of people. [3]
(b) Work out an estimate for the number of people who waited more than 25 minutes. [2]
(a) $30 \div 5 = 6$, so the $15 \lt t \le 20$ bar has frequency density 6 (each small square is 0.2 in height), or 150 small squares represent 30 people so 5 small squares represent 1 person (oe scale found)M1
(a) a correct method for the frequency of at least one other bar (16, 22, 30 or 10)M1
(a) 108A1
(b) $\frac{1}{2} \times 30$ (= 15) for $25 \lt t \le 30$M1
(b) 25A1
Answer: (a) 108 (b) 25
Examiner tip: When the scale is missing, use the given bar to work out what one small square (or one unit of height) is worth.
4Grade 8–9HigherCalculator5 marks
The histogram shows information about the masses, $m$ kg, of some parcels.
(a) Show that there are 100 parcels. [2]
(b) Work out an estimate for the median mass of the parcels. [3]
(a) frequency density × class width for at least three classes (10, 22, 30, 24, 14)M1
(a) all five frequencies with total 100 shownA1
(b) identifies the median as the 50th (or 50.5th) value, in the $30 \lt m \le 40$ class (cumulative 32 before it)M1
(b) $30 + \frac{18}{30} \times 10$ or $18 \div 3 = 6$ kg into the classM1
Examiner tip: Count how many values come before the median class, then find how far into the class you need to go: (values needed) ÷ (frequency density).
5Grade 6–7HigherCalculator4 marks
The table and the histogram show some information about the distances, $d$ km, that some people drove one day.
Distance ($d$ km)
Frequency
$0 \lt d \le 10$
20
$10 \lt d \le 20$
$20 \lt d \le 25$
24
$25 \lt d \le 40$
$40 \lt d \le 60$
10
(a) Use the histogram to complete the table. [2]
(b) Use the table to work out the heights of the two missing bars on the histogram. [2]
(a) 36 for $10 \lt d \le 20$B1
(a) 18 for $25 \lt d \le 40$B1
(b) bar for $20 \lt d \le 25$ at frequency density 4.8B1
(b) bar for $40 \lt d \le 60$ at frequency density 0.5B1
Answer: (a) 36 and 18 (b) heights 4.8 and 0.5
Examiner tip: Check one complete row first: 20 ÷ 10 = 2 matches the first bar, which confirms frequency = frequency density × width.
6Grade 6–7HigherNon-calculator2 marks
The histogram shows information about the ages, $a$ years, of the people at a party.
Sam says, “The same number of people are in the $0 \lt a \le 10$ class as in the $10 \lt a \le 40$ class, because the bars are the same height.”
No, with both frequencies 15 and 45 (oe, e.g. the second bar is 3 times as wide so has 3 times as many people)C1
Answer: No – there are 15 people aged 0–10 but 45 people aged 10–40.
Examiner tip: Bars of equal height only have equal frequencies if they are also equal in width.
7Grade 8–9HigherCalculator5 marks
The histogram shows information about the speeds, $s$ mph, of some cars on a road.
Work out an estimate for the mean speed of the cars. [5]
frequency density × class width for at least three classesM1
all frequencies correct: 12, 26, 32, 22, 8 (total 100)A1
products of frequency and midpoint (25, 32.5, 37.5, 45, 60), at least three correctM1
(dep) $\Sigma fx \div \Sigma f$, e.g. $3815 \div 100$M1
38.15 (accept 38.1 or 38.2)A1
Answer: 38.15 mph
Examiner tip: The two hidden steps: first turn each bar into a frequency (area), then use midpoints exactly as for a grouped frequency table.
8Grade 4–5HigherNon-calculator4 marks
The table gives some information about the lengths, $l$ cm, of some fish caught in a lake.
Length ($l$ cm)
Frequency
Frequency density
$20 \lt l \le 25$
14
$25 \lt l \le 40$
1.6
$40 \lt l \le a$
9
0.45
(a) Work out the frequency density for the class $20 \lt l \le 25$. [1]
(b) Work out the frequency for the class $25 \lt l \le 40$. [1]
(c) Work out the value of $a$. [2]
(a) 2.8 (from $14 \div 5$)B1
(b) 24 (from $1.6 \times 15$; do not accept 64 from $1.6 \times 40$)B1
(c) $9 \div 0.45$ (= 20) seen, or class width = frequency ÷ frequency density usedM1
(c) 60 (do not accept 20 alone as the final answer)A1
Answer: (a) 2.8 (b) 24 (c) $a = 60$
Examiner tip: In (c), $9 \div 0.45 = 20$ is the class WIDTH, not the upper bound – add it on to 40.
