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7.16 Histograms

EDEXCEL 1MA1 · Calculator allowed · about 75 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A frequency table shows the delivery times, t minutes, for 100 parcels sent by a courier company. Some values have already been filled in.

Time, t (minutes)	Class width	Frequency	Frequency density
0 < t ≤ 10	10	20	2.0
10 < t ≤ 15	5	...	4.4
15 < t ≤ 25	10	45	...
25 < t ≤ 40	15	...	0.6
40 < t ≤ 60	20	4	...

(a)Complete the table above.(4)

(b)The class widths in the table are not all equal. Explain why frequency density, rather than frequency, should be used for the vertical axis when drawing a histogram for this data.(1)

(Total for Question 1 is 5 marks)

Which of the following is the correct formula for the frequency density of a class in a histogram?

A) frequency density = frequency x class width
B) frequency density = frequency / class width
C) frequency density = class width / frequency
D) frequency density = frequency + class width

(Total for Question 2 is 1 mark)

Using your completed table from Question 1, draw a histogram to represent the delivery times of the 100 parcels. Use the grid described below.

(Total for Question 3 is 3 marks)

The histogram below shows the masses, in kg, of 100 parcels handled by a warehouse in one day.

(a)Write down the frequency density of the class 10 < m ≤ 20 kg.(1)

A) 0.3
B) 3
C) 30
D) 300

(b)Work out the number of parcels with mass 5 < m ≤ 10 kg.(2)

(Total for Question 4 is 3 marks)

Using the histogram of parcel masses shown again below (100 parcels in total), work out the probability that a parcel chosen at random has a mass greater than 20 kg.

(Total for Question 5 is 2 marks)

A histogram is used to show the heights, in cm, of 80 sunflowers grown in a trial plot. Three of the four bars have been drawn, but the bar for the class 130 ≤ h < 150 is missing.

(a)Given that there are 80 sunflowers in total, work out the frequency for the class 130 ≤ h < 150.(3)

(b)Hence state the height of the missing bar (its frequency density).(1)

(Total for Question 6 is 4 marks)

Using the completed sunflower height data (100 ≤ h < 120: frequency 12; 120 ≤ h < 130: frequency 24; 130 ≤ h < 150: frequency 32; 150 ≤ h < 180: frequency 12; 80 sunflowers in total), estimate the median height of the sunflowers.

(Total for Question 7 is 3 marks)

Using the same completed sunflower height data as Question 7, estimate the interquartile range of the sunflower heights.

(Total for Question 8 is 5 marks)

A histogram shows the ages of visitors to a museum. The class 20 ≤ a < 30 has frequency density 3.2, and the class 30 ≤ a < 50 has frequency density 1.5.

James says: 'More visitors were aged 30-50 than 20-30, because that class covers a bigger range of ages.'

(Total for Question 9 is 3 marks)

The frequency table shows the speeds, in mph, of 90 cars recorded on a road.

Speed, v (mph)	Class width	Frequency
0 < v ≤ 20	20	10
20 < v ≤ 30	10	24
30 < v ≤ 40	10	36
40 < v ≤ 50	10	15
50 < v ≤ 70	20	5

(Total for Question 10 is 4 marks)

Using the car speed data from Question 10 (90 cars in total), work out the probability that a car chosen at random was travelling faster than 30 mph.

(Total for Question 11 is 2 marks)

Using the car speed data from Question 10 (90 cars in total), estimate the mean speed of the cars.

(Total for Question 12 is 3 marks)

Using the car speed data from Question 10, estimate the number of cars travelling faster than 35 mph. Assume speeds are evenly distributed within each class.

(Total for Question 13 is 3 marks)

A histogram has two adjacent classes. The class 10 < x ≤ 25 has frequency density 4. The next class, 25 < x ≤ (25 + w), has frequency density 6 and a frequency three times that of the first class.

(Total for Question 14 is 3 marks)

Two groups of 50 runners, Group P and Group Q, complete a fun run. Group P has a median completion time of 27.5 minutes and an interquartile range of 9 minutes. The frequency table for Group Q's times, t minutes, is shown below.

Time, t (minutes)	Class width	Frequency
20 ≤ t < 25	5	5
25 ≤ t < 30	5	20
30 ≤ t < 40	10	20
40 ≤ t < 50	10	5

(a)Show that the median completion time for Group Q is 30 minutes.(3)

(b)Group Q has an interquartile range of 10 minutes. Compare the completion times of Group P and Group Q, using the median and the interquartile range.(2)

(Total for Question 15 is 5 marks)

Using the frequency table for Group Q's completion times from Question 15 (50 runners), two runners are selected at random, without replacement, from those who took more than 30 minutes. Find the probability that both selected runners took between 30 and 40 minutes.

(Total for Question 16 is 3 marks)

A histogram represents the finishing times, in minutes, of 200 athletes in a marathon, grouped into three classes: 180 ≤ t < 190 (width 10), 190 ≤ t < 210 (width 20), and 210 ≤ t < 240 (width 30). The heights of the three bars are in the ratio 3 : 2 : 1 respectively.

(Total for Question 17 is 3 marks)

The incomplete histogram and table give information about the time, h hours, spent on homework in one week by 120 students. The bar for the class 5 ≤ h < 10 has not been drawn.

Time, h (hours)	Class width	Frequency density
0 ≤ h < 5	5	3
5 ≤ h < 10	5	(bar not drawn)
10 ≤ h < 20	10	5
20 ≤ h < 30	10	2
30 ≤ h < 50	20	0.5

(a)Find the frequency for the class 5 ≤ h < 10, and state the frequency density that should be used for its missing bar.(3)

(b)Estimate the median amount of time spent on homework.(3)

(c)A student who spent at least 20 hours on homework is chosen at random. Find the probability that this student spent at least 30 hours on homework.(2)

(Total for Question 18 is 8 marks)

Mark scheme · 7.16 Histograms

Question 1

(a) B1 frequency for 10 < t ≤ 15 is 22 cao
(a) B1 frequency density for 15 < t ≤ 25 is 4.5 cao
(a) B1 frequency for 25 < t ≤ 40 is 9 cao
(a) B1 frequency density for 40 < t ≤ 60 is 0.2 cao
(a) Answer: 22, 4.5, 9, 0.2
(b) B1 correct explanation, e.g. because the class widths are unequal, plotting frequency directly would make wider classes look like they contain more data than they really do; frequency density (= frequency / class width) means the area of each bar represents the frequency fairly oe
(b) Answer: Because the class widths are unequal, frequency density must be used so that the area of each bar (not just its height) is proportional to the frequency it represents.

Question 2

B1 B cao
Answer: B

Question 3

B1 ft: correct axes drawn and labelled with a suitable linear scale
B1 ft: at least 3 bars drawn with correct frequency density heights from Question 1(a)
B1 ft: all 5 bars drawn correctly with heights 2.0, 4.4, 4.5, 0.6, 0.2 and no gaps between adjacent bars
Answer: Bars of height (frequency density) 2.0, 4.4, 4.5, 0.6 and 0.2 drawn over the classes 0-10, 10-15, 15-25, 25-40 and 40-60 minutes respectively.

Question 4

(a) B1 B (3) cao
(a) Answer: B (3)
(b) M1 8 x 5 oe
(b) A1 40 cao
(b) Answer: 40 parcels

Question 5

M1 frequency for 20 < m ≤ 30 = 2 x 10 = 20 oe
A1 20/100 = 0.2 (or 1/5) oe
Answer: 0.2 (or 1/5)

Question 6

(a) M1 sum of known frequencies: (0.6 x 20) + (2.4 x 10) + (0.4 x 30) = 12 + 24 + 12 = 48
(a) M1 80 - 48 oe
(a) A1 32 cao
(a) Answer: 32 sunflowers
(b) B1 ft: 32 / 20 = 1.6 cao
(b) Answer: 1.6

Question 7

M1 cumulative frequencies 12, 36, 68, 80 and identifying the median class as 130 ≤ h < 150 (since n/2 = 40 lies between 36 and 68)
M1 correct interpolation: 130 + ((40 - 36) / 32) x 20 oe
A1 awrt 132.5 cm (accept 132-133)
Answer: awrt 132.5 cm

Question 8

M1 identifying lower quartile position n/4 = 20, lying in class 120 ≤ h < 130 (cumulative 12 to 36)
A1 lower quartile = 120 + ((20 - 12)/24) x 10 = awrt 123.3 cm (accept 122-125)
M1 identifying upper quartile position 3n/4 = 60, lying in class 130 ≤ h < 150 (cumulative 36 to 68)
A1 upper quartile = 130 + ((60 - 36)/32) x 20 = 145 cao
A1 ft: IQR = 145 - 123.3 = awrt 21.7 cm (accept 20-23)
Answer: awrt 21.7 cm

Question 9

M1 frequency for 20 ≤ a < 30 = 3.2 x 10 = 32
M1 frequency for 30 ≤ a < 50 = 1.5 x 20 = 30
B1 correct conclusion with comparison shown: since 32 > 30, James is wrong oe
Answer: James is wrong: there were 32 visitors aged 20-30 but only 30 aged 30-50, even though that class covers a bigger age range.

