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Year 9 Maths Probability and Statistics

Grouped Data

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Theory

A histogram shows how often each score, or each group of scores, occurs. This Year 9 maths page shows how to read and draw histograms, group data into class intervals, find the modal class and the class centres, draw a frequency polygon, find the mean and median from a histogram, and estimate the mean when only the classes are known.

A histogram shows numerical data. The scores run along the horizontal axis and the frequency (how many times each score occurs) runs up the vertical axis. The columns touch, because the horizontal axis is a number line. For single scores, each column is centred on its score.

A column graph shows categories, such as favourite sport, so its columns have gaps.

For data with many different values, the scores are grouped into class intervals of equal width. The class \(50\)–\(55\) usually means from \(50\) up to, but not including, \(55\), so \(55\) belongs to the next class. For whole numbers the classes may be written \(0\)–\(9\), \(10\)–\(19\) and so on.

The class centre is the middle of a class. The modal class is the class with the highest frequency, the tallest column.

A frequency polygon joins the midpoints of the tops of the columns with straight lines. It is closed off on the horizontal axis half a class width beyond each end.

Once data are grouped, the exact scores are lost. The histogram still gives totals and percentages for whole classes, but values inside a class can only be estimated.

Histogram of 25 scoresA histogram with columns of height 2, 6, 9, 5 and 3 centred on the scores 1 to 5. The columns touch. The tallest column, above 3, is shaded red. 0 2 4 6 8 10 1 2 3 4 5 Score Frequency
Each column is centred on its score. Adding the heights gives \(2 + 6 + 9 + 5 + 3 = 25\) scores, and the tallest column makes the mode \(3\).
Grouped histogram with a frequency polygonA histogram of 28 masses in the classes 40 to 50, 50 to 60, 60 to 70, 70 to 80 and 80 to 90 kilograms, with frequencies 3, 7, 10, 6 and 2. A red frequency polygon joins the midpoints of the column tops and meets the horizontal axis at 35 and 95. 0 2 4 6 8 10 40 50 60 70 80 90 Mass (kg) Frequency
The polygon passes through the class centres \(45\), \(55\), \(65\), \(75\), \(85\) and meets the axis at \(35\) and \(95\). The modal class is \(60\)–\(70\) kg. \(8\) of the \(28\) masses are \(70\) kg or more, about \(28.6\%\).
\[\text{number of scores} = \sum f \qquad \text{mean} = \dfrac{\sum fx}{\sum f}\]
mean=βˆ‘fxΓ·βˆ‘f
\[\text{class centre} = \dfrac{\text{lower end} + \text{upper end}}{2}\]
\[\text{estimated mean} = \dfrac{\sum (f \times \text{centre})}{\sum f}\]
class centre=(lower end + upper end) Γ· 2

For the class \(60\)–\(65\) cm the centre is \(62.5\) cm. For whole-number classes such as \(10\)–\(19\), the centre is \(\dfrac{10 + 19}{2} = 14.5\).

Reading a histogram. The height of each column is a frequency. Add the heights for the total. For β€œless than” or β€œat least” a value, add the columns on that side, and for a percentage divide by the total and multiply by \(100\). The mode, or modal class, is under the tallest column: give the score, not the height.
The median from a histogram. Find position \(\dfrac{n+1}{2}\) and add up the column heights from the left until you reach it. For grouped data this gives the class that contains the median.
Smallest and largest possible mean. For whole-number data in classes such as \(50\)–\(54\), the smallest possible mean puts every score at the bottom of its class (\(50\)), and the largest puts every score at the top (\(54\)). The true mean lies between the two.

How to work with a histogram

  1. Read the axes. Are the columns single scores or classes? What does a class such as \(50\)–\(55\) include?
  2. Count: add the column heights.
  3. Mode: the score or class under the tallest column.
  4. Mean: multiply each score, or each class centre, by its frequency, add, and divide by the total. With classes, call it an estimate.
  5. Median: add the heights from the left to position \(\dfrac{n+1}{2}\).
  6. Drawing: use an even scale on both axes, make the columns touch, and label both axes. For a frequency polygon, mark the midpoint of each column top, join them, and close off at the class centres either side.
Example 1 β€” The mean and median from a histogram
A histogram shows the number of goals a team scored in each game: \(0\) goals in \(3\) games, \(1\) in \(7\), \(2\) in \(6\), \(3\) in \(3\) and \(4\) in \(1\). Find the mode, the mean and the median.
Solution

The tallest column is above \(1\), so the mode is \(1\) goal. For the mean, multiply each score by its frequency.

\(\sum f = 3 + 7 + 6 + 3 + 1 = 20\)

\(\sum fx = 0 + 7 + 12 + 9 + 4 = 32\)

\(\text{mean}\)\(=\)\(\dfrac{32}{20} = \mathbf{1.6}\)
mean=32Γ·20=1.6

With \(20\) scores the median is between the \(10\)th and \(11\)th. The \(0\)s fill places \(1\)–\(3\) and the \(1\)s places \(4\)–\(10\), so the \(10\)th is \(1\) and the \(11\)th is \(2\).

\(\text{median}\)\(=\)\(\dfrac{1 + 2}{2} = \mathbf{1.5}\)
Example 2 β€” Classes and a frequency polygon
The heights of \(30\) seedlings are grouped in the table. The class \(10\)–\(15\) holds heights from \(10\) cm up to, but not including, \(15\) cm.
Height (cm)\(10\)–\(15\)\(15\)–\(20\)\(20\)–\(25\)\(25\)–\(30\)\(30\)–\(35\)
Frequency\(3\)\(8\)\(12\)\(5\)\(2\)
(i) Which is the modal class?
(ii) Where does a frequency polygon meet the horizontal axis?
(iii) What percentage of the seedlings are \(25\) cm or taller?
Solution

(i) The highest frequency is \(12\), so the modal class is \(20\)–\(25\) cm.

(ii) The polygon is closed off half a class width (\(2.5\) cm) beyond each end, at the centres of the empty classes \(5\)–\(10\) and \(35\)–\(40\): at \(7.5\) cm and \(37.5\) cm.

(iii) Heights of \(25\) cm or more are in the last two classes.

\(5 + 2\)\(=\)\(7\)
\(\dfrac{7}{30} \times 100\%\)\(\approx\)\(\mathbf{23.3\%}\)
7Γ·30Γ—100β‰ˆ23.3
Example 3 β€” Estimating the mean
Estimate the mean height of the seedlings in Example 2. Which class contains the median? Why is the mean only an estimate?
Solution

Replace every height by its class centre, then multiply by the frequency.

ClassCentre \(c\)\(f\)\(f \times c\)
\(10\)–\(15\)\(12.5\)\(3\)\(37.5\)
\(15\)–\(20\)\(17.5\)\(8\)\(140\)
\(20\)–\(25\)\(22.5\)\(12\)\(270\)
\(25\)–\(30\)\(27.5\)\(5\)\(137.5\)
\(30\)–\(35\)\(32.5\)\(2\)\(65\)
\(\sum (f \times \text{centre})\)\(=\)\(650\)
\(\text{estimated mean}\)\(=\)\(\dfrac{650}{30}\)
\(\)\(\approx\)\(\mathbf{21.67}\text{ cm}\)
estimated mean=650Γ·30β‰ˆ21.67

The median is between the \(15\)th and \(16\)th heights. Running totals are \(3\), \(11\), \(23\), so both are in the class \(20\)–\(25\) cm.

It is an estimate because the exact heights inside each class are not known. Each one has been replaced by its class centre.

Example 4 β€” Smallest and largest possible mean
Twenty test marks, all whole numbers, are grouped: \(2\) in \(40\)–\(49\), \(5\) in \(50\)–\(59\), \(8\) in \(60\)–\(69\) and \(5\) in \(70\)–\(79\). Find the smallest and largest possible mean.
Solution

Smallest: put every mark at the bottom of its class. Largest: put every mark at the top.

Smallest total: \(2 \times 40\) \(+ \, 5 \times 50\) \(+ \, 8 \times 60\) \(+ \, 5 \times 70\)

Largest total: \(2 \times 49\) \(+ \, 5 \times 59\) \(+ \, 8 \times 69\) \(+ \, 5 \times 79\)

\(\text{smallest}\)\(=\)\(80 + 250 + 480\)
\(\)\(\)\({} + 350 = 1160\)
\(\text{largest}\)\(=\)\(98 + 295 + 552\)
\(\)\(\)\({} + 395 = 1340\)
\(\text{smallest mean}\)\(=\)\(\dfrac{1160}{20} = \mathbf{58}\)
\(\text{largest mean}\)\(=\)\(\dfrac{1340}{20} = \mathbf{67}\)
smallest mean=58,largest mean=67

Every mark moved up by \(9\), so the mean moved up by \(9\) too. The true mean is somewhere from \(58\) to \(67\).

Common pitfalls

The mode is the score, not the height. A tallest column of height \(9\) above the score \(3\) makes the mode \(3\).
Check the scale. The vertical scale must go up in equal steps. A scale that jumps from \(8\) to \(12\) makes some columns look the wrong size.
A class boundary goes up. With classes \(50\)–\(55\) and \(55\)–\(60\), a value of exactly \(55\) belongs to \(55\)–\(60\).
Polygon points go at the centre. Mark each point at the middle of the column top, above the class centre, not at a corner.
Grouped data hide the exact values. A histogram cannot tell you how many scores are above \(47\) when \(47\) is inside a class. It can only give an estimate.

Frequently asked questions

What is the difference between a histogram and a column graph?

A histogram shows numerical data on a number line, so its columns touch. A column graph shows categories, such as favourite sports, so its columns have gaps.

What is a class centre?

The middle of a class interval: add the two ends and divide by \(2\). The class \(60\)–\(65\) has a centre of \(62.5\).

What is the modal class?

The class with the highest frequency, shown by the tallest column of a grouped histogram.

How do you draw a frequency polygon?

Mark a point at the middle of the top of each column, join the points with straight lines, and close the polygon on the horizontal axis half a class width beyond each end.

How do you estimate the mean of grouped data?

Multiply each class centre by its frequency, add the results, and divide by the total frequency. It is only an estimate because the exact values in each class are unknown.