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Calculator 4 marks
The table gives information about the heights of 190 plants.
Height (cm) Frequency \(0 < h \le 10\) 35 \(10 < h \le 20\) 5 \(20 < h \le 50\) 90 \(50 < h \le 70\) 60 The histogram has been started. Two of the bars are missing.
(a)Find the frequency density of the bar for \(10 < h \le 20\).
(b)Find the frequency density of the bar for \(50 < h \le 70\).
Show answer
(a) \(0.5\)
(b) \(3\)The height of a bar on a histogram is the frequency density.
\(\text{frequency density} = \frac{\text{frequency} }{\text{class width} } = \frac{ 5 }{ 10 } = 0.5\)
This class is \(20\) cm wide and holds 60 plants.
\(\text{frequency density} = \frac{ 60 }{ 20 } = 3\)
- (a) M1frequency divided by class width \(\frac{ 5 }{ 10 }\)
- (a) A1frequency density \(0.5\)
- (b) M1frequency divided by class width \(\frac{ 60 }{ 20 }\)
- (b) A1frequency density \(3\)
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Calculator 2 marks
The table shows information about the lengths of 135 journeys.
Length (km) Frequency \(0 < d \le 10\) 20 \(10 < d \le 20\) 15 \(20 < d \le 40\) 60 \(40 < d \le 60\) 40 The histogram has been started. Two of the bars are missing.
Complete the histogram.
You cannot draw on this page. Draw this on paper, then show the answer and check your drawing against the model.
Show answer
The bar for \(10 < d \le 20\) drawn to \(1.5\) and the bar for \(40 < d \le 60\) drawn to \(2\)The height of a bar on a histogram is the frequency density, not the frequency.
\(\text{frequency density} = \frac{\text{frequency} }{\text{class width} }\)
For \(10 < d \le 20\): \(\frac{ 15 }{ 10 } = 1.5\), so that bar is drawn to a height of \(1.5\).
For \(40 < d \le 60\): \(\frac{ 40 }{ 20 } = 2\), so that bar is drawn to a height of \(2\).
Each bar is as wide as its class.
- B1one of the two missing bars drawn to its frequency density
- B1both missing bars drawn to their frequency densities, \(1.5\) and \(2\)
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Calculator 4 marks
The histogram gives information about the ages of some members of a sports club.
(a)Work out the number of members of a sports club whose age was between \(40\) and \(60\) years.
(b)Work out the total number of members of a sports club.
Show answer
(a) \(40\)
(b) \(155\)On a histogram the frequency is the area of the bar, not its height.
The bar from \(40\) to \(60\) is \(20\) years wide and its height is a frequency density of \(2\).
\(\text{frequency} = \text{class width} \times \text{frequency density} = 20 \times 2 = 40\)
Work out the area of each bar in turn.
\(0\) to \(10\): \(10 \times 2.5 = 25\)
\(10\) to \(40\): \(30 \times 2 = 60\)
\(40\) to \(60\): \(20 \times 2 = 40\)
\(60\) to \(80\): \(20 \times 1.5 = 30\)
Total: \(25 + 60 + 40 + 30 = 155\)
- (a) M1class width \(20\) times frequency density \(2\)
- (a) A1frequency \(40\)
- (b) M1area of each bar: \(25\), \(60\), \(40\), \(30\)
- (b) A1total \(155\)
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Calculator 3 marks
The histogram gives information about the durations (\(c\) minutes) of some phone calls.
(a)Which class contains the most phone calls? Show working to explain your answer.
(b)Calculate how many more phone calls are in the class you found in part (a) than in the class \(80 < c \le 100\) minutes.
Show answer
(a) \(50 < c \le 80\) minutes
(b) \(20\)The frequency of a class is its area, not the height of its bar — compare areas, not heights.
\(0 < c \le 40\) minutes: \(40 \times 1.5 = 60\) phone calls
\(40 < c \le 50\) minutes: \(10 \times 0.5 = 5\) phone calls
\(50 < c \le 80\) minutes: \(30 \times 3 = 90\) phone calls
\(80 < c \le 100\) minutes: \(20 \times 3.5 = 70\) phone calls
The tallest bar is \(80 < c \le 100\) minutes, with a frequency density of \(3.5\). Its area is only \(70\) phone calls, which is not the greatest.
The greatest frequency is \(90\) phone calls, in the class \(50 < c \le 80\) minutes.
\(90 - 70 = 20\)
- (a) M1frequencies found as width times frequency density: \(60\), \(5\), \(90\), \(70\)
- (a) A1class \(50 < c \le 80\) with the greatest frequency \(90\)
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Calculator 4 marks
The incomplete histogram and the incomplete table give information about the lengths of some journeys.
Length (km) Frequency \(10 < d \le 20\) ? \(20 < d \le 26\) ? \(26 < d \le 38\) 42 \(38 < d \le 58\) 90 (a)Find the frequency for the class \(10 < d \le 20\).
(b)Find the frequency for the class \(20 < d \le 26\).
(c)The bar for the class \(26 < d \le 38\) is missing from the histogram. Draw the missing bar on the grid.
You cannot draw on this page. Draw (c) on paper, then show the answer and check your drawing against the model.
Show answer
(a) \(45\)
(b) \(21\)
(c) The bar for \(26 < d \le 38\) drawn to a height of \(3.5\)(c)On a histogram, area represents frequency.
\(\text{frequency} = \text{frequency density} \times \text{class width} = 4.5 \times 10 = 45\)
\(\text{frequency} = 3.5 \times 6 = 21\)
The height of a bar on a histogram is the frequency density, not the frequency.