9Grade 6–7HigherCalculator3 marks
The table shows information about the times, $t$ seconds, that 70 people took to solve a puzzle.
Time ($t$ seconds)
Frequency
$0 \lt t \le 20$
16
$20 \lt t \le 30$
18
$30 \lt t \le 35$
13
$35 \lt t \le 50$
15
$50 \lt t \le 90$
8
On the grid, draw a histogram for this information. [3]
frequency density (frequency ÷ class width) worked out for at least 2 classes, e.g. $16 \div 20 = 0.8$, or 2 bars of the correct width and heightB1
3 or 4 bars of the correct width and heightB1
fully correct histogram: bars at heights 0.8, 1.8, 2.6, 1, 0.2 over 0–20, 20–30, 30–35, 35–50, 50–90 with no gaps (ignore shading)B1
Answer: Bars at frequency densities 0.8 (0–20), 1.8 (20–30), 2.6 (30–35), 1 (35–50), 0.2 (50–90)
Examiner tip: Plot frequency density, not frequency – a bar 18 squares high for 20–30 seconds would score nothing.
10Grade 6–7HigherNon-calculator3 marks
The table shows information about the journey times, $t$ minutes, of 65 workers.
Journey time ($t$ minutes)
Frequency
$0 \lt t \le 10$
12
$10 \lt t \le 15$
14
$15 \lt t \le 20$
11
$20 \lt t \le 40$
22
$40 \lt t \le 60$
6
Ellie is going to draw a histogram for this information.
She says, “The tallest bar will be for the class $20 \lt t \le 40$ because it has the highest frequency.”
(a) Is Ellie correct? You must show how you get your answer. [2]
(b) Ellie's friend draws a diagram for the same data using frequency on the vertical axis and the same class widths. Explain why this diagram would be misleading. [1]
(a) frequency density worked out for $20 \lt t \le 40$ ($22 \div 20 = 1.1$) and at least one other class, e.g. $14 \div 5 = 2.8$M1
(a) No, with the tallest bar for $10 \lt t \le 15$ (frequency density 2.8, greater than 1.1); accept 'No' with correct comparison of frequency densities that shows another bar is tallerC1
(b) the classes are different widths so the area of a bar would not be proportional to the frequency, e.g. 'the 20–40 bar would be 4 times as wide as the 10–15 bar so it would look as if there were far more people' (oe); do not accept 'it should be frequency density' with no reasonB1
Answer: (a) No – the tallest bar is $10 \lt t \le 15$ (FD 2.8; the $20 \lt t \le 40$ bar has FD 1.1) (b) wide classes would look too big because area would not represent frequency
Examiner tip: With unequal class widths, the bar height depends on frequency ÷ width, so a big frequency spread over a wide class can give a short bar.
11Grade 8–9HigherCalculator4 marks
The histogram shows information about the lifetimes, $h$ hours, of some batteries.
One of these batteries is chosen at random.
Work out an estimate for the probability that its lifetime is between 12 hours and 25 hours. [4]
a correct method for the part of a class inside the range, e.g. $3 \times 3$ (= 9) for $12 \lt h \le 15$ or $\frac{1}{2} \times 28$ (= 14) for $20 \lt h \le 25$M1
total number of batteries found: $6 + 15 + 23 + 28 + 8 = 80$ (at least 4 frequencies correct with a total attempted)M1
$\dfrac{9 + 23 + 14}{80}$ or $\dfrac{46}{\text{their total}}$M1
Examiner tip: Only the part of each end class inside the range counts – use frequency density × the width of that part, not the whole bar.
12Grade 6–7HigherNon-calculator4 marks
The incomplete histogram shows information about the heights, $h$ cm, of some seedlings.
The bar for the class $20 \lt h \le 30$ is missing.
There are twice as many seedlings in the class $20 \lt h \le 30$ as in the class $30 \lt h \le 50$.
