Histograms and cumulative frequency Worked Solutions — GCSE Maths

Fully worked, step-by-step solutions to GCSE Histograms and cumulative frequency questions. See exactly how to solve problems on frequency density, class width, grouped data, area of a bar.

frequency densityclass widthgrouped dataarea of a barreading a histogramcumulative frequency
GCSE Higher70 questionsStep-by-step solutions
Question 1
1 markeasy
The table shows the waiting time, in minutes, of 160 customers at a post office. 0<t≤100 < t \le 10: 1010; 10<t≤2010 < t \le 20: 3030; 20<t≤3020 < t \le 30: 4040; 30<t≤4030 < t \le 40: 4040; 40<t≤5040 < t \le 50: 2020; 50<t≤7050 < t \le 70: 2020. Work out the frequency density for the class 30<t≤4030 < t \le 40.

Worked solution

  1. Recall the formula for frequency density

    frequency density=frequencyclass width\text{frequency density} = \frac{\text{frequency}}{\text{class width}}

    In a histogram the height of a bar is the frequency density, not the frequency. It is the frequency shared out over the width of the class.

  2. Work out the class width

    class width=40−30=10\text{class width} = 40 - 30 = 10

    The class 30<t≤4030 < t \le 40 runs from 3030 to 4040, so it is 1010 minutes wide.

  3. Divide the frequency by the class width

    frequency density=4010=4\text{frequency density} = \frac{40}{10} = 4

    There are 4040 customers in a class of width 1010, so the frequency density is 44.

Answer
frequency density=4\text{frequency density} = 4
Question 2
1 markeasy
The table shows the finishing time, in minutes, of 140 runners in a fun run. 0<t≤200 < t \le 20: 3030; 20<t≤3020 < t \le 30: 2525; 30<t≤4030 < t \le 40: 5050; 40<t≤4540 < t \le 45: 3030; 45<t≤5045 < t \le 50: 22; 50<t≤6050 < t \le 60: 33. Work out the frequency density for the class 30<t≤4030 < t \le 40.

Worked solution

  1. Recall the formula for frequency density

    frequency density=frequencyclass width\text{frequency density} = \frac{\text{frequency}}{\text{class width}}

    In a histogram the height of a bar is the frequency density, not the frequency. It is the frequency shared out over the width of the class.

  2. Work out the class width

    class width=40−30=10\text{class width} = 40 - 30 = 10

    The class 30<t≤4030 < t \le 40 runs from 3030 to 4040, so it is 1010 minutes wide.

  3. Divide the frequency by the class width

    frequency density=5010=5\text{frequency density} = \frac{50}{10} = 5

    There are 5050 runners in a class of width 1010, so the frequency density is 55.

Answer
frequency density=5\text{frequency density} = 5
Question 3
1 markeasy
The table shows the mass, in grams, of 140 letters posted in one day. 0<m≤200 < m \le 20: 1010; 20<m≤4020 < m \le 40: 5050; 40<m≤5040 < m \le 50: 5050; 50<m≤6050 < m \le 60: 1010; 60<m≤8060 < m \le 80: 88; 80<m≤12080 < m \le 120: 1212. Work out the frequency density for the class 20<m≤4020 < m \le 40.

Worked solution

  1. Recall the formula for frequency density

    frequency density=frequencyclass width\text{frequency density} = \frac{\text{frequency}}{\text{class width}}

    In a histogram the height of a bar is the frequency density, not the frequency. It is the frequency shared out over the width of the class.

  2. Work out the class width

    class width=40−20=20\text{class width} = 40 - 20 = 20

    The class 20<m≤4020 < m \le 40 runs from 2020 to 4040, so it is 2020 grams wide.

  3. Divide the frequency by the class width

    frequency density=5020=2.5\text{frequency density} = \frac{50}{20} = 2.5

    There are 5050 letters in a class of width 2020, so the frequency density is 2.52.5.

Answer
frequency density=2.5\text{frequency density} = 2.5
Question 4
1 markeasy
The table shows the height, in cm, of 140 plants in a greenhouse. 0<h≤100 < h \le 10: 3535; 10<h≤1510 < h \le 15: 2020; 15<h≤2015 < h \le 20: 3030; 20<h≤2520 < h \le 25: 2525; 25<h≤3025 < h \le 30: 1515; 30<h≤4030 < h \le 40: 1515. Work out the frequency density for the class 30<h≤4030 < h \le 40.

Worked solution

  1. Recall the formula for frequency density

    frequency density=frequencyclass width\text{frequency density} = \frac{\text{frequency}}{\text{class width}}

    In a histogram the height of a bar is the frequency density, not the frequency. It is the frequency shared out over the width of the class.

  2. Work out the class width

    class width=40−30=10\text{class width} = 40 - 30 = 10

    The class 30<h≤4030 < h \le 40 runs from 3030 to 4040, so it is 1010 cm wide.

  3. Divide the frequency by the class width

    frequency density=1510=1.5\text{frequency density} = \frac{15}{10} = 1.5

    There are 1515 plants in a class of width 1010, so the frequency density is 1.51.5.

Answer
frequency density=1.5\text{frequency density} = 1.5
Question 5
2 markseasy
A histogram shows the speed, in km/h, of some cars passing a checkpoint. The bar for the class 30<v≤4030 < v \le 40 has a frequency density of 55. Work out the frequency for this class.

Worked solution

  1. Recall that the frequency is the area of the bar

    frequency=frequency density×class width=area of the bar\text{frequency} = \text{frequency density} \times \text{class width} = \text{area of the bar}

    Rearranging frequency density = frequency / class width gives frequency = frequency density x class width. That product is exactly the area of the bar.

  2. Work out the class width

    class width=40−30=10\text{class width} = 40 - 30 = 10

    The class 30<v≤4030 < v \le 40 is 1010 km/h wide.

  3. Multiply the frequency density by the class width

    frequency=5×10=50\text{frequency} = 5 \times 10 = 50

    A bar of height 55 and width 1010 has area 5050, so 5050 cars are in this class.

Answer
frequency=50\text{frequency} = 50

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