\(\text{frequency density} = \frac{\text{frequency} }{\text{class width} } = \frac{ 42 }{ 12 } = 3.5\)
The bar is drawn across the class \(26\) to \(38\), to a height of \(3.5\).
- (c) B2a bar drawn for the class \(26 < d \le 38\) at height \(3.5\), the frequency \(42\) divided by the width \(12\)
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Calculator 3 marks
The table shows information about the lengths of 130 phone calls.
Length (seconds) Frequency \(50 < t \le 80\) 60 \(80 < t \le 100\) 50 \(100 < t \le 110\) 10 \(110 < t \le 130\) 10 One bar of the histogram has been drawn.
Complete the histogram.
You cannot draw on this page. Draw this on paper, then show the answer and check your drawing against the model.
Show answer
Bars of height \(2\), \(2.5\), \(1\) and \(0.5\) for the four classes in order, one of them already drawnThe height of a bar on a histogram is the frequency density, not the frequency. The classes are not all the same width, so each frequency is divided by its own class width. \(\text{frequency density} = \frac{\text{frequency} }{\text{class width} }\)
The bar already drawn has a height of \(2\), which agrees with its row, so the scale is right. Each bar is as wide as its class and is drawn to its frequency density.Length (seconds) Frequency Class width Frequency density \(50 < t \le 80\) 60 30 2 \(80 < t \le 100\) 50 20 2.5 \(100 < t \le 110\) 10 10 1 \(110 < t \le 130\) 10 20 0.5 - B1one of the three missing bars drawn to its frequency density
- B1two of the three missing bars drawn to their frequency densities
- B1all three missing bars drawn to their frequency densities, \(2\), \(2.5\), \(1\) and \(0.5\) in class order
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Calculator 4 marks
The histogram gives information about the times of some visitors.
Two of the frequencies are missing from the table.
Time in the museum (minutes) Frequency \(0 < t \le 20\) 10 \(20 < t \le 30\) \(30 < t \le 40\) 10 \(40 < t \le 60\) \(60 < t \le 100\) 60 (a)Use the histogram to work out the frequency for \(20 < t \le 30\).
(b)Find the total number of visitors.
Show answer
(a) \(10\)
(b) \(130\)Read the height of that bar: the frequency density is \(1\).
The class is \(10\) minutes wide, and the frequency is the area of the bar:
\(10 \times 1 = 10\)
The other missing frequency is for \(40 < t \le 60\):
\(20 \times 2 = 40\)
The five frequencies are 10, 10, 10, 40 and 60, so the total is
\(10 + 10 + 10 + 40 + 60 = 130\)
- (a) M1frequency is class width times density \(10 \times 1\)
- (a) A1frequency \(10\)
- (b) M1other missing frequency \(20 \times 2 = 40\)
- (b) A1total \(130\)
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Calculator 3 marks
The histogram gives information about the lengths of stay of 155 visitors to a museum.
One of the bars has not been drawn.
Calculate the frequency density of the missing bar.
Show answer
\(1.5\)The frequency of a class on a histogram is the area of its bar: class width \(\times\) frequency density.
Work out the frequency of each bar that is drawn.
\(10\) to \(20\) minutes: \(10 \times 3 = 30\)
\(20\) to \(30\) minutes: \(10 \times 2 = 20\)
\(30\) to \(60\) minutes: \(30 \times 3 = 90\)
These three bars add up to \(140\) visitors to a museum.
The total is \(155\), so the missing bar's frequency is
\(155 - 140 = 15\)
This bar covers \(0 < t \le 10\) minutes, which is \(10\) minutes wide:
\(\text{frequency density} = \frac{ 15 }{ 10 } = 1.5\)
- M1frequencies of the drawn bars, total \(140\)
- M1missing frequency \(155 - 140 = 15\) divided by width \(10\)
- A1frequency density \(1.5\)
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Calculator 3 marks
The histogram gives information about the charging times of some phones.
The number of phones with a charging time between \(30\) and \(50\) minutes is the same as the number with a charging time between \(50\) and \(60\) minutes.
The frequency density of this last class is \(x\).
Calculate the value of \(x\).
Show answer
\(3\)The two classes hold the same number of phones, so their areas (frequencies) are equal.
The class from \(30\) to \(50\) minutes is drawn, so its frequency is
\(20 \times 1.5 = 30\)
The class from \(50\) to \(60\) minutes must also have a frequency of \(30\), and its frequency density is \(x\).
So its width is \(60 - 50 = 10\) minutes, so its frequency density is \(30 \div 10 = 3\).
\(x = 3\)
- M1frequency of the drawn class \(20 \times 1.5 = 30\)
- M1equal frequencies, width of the other class \(60 - 50 = 10\)
- A1\(x = 3\)
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Calculator 3 marks
The histogram gives information about the heights of 145 trees.
Estimate the number of trees whose height was between \(20\) and \(40\) metres.
Show answer
\(45\)Neither \(20\) nor \(40\) is a class boundary, so each end takes part of a bar. Assume the heights are spread evenly across each class.
From \(20\) to \(30\) is \(10\) of the \(20\) in the class \(10\) to \(30\), and that bar has a frequency density of \(1\):
\(10 \times 1 = 10\)
From \(30\) to \(40\) is \(10\) of the \(20\) in the class \(30\) to \(50\), at a frequency density of \(3.5\):
\(10 \times 3.5 = 35\)
\(10 + 35 = 45\)
- M1part of the class \(10\) to \(30\): \(10 \times 1 = 10\)
- M1part of the class \(30\) to \(50\): \(10 \times 3.5 = 35\)
- A1estimate \(45\)