A grouped frequency table records heights of 50 students in cm: 140≤h<150: 5, 150≤h<160: 12, 160≤h<170: 18, 170≤h<180: 10, 180≤h<190: 5. Calculate the frequency density for the class 150≤h<160.
(Total for Question 1 is 1 mark)
2
From the same grouped table as Question 1, calculate the frequency density for the class 160≤h<170.
(Total for Question 2 is 1 mark)
3
A histogram will be drawn for the distribution: class 0≤t<5 freq 8, 5≤t<15 freq 20, 15≤t<20 freq 12. Calculate the frequency density for 5≤t<15 (note class width is not 10 for every class).
(Total for Question 3 is 2 marks)
4
The histogram below (data described) has a bar 0≤z<4 with height 1.5, a bar 4≤z<7 with height 2.4, and a bar 7≤z<10 with height 1.2. Use these to complete the frequency column of the table: classes 0-4, 4-7, 7-10. Give each frequency as an integer.
(Total for Question 4 is 3 marks)
5
A student draws a histogram for daily rainfall (mm) with unequal widths. The class 0≤r<2 has area 4 on the histogram. The class 2≤r<5 has frequency 9 and width 3. Calculate the frequency density for 2≤r<5 and then the height of the bar. Show your working.
(Total for Question 5 is 3 marks)
6
A histogram shows the following bars (frequency densities shown): class 10≤x<20 height 0.5, class 20≤x<30 height 1.0, class 30≤x<50 height 0.6. Calculate the frequency in the class 30≤x<50 by using area of the bar.
(Total for Question 6 is 3 marks)
7
A histogram representing age groups has classes 0≤a<10, 10≤a<30, 30≤a<60. The heights are 0.8, 0.6 and 0.4 respectively. Estimate the total population represented by the histogram and explain briefly why heights decrease as class width increases in some histograms.
(Total for Question 7 is 4 marks)
8
Complete the following: a histogram for exam marks has bars for classes 0≤m<10, 10≤m<30, 30≤m<50. The class 10≤m<30 has frequency 40 and width 20. The histogram shows the height of this bar as 2.0. Using this, calculate the total number of students represented by the three classes if the other bars have heights 1.5 (width 10) and 1.8 (width 20) respectively. Show working for each class area.
(Total for Question 8 is 4 marks)
9
A histogram bar for the interval 50≤y<70 has height 0.9. Estimate the frequency in the sub-interval 60≤y<70 (the upper half of that class) assuming uniform distribution within the class. State your method.
(Total for Question 9 is 3 marks)
10
A histogram is given for weights of parcels with unequal class widths. The class 0≤w<2 has frequency 8, 2≤w<5 has unknown frequency f, and 5≤w<10 has frequency 30. The histogram shows heights: 0≤w<2 height 4, 2≤w<5 height 3, 5≤w<10 height 6. Complete the frequency f using areas and show steps.
(Total for Question 10 is 4 marks)
11
A histogram shows a bar for 100≤x<140 with height 0.75. A researcher asks for the estimated number in 110≤x<120. Explain how to estimate this number and calculate it, assuming uniform distribution within the class.
(Total for Question 11 is 4 marks)
Mark scheme · 4.1 Histograms and Frequency Density
Question 1
B1 frequency density = frequency / class width = 12 / 10 = 1.2 cao
Answer: 1.2
Question 2
B1 frequency density = 18 / 10 = 1.8 cao
Answer: 1.8
Question 3
M1 identifies correct class width = 10 and divides 20 by 10
M1 gives a correct reason linking height to frequency density and class width, e.g. density = frequency/width so wider classes often have lower heights for same frequency, or to keep area equal to frequency
A1 total = 32 and sensible explanation about density and width, cao
Answer: Total = 32, explanation: heights are frequency densities; density = frequency/width so if class width increases for similar frequencies the height falls to keep area equal to frequency
Question 8
M1 calculates frequency for first class from area: 1.5 * 10 = 15
M1 uses given for middle class: 2.0 * 20 = 40
M1 calculates frequency for third class: 1.8 * 20 = 36
A1 total = 15 + 40 + 36 = 91 cao
Answer: frequencies: 15, 40, 36. Total = 91
Question 9
M1 uses area method and recognises width of full class = 20 and sub-interval width = 10
M1 calculates full class frequency = 0.9 * 20 = 18, then takes half for 10 width = 9
A1 estimate = 9 cao
Answer: 9
Question 10
M1 uses area = height * width to find frequencies for known bars: first and third
M1 calculates first area = 4 * 2 = 8 and third gives 6 * 5 = 30
M1 sets f = height * width for middle class: f = 3 * (5 - 2) = 3 * 3
A1 f = 9 cao
Answer: f = 9
Question 11
M1 states method: find full class frequency by area then take proportion corresponding to sub-interval width
M1 calculates full class width = 40 and full frequency = 0.75 * 40 = 30
M1 calculates sub-interval width = 10 and proportional frequency = 30 * (10/40) = 7.5
A1 gives estimate 7.5 and comment that this may be rounded or reported as 7 or 8, cao
Answer: Estimate = 7.5 (about 8 people if rounded)