5.1.5 Cumulative Frequency Graphs

In this topic we will learn how to:

draw and interpret cumulative frequency graphs

A cumulative frequency graph is used to represent grouped continuous data. It is represented in the form of an $s$ -shaped curve.

Cumulative frequency is the total frequency of the previous classes up to and including the present class.

To draw a cumulative frequency graph you need the following information:

upper bound of a class
upper bound of a class

To calculate the cumulative frequency, add the frequencies of previous classes together with that of the current class. $\large\textcolor{gray}{\textbf{Variation and Measures of Central Tendency}\\ \textbf{ for Cumulative Frequency Graph}}$ $\textbf{\textcolor{gray}{Mode}}$ A cumulative frequency curve represents grouped data, so we cannot find the mode but instead we can find the modal class. The modal class is the class with the highest frequency. $\textcolor{gray}{n\textbf{-th percentile}}$ To calculate the $n$ -th percentile, we use the formula $n\textmd{-th percentile} = \frac{xn}{100}$ Where $x$ represents the percentile and $n$ represents the sample size. $\textbf{\textcolor{gray}{Lower Quartile}}$ To calculate the lower quartile, we use the formula,
$q_{1} = \frac{1}{4}n$ Where $q_{1}$ represents the lower quartile. $\textbf{\textcolor{gray}{Upper Quartile}}$ To calculate the upper quartile, we use the formula,
$q_{3} = \frac{3}{4}n$ Where $q_{3}$ represents the upper quartile.
$\textbf{\textcolor{gray}{Intequartile Range}}$ To calculate the interquartile range, we use the formula,
$IQR = q_{3} - q_{1}$ Where $IQR$ represents the interquartile range, $q_{3}$ represents the upper quartile, $q_{1}$ represents the lower quartile.
$\textbf{\textcolor{gray}{Mean}}$ To calculate the mean when data is displayed in the form of a cumulative frequency curve, we need to first find the mid interval. This is the middle value for each value. We use the formula,
$\overline{x} = \frac{\Sigma xf}{\Sigma f}$ Where $\overