Data presentation Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Data presentation questions. See exactly how to solve problems on histogram, frequency density, frequency from area, box plot.

histogramfrequency densityfrequency from areabox plotIQRrange
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
In a histogram of the mass, in grams, the class 10≤m<2010\le m<20 contains 1515 items. Find the frequency density of this class.

Worked solution

  1. Recall the frequency density formula

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

    On a histogram the bar height is the frequency density, not the frequency.

  2. Substitute the frequency and the class width

    =1520−10=1510= \frac{15}{20 - 10} = \frac{15}{10}

    The class width is the difference of the upper and lower boundaries.

  3. Evaluate the frequency density

    =1.5= 1.5

    This is the height of the bar for this class.

Answer
frequency density=1.5\text{frequency density} = 1.5
Question 2
2 markseasy
In a histogram of the time, in minutes, the class 0≤t<100\le t<10 contains 2424 items. Find the frequency density of this class.

Worked solution

  1. Recall the frequency density formula

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

    On a histogram the bar height is the frequency density, not the frequency.

  2. Substitute the frequency and the class width

    =2410−0=2410= \frac{24}{10 - 0} = \frac{24}{10}

    The class width is the difference of the upper and lower boundaries.

  3. Evaluate the frequency density

    =2.4= 2.4

    This is the height of the bar for this class.

Answer
frequency density=2.4\text{frequency density} = 2.4
Question 3
2 markseasy
In a histogram of the height, in cm, the class 20≤h<4020\le h<40 contains 1818 items. Find the frequency density of this class.

Worked solution

  1. Recall the frequency density formula

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

    On a histogram the bar height is the frequency density, not the frequency.

  2. Substitute the frequency and the class width

    =1840−20=1820= \frac{18}{40 - 20} = \frac{18}{20}

    The class width is the difference of the upper and lower boundaries.

  3. Evaluate the frequency density

    =0.9= 0.9

    This is the height of the bar for this class.

Answer
frequency density=0.9\text{frequency density} = 0.9
Question 4
2 markseasy
In a histogram of the length, in mm, the class 5≤x<105\le x<10 contains 2020 items. Find the frequency density of this class.

Worked solution

  1. Recall the frequency density formula

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

    On a histogram the bar height is the frequency density, not the frequency.

  2. Substitute the frequency and the class width

    =2010−5=205= \frac{20}{10 - 5} = \frac{20}{5}

    The class width is the difference of the upper and lower boundaries.

  3. Evaluate the frequency density

    =4= 4

    This is the height of the bar for this class.

Answer
frequency density=4\text{frequency density} = 4
Question 5
2 markseasy
In a histogram of the volume, in ml, the class 30≤v<5030\le v<50 contains 1616 items. Find the frequency density of this class.

Worked solution

  1. Recall the frequency density formula

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

    On a histogram the bar height is the frequency density, not the frequency.

  2. Substitute the frequency and the class width

    =1650−30=1620= \frac{16}{50 - 30} = \frac{16}{20}

    The class width is the difference of the upper and lower boundaries.

  3. Evaluate the frequency density

    =0.8= 0.8

    This is the height of the bar for this class.

Answer
frequency density=0.8\text{frequency density} = 0.8

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