Interpreting graphs of grouped data – Interpretation of Statistical Graphs (Senior 3 Mathematics) | Ganzaa

Interpretation of Statistical Graphs

Lessons

Senior 3 · Mathematics

Interpretation of Statistical Graphs

3 lessons

Lesson 3 of 3

Interpreting graphs of grouped data

25 min

Count the area, not the height

One hundred traders at a market were asked how far they had travelled that morning (illustrative data). The classes are not all the same width: the first three are 2 km wide, the next is 4 km and the last is 10 km. On a histogram of such data, the vertical axis is not "number of traders" but frequency density, and the number of traders in a class is the area of its bar.

Look at the bar for 6 to 10 km. It is lower than the bar for 2 to 4 km, yet it holds more traders, because it is twice as wide. Reading a histogram of unequal classes means multiplying, never just looking.

Definition

A histogram is a graph of grouped continuous data made of touching bars, one per class, drawn between the class boundaries. The area of each bar is proportional to the frequency of its class.

When all classes have the same width, the height can simply be the frequency. When the widths are unequal, the height must be the frequency density, $\frac{\text{frequency}}{\text{class width}}$ , so that area still stands for frequency.

Important

Reading a histogram with frequency density on the vertical axis:

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

To estimate how many values lie in part of a class, take the same fraction of the bar's area; this assumes the values are spread evenly across the class.

Worked example 1: reading the traders' histogram

(a) How many traders travelled 6 to 10 km? (b) Which class holds the most traders, and which bar is tallest? (c) Estimate how many traders travelled between 5 km and 8 km. (d) Check that the bars account for all 100 traders.

Solution

(a) The bar is at height 7 and is $10 - 6 = 4$ km wide: $7 \times 4 = 28$ traders.

(b) The other frequencies are $9 \times 2 = 18$ , $13 \times 2 = 26$ , $10 \times 2 = 20$ and $0.8 \times 10 = 8$ . So the class 6 to 10 km holds the most traders (28), but the tallest bar is 2 to 4 km (26 traders, at 13 per km). The tallest bar is the most crowded class, not the largest one.

(c) From 5 to 6 km is the last 1 km of the 4 to 6 km bar: $10 \times 1 = 10$ traders. From 6 to 8 km is the first 2 km of the 6 to 10 km bar: $7 \times 2 = 14$ . Estimate: $10 + 14 = 24$ traders.

(d) $18 + 26 + 20 + 28 + 8 = 100$ . ✓

Warning

The unequal-width trap. If the same data are drawn with height equal to frequency, every wide class is inflated. Below, the 10 to 20 km bar now covers more area than the 0 to 2 km bar, as if many more traders had come from far away, when really only 8 did, against 18. Before reading a histogram, check the label on the vertical axis: if the classes are unequal, it must say frequency density.

Frequency polygons: comparing two distributions

A school farm grew 60 maize plants without fertiliser and 60 with fertiliser, and measured every plant four weeks after planting (illustrative data). Two histograms on one grid would hide each other, so each set is drawn as a frequency polygon. The classes are 10 to 20 cm, 20 to 30 cm and so on, each 10 cm wide.

Definition

A frequency polygon joins with straight lines the points (class mid-point, frequency) of a grouped distribution. It is closed at both ends by a point of frequency 0 at the mid-point of the empty class on each side, here 5 cm and 75 cm.

Its peak marks the modal class, its position along the axis shows where the values are centred, and its width shows how spread out they are. Two polygons on one grid can be compared directly, provided the totals are equal.

Worked example 2: comparing the maize polygons

(a) Read the modal class of each set. (b) How many plants in each set were taller than 40 cm? (c) Estimate the mean height of each set. (d) What do the graphs suggest?

Solution

(a) Without fertiliser the peak is at mid-point 35, so the modal class is 30 to 40 cm (20 plants). With fertiliser the peak is at 45: modal class 40 to 50 cm (20 plants).

(b) Read the points at mid-points 45, 55 and 65. Without: $10 + 4 + 0 = 14$ plants. With: $20 + 12 + 6 = 38$ plants.

(c) Using mid-points, $\bar{x} \approx \frac{\sum fx}{\sum f}$ : $\text{without: } \frac{15(8) + 25(18) + 35(20) + 45(10) + 55(4)}{60} = \frac{1940}{60} \approx 32.3 \text{ cm}$ $\text{with: } \frac{15(2) + 25(6) + 35(14) + 45(20) + 55(12) + 65(6)}{60} = \frac{2620}{60} \approx 43.7 \text{ cm}$

(d) The fertiliser polygon has a similar shape but sits about 10 cm further to the right: the fertilised plants were about 11 cm taller on average. The comparison is only fair if both sets had the same seed, soil, water and planting date.

Cumulative frequency curves: medians and quartiles from a graph

The heights of 80 students were grouped in classes 5 cm wide (illustrative data). The cumulative frequency at each upper boundary counts the students below that height.

Height $h$ (cm)	Frequency	Upper boundary	Cumulative frequency
$145 \le h < 150$	4	150	4
$150 \le h < 155$	11	155	15
$155 \le h < 160$	10	160	25
$160 \le h < 165$	25	165	50
$165 \le h < 170$	20	170	70
$170 \le h < 175$	7	175	77
$175 \le h < 180$	3	180	80

Each point (upper boundary, cumulative frequency) is plotted, starting from (145, 0), and the points are joined. Here they are joined with straight segments, which is the same as assuming the heights are spread evenly within each class; a smooth curve through the same points gives almost the same readings.