Question 10

B1 correct frequency densities calculated and shown: 0.5, 2.4, 3.6, 1.5, 0.25
B1 correct axes drawn and labelled with a suitable linear scale
B1 at least 3 bars drawn with correct heights
B1 all 5 bars drawn correctly with no gaps between adjacent bars
Answer: Frequency densities: 0.5, 2.4, 3.6, 1.5, 0.25 for the five classes in order.

Question 11

M1 frequency for v > 30 = 36 + 15 + 5 = 56
A1 56/90 = 28/45 = awrt 0.622 oe
Answer: 28/45 (awrt 0.622)

Question 12

M1 using midpoints 10, 25, 35, 45, 60 and finding sum of (frequency x midpoint) = 100 + 600 + 1260 + 675 + 300 = 2935
M1 dividing by the total frequency, 2935 / 90
A1 awrt 32.6 mph (accept 32.5-32.7)
Answer: awrt 32.6 mph

Question 13

M1 recognising 35 mph lies inside the class 30 < v ≤ 40, and finding the proportion of that class above 35: (40-35)/10 x 36 = 18
M1 adding the frequencies of the classes fully above 35: 15 + 5 = 20
A1 18 + 20 = 38 cars (accept awrt 38)
Answer: awrt 38 cars

Question 14

M1 frequency of first class = 15 x 4 = 60
M1 frequency of second class = 3 x 60 = 180
A1 w = 180 / 6 = 30 cao
Answer: w = 30

Question 15

(a) M1 cumulative frequencies 5, 25, 45, 50 and identifying n/2 = 25
(a) M1 correct interpolation: 25 + ((25 - 5)/20) x 5 oe
(a) A1 cso: 30 minutes
(a) Answer: 30 minutes
(b) B1 median comparison in context, e.g. Group P has the lower median (27.5 minutes compared with 30 minutes), so Group P runners were faster on average oe
(b) B1 IQR/spread comparison in context, e.g. Group P has the smaller interquartile range (9 minutes compared with 10 minutes), so Group P's times were slightly more consistent oe
(b) Answer: Group P was faster on average (lower median, 27.5 < 30 minutes) and slightly more consistent (smaller IQR, 9 < 10 minutes).

Question 16

M1 identifying 25 runners took more than 30 minutes (20 + 5), of whom 20 took between 30 and 40 minutes
M1 (20/25) x (19/24) oe
A1 19/30 (awrt 0.633) oe
Answer: 19/30 (awrt 0.633)

Question 17

M1 expressing frequencies in terms of k: 10 x 3k = 30k, 20 x 2k = 40k, 30 x k = 30k
M1 30k + 40k + 30k = 200, so 100k = 200, k = 2
A1 frequency for 190 ≤ t < 210 = 40k = 80 cao
Answer: 80 athletes

Question 18

(a) M1 sum of known frequencies: (3x5) + (5x10) + (2x10) + (0.5x20) = 15 + 50 + 20 + 10 = 95
(a) A1 120 - 95 = 25 cao
(a) B1 ft: frequency density = 25 / 5 = 5
(a) Answer: 25 students; frequency density 5
(b) M1 ft: cumulative frequencies 15, 40, 90, 110, 120 and identifying n/2 = 60 lies in the class 10 ≤ h < 20 (cumulative 40 to 90)
(b) M1 correct interpolation: 10 + ((60 - 40)/50) x 10 oe
(b) A1 14 hours cao
(b) Answer: 14 hours
(c) M1 identifying 30 students spent at least 20 hours (20 + 10) and 10 of these spent at least 30 hours
(c) A1 10/30 = 1/3 oe (accept 0.333 or 33.3%)
(c) Answer: 1/3 (awrt 0.333)

Histograms - Worksheets, Questions and Revision