Grouped Data

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Grouped Data

Grouped data is represented in a histogram or frequency polygon. We can use histograms to estimate the mean, median and standard deviation of data sets.

Make sure you are happy with the following topics before continuing.

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Estimation from Histograms

(Note: for guidance on how to draw histograms, see Presenting Data.)

Since histograms collate data, it may seem impossible to answer questions such like how many data points are greater than 99, unless 99 is a class boundary. We can, however, estimate the answers to these questions by assuming frequency is evenly distributed across an entire class. Here is how to do it:

Example: Approximately how many values are greater than 1212 in this histogram?

Draw a line at 1212 on the xx axis. This will split the second block. Then, the area of the graph to the right of the line is our estimate. In this case, the second block now extends from 1212 to 2020, with a height of 22, so a frequency of 2×8=162times 8=16 comes from the second block. The third block has a length of 1010 and a height of 33, so gives 3030 frequency. In total, there are 16+30=4616+30=46 values larger than 1212.

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Frequency Polygon

A frequency polygon is another way to represent grouped data. It is a line graph joining the points with co-ordinates (midpoint of class, frequency).

Example:

The midpoints are 5,14,21,25,335,14,21,25,33, so we plot the points:

(5,9)(5,9)

(14,15)(14,15)

(21,17)(21,17)

(25,9)(25,9)

(33,4)(33,4)

and connect them with straight lines.

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Estimating the Mean and Standard Deviation from a Histogram

Previously, when we used frequency tables to find the mean and standard deviation, we looked at xx, fxfx and fx2fx^{2}. While we clearly still have ff, it is not obvious how we should get xx. This is where the idea of midpoints comes in again.

To estimate the mean and standard deviation from a histogram, first turn the histogram into a table, then add a column of the midpoints of each class labelled xx. Then, create columns fxfx and fx2fx^{2} and find the totals of all of the columns. Finally, use these totals in the formulas for mean and standard deviation.

Recall: The formulas:

mean=fxftext{mean}=dfrac{sum{fx}}{sum{f}}

variance=fx2fmean2text{variance}=dfrac{sum{fx^{2}}}{sum{f}}-text{mean}^{2}

standard deviation=variancetext{standard deviation}=sqrt{text{variance}}

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Estimating the Median from a Histogram

To estimate the median from a histogram we use linear interpolation. This is where we assume that within each block, the frequency is evenly spaced.

To find the median, first find fsum{f} and divide it by 22 to find the position of the median (since this is an estimate, if we obtain a decimal we can treat it as if it is a whole number position). Then, find which block the position falls into. Then, within that block, find where it lies.

For example, if the median is the 77th position of a block with 1010 values of length 55, then you would add 7×510=3.5dfrac{7times 5}{10}=3.5 to the lower bound of the block to find the median.

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Example 1: Estimating the Mean and Standard Deviation from a Histogram

Find the mean and standard deviation of the data in the histogram below.

[6 marks]

Step 1: Create a table of the data from the histogram.

Step 2: Add columns for the midpoint (xx), fxfx and fx2fx^{2}.

Step 3: Use the formulas to find the mean and standard deviation.

mean=fxf=138.531=4.47variance=fx2fmean2=759.75314.472=4.55standard deviation=variance=4.55=2.13begin{aligned}text{mean}&=dfrac{sum{fx}}{sum{f}}[1.2em]&=dfrac{138.5}{31}=4.47[1.2em]text{variance}&=dfrac{sum{fx^{2}}}{sum{f}}-text{mean}^{2}[1.2em]&=dfrac{759.75}{31}-4.47^{2}[1.2em]&=4.55[1.2em]text{standard deviation}&=sqrt{text{variance}}[1.2em]&=sqrt{4.55}[1.2em]&=2.13end{aligned}

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Example 2: Estimating the Median from a Histogram

Find the median of the data in the histogram from the previous example.

[3 marks]

There are 3131 data points, so the median is the 15.515.5th data point. We can treat the decimal like it is a whole number position for our estimate. There are 1515 data points in the first two blocks, so this falls 0.50.5 data points into the third block. Said block contains 88 data points and has a width of 11. So we are 1×0.58=0.0625dfrac{1times 0.5}{8}=0.0625, so we are 0.06250.0625 into the block. The block starts at 55, so the median is 5.06255.0625.

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Grouped Data Example Questions

Question 1: Create a histogram from the following table.

[4 marks]

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Question 2: If f=18sum{f}=18, fx=162sum{fx}=162 and fx2=2430sum{fx^{2}}=2430, what is the variance?

[2 marks]

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mean=fxf=16218=9variance=fx2fmean2=fx2f92=24301881=13581=54begin{aligned}text{mean}&=dfrac{sum{fx}}{sum{f}}[1.2em]&=dfrac{162}{18}[1.2em]&=9[1.2em]text{variance}&=dfrac{sum{fx^{2}}}{sum{f}}-text{mean}^{2}[1.2em]&=dfrac{sum{fx^{2}}}{sum{f}}-9^{2}[1.2em]&=dfrac{2430}{18}-81[1.2em]&=135-81[1.2em]&=54end{aligned}

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Question 3: Consider this histogram.

a) Estimate how many values are greater than 1515.

b) Turn the values in the histogram into a frequency table.

c) What is the mean and standard deviation of the data in the histogram?

d) What is the median of the data in the histogram?

[10 marks]

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a) A line at 1515 would split the second block. To the right of 1515 in this block is a width of 55 and a height of 2828, for a total of 5×28=1405times 28=140 frequency. The third block has a width of 1010 and a height of 1212, for a total of 10×12=12010times 12=120 frequency. Overall, the number of values greater than 1515 is 140+120=260140+120=260

 

b)

 

c) Step 1: Using the table from the second question, create a table containing totals, midpoints, fxfx and fx2fx^{2}.

Step 2: Use the formulas to find the mean and standard deviation.

mean=fxf=8650700=12.4text{mean}=dfrac{sum{fx}}{sum{f}}=dfrac{8650}{700}=12.4

variance=fx2fmean2=14162570012.42=49.6text{variance}=dfrac{sum{fx^{2}}}{sum{f}}-text{mean}^{2}=dfrac{141625}{700}-12.4^{2}=49.6

standard deviation=variance=49.6=7.04text{standard deviation}=sqrt{text{variance}}=sqrt{49.6}=7.04

 

d) The median is the 350350th value, which falls within the second block. Since 160160 values are in the first block, this is the 190190th value of the second block. The second block has a width of 1515 and a frequency of 420420. So position 190190 is

190×15420=9514dfrac{190times 15}{420}=dfrac{95}{14}

Adding on the original 55 from the width of the first block gives a value of 16514dfrac{165}{14}, which is our median.

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Additional Resources

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Exam Tips Cheat Sheet

A Level
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Formula Booklet

A Level

Specification Points Covered

L1 – Interpret diagrams for single-variable data, including understanding that area in a histogram represents frequency
L3 – Interpret measures of central tendency and variation, extending to standard deviation

Grouped Data Worksheet and Example Questions

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Single Variable Data and Histograms

A Level

Related Topics

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Mean and Standard Deviation

A Level
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Presenting Data

A Level