Definition

A cumulative frequency curve (ogive) plots cumulative frequency against the upper class boundaries. For $n$ values read across from these heights on the vertical axis:

the lower quartile $Q_{1}$ at $\frac{n}{4}$ : a quarter of the values lie below it;
the median at $\frac{n}{2}$ : half lie below it;
the upper quartile $Q_{3}$ at $\frac{3n}{4}$ : three quarters lie below it;
the $k$ th percentile at $\frac{k}{100} \times n$ .

The interquartile range $\text{IQR} = Q_3 - Q_1$ is the width of the middle half of the data. Unlike the range, one very tall or very short student cannot change it.

Worked example 3: median and quartiles from the curve

Find the median, the quartiles and the interquartile range of the 80 heights, then check the median by calculation.

Solution Here $n = 80$ , so read across at $\frac{80}{4} = 20$ , $\frac{80}{2} = 40$ and $\frac{3 \times 80}{4} = 60$ (the dashed lines).

At 20 the curve is above 157.5 cm: $Q_{1} = 157.5$ cm.
At 40 it is above 163 cm: median $= 163$ cm.
At 60 it is above 167.5 cm: $Q_{3} = 167.5$ cm.

$\text{IQR} = 167.5 - 157.5 = 10$ cm: the middle half of the students are within 10 cm of each other.

Check: the 40th student is in the class $160 \le h < 165$ , which starts after 25 students and holds 25. Going $40 - 25 = 15$ students into it covers $\frac{15}{25}$ of its 5 cm width: $\text{median} = 160 + \frac{40 - 25}{25} \times 5 = 163 \text{ cm.}$

Worked example 4: reading the curve both ways

(a) What percentage of the students are shorter than 160 cm? (b) About how many are taller than 172 cm? (c) Above what height are the tallest 10%?

Solution

(a) Read up from 160 cm: cumulative frequency 25. $\frac{25}{80} \times 100 = 31.25\%$ .

(b) Read up from 172 cm: the curve is at about 73 (it rises from 70 at 170 cm by $\frac{2}{5} \times 7 = 2.8$ ). So about $80 - 73 = 7$ students are taller than 172 cm.

(c) The tallest 10% begin at the 90th percentile, at $\frac{90}{100} \times 80 = 72$ on the vertical axis. Reading across: $170 + \frac{72 - 70}{7} \times 5 \approx 171.4$ , so about 171 cm.

Warning

Three ways to lose the marks. (1) Plotting at the class mid-points: cumulative frequency counts values below a height, so it belongs at the upper boundary. Plotting at mid-points moves the whole curve half a class to the left. (2) Forgetting the starting point (145, 0) at the lower boundary of the first class. (3) Reading the median at $\frac{n+1}{2}$ : for a curve of grouped data use $\frac{n}{2}$ .

Tip

On every cumulative frequency question, write the three reading heights first: $\frac{n}{4}$ , $\frac{n}{2}$ , $\frac{3n}{4}$ . Then draw the reading lines in pencil, across to the curve and straight down, and read the scale of one small square before writing the answer.

Corbettmaths: Reading Cumulative Frequency Graphs. Watch how the median, the quartiles and the interquartile range are read from the curve, then repeat the method on the heights curve above.

✦

Practice

Questions 2 to 5 use the graphs in this lesson.

In a histogram, the bar for the class $20 \le t < 30$ minutes has frequency density 2.5. How many values are in the class?
From the traders' histogram, estimate how many traders travelled less than 3 km.
From the maize polygons, how many plants grown without fertiliser were shorter than 30 cm?
From the heights curve, how many students are between 155 cm and 165 cm tall?
From the heights curve, estimate the 30th percentile.
Sixty pupils timed how long, in minutes, they took to solve a puzzle: $0 \le t < 4$ : 6; $4 \le t < 8$ : 14; $8 \le t < 12$ : 22; $12 \le t < 16$ : 12; $16 \le t < 20$ : 6. Build the cumulative frequency table, then find the median, the quartiles and the interquartile range by calculation.
Another group of 80 students has median 160 cm and interquartile range 14 cm. Compare its heights with the group in this lesson.
A pupil plotted the heights' cumulative frequencies at the class mid-points (147.5, 152.5, …). What median would the pupil read, and why is it wrong?

Show answersHide answers

$2.5 \times 10 = 25$ .
$18 + 13 \times 1 = 31$ traders (all of 0 to 2 km, plus the first 1 km of the 2 to 4 km bar).
$8 + 18 = 26$ plants (the points at mid-points 15 and 25).
$50 - 15 = 35$ students.
Read across at $0.3 \times 80 = 24$ : $155 + \frac{24 - 15}{10} \times 5 = 159.5$ cm.
Cumulative frequencies at 4, 8, 12, 16, 20 minutes: 6, 20, 42, 54, 60. Median at 30: $8 + \frac{30 - 20}{22} \times 4 \approx 9.8$ min. $Q_{1}$ at 15: $4 + \frac{15 - 6}{14} \times 4 \approx 6.6$ min. $Q_{3}$ at 45: $12 + \frac{45 - 42}{12} \times 4 = 13$ min. IQR $\approx 13 - 6.6 = 6.4$ min.
The other group is shorter on average (median 160 cm against 163 cm) and its heights are more spread out (IQR 14 cm against 10 cm).
Every point moves 2.5 cm to the left, so the pupil reads $163 - 2.5 = 160.5$ cm. It is wrong because the cumulative frequency 50 counts students below 165 cm, the upper boundary, not below 162.5 cm.

1Count the area, not the height

2Frequency polygons: comparing two distributions

3Cumulative frequency curves: medians and quartiles from a graph

Count the area, not the height

Frequency polygons: comparing two distributions

Cumulative frequency curves: medians and quartiles from a graph