(a) Work out the height of the missing bar. [3]
(b) What fraction of the seedlings are taller than 30 cm? [1]
(a) $0.8 \times 20$ (= 16) for $30 \lt h \le 50$M1
(a) $2 \times 16$ (= 32) then $32 \div 10$, or $2 \times 16 \div 10$M1
(a) 3.2 (accept a correctly drawn bar at frequency density 3.2 from 20 to 30)A1
(b) $\dfrac{16}{106}$ oe, e.g. $\dfrac{8}{53}$ (ft $\frac{16}{\text{their total}}$ where their total uses their 32 and 12 + 20 + 26)B1
Answer: (a) frequency density 3.2 (b) $\dfrac{16}{106} = \dfrac{8}{53}$
Examiner tip: “Twice as many seedlings” means twice the frequency (area), not twice the bar height – the 20–30 bar is half as wide, so it is NOT 2 × 0.8.
13Grade 6–7HigherCalculator4 marks
The histogram shows information about the daily screen time, $t$ hours, of 120 adults.
The table shows information about the daily screen time, $t$ hours, of 80 teenagers.
Screen time ($t$ hours)
Frequency
$0 \lt t \le 2$
28
$2 \lt t \le 4$
30
$4 \lt t \le 8$
22
Kai says, “A greater proportion of the adults than the teenagers had more than 2 hours of screen time.”
Is Kai correct? You must show your working. [4]
frequency density × class width for at least one adult class, e.g. $21 \times 2$ (= 42)M1
72 adults with more than 2 hours (42 + 18 + 12), or 48 with 2 hours or lessA1
both proportions in a form that can be compared, e.g. $\frac{72}{120}$ and $\frac{52}{80}$, or 60% and 65%, or 0.6 and 0.65M1
No, with 60% (adults) and 65% (teenagers) oe correctly comparedC1
Answer: No – 60% of adults ($\frac{72}{120}$) but 65% of teenagers ($\frac{52}{80}$) had more than 2 hours.
Examiner tip: The groups are different sizes, so compare proportions (percentages), not the raw numbers 72 and 52.
14Grade 8–9HigherCalculator5 marks
The histogram shows information about the lengths, $m$ minutes, of 120 phone calls.
(a) Work out an estimate for the interquartile range of the lengths of the calls. [4]
(b) Explain why your answer to part (a) can only be an estimate. [1]
(a) frequencies from frequency density × class width, at least 3 correct from 10, 24, 32, 30, 24M1
(a) lower quartile identified as the 30th (or 30.25th) value, in $2 \lt m \le 5$, or upper quartile as the 90th (or 90.75th), in $10 \lt m \le 15$M1
(a) a correct method to estimate both quartiles, e.g. $2 + \frac{20}{8}$ (= 4.5) and $10 + \frac{24}{6}$ (= 14)M1
(a) 9.5 (accept 9.5 to 9.6)A1
(b) we do not know the exact lengths within each class (the original data is grouped, so values are assumed to be spread evenly across a class) oeB1
Answer: (a) $14 - 4.5 = 9.5$ minutes (b) the exact values in each class are not known
Examiner tip: To find a quartile inside a bar, divide the number of calls you still need by the frequency density to get the distance into the class.
15Grade 4–5HigherNon-calculator3 marks
The table shows information about the masses, $m$ grams, of 50 apples.
Mass ($m$ grams)
Frequency
$100 \lt m \le 120$
8
$120 \lt m \le 130$
14
$130 \lt m \le 140$
16
$140 \lt m \le 170$
12
Ravi drew this diagram to show the information as a histogram.
(a) Write down two mistakes Ravi has made. [2]
(b) What should the height of the bar for $140 \lt m \le 170$ be? [1]
(a) the heights of the bars show frequency instead of frequency density (oe, e.g. 'the vertical axis should be frequency density')B1
(a) the bar for $140 \lt m \le 170$ stops at 160 instead of 170 (oe, e.g. 'the last bar is too narrow'); ignore any other incorrect commentsB1
(b) 0.4 (from $12 \div 30$)B1
Answer: (a) frequency used instead of frequency density; last bar drawn from 140 to 160 instead of 140 to 170 (b) 0.4
Examiner tip: Check the axis label and every bar's width against the table – a histogram needs frequency density AND the true class boundaries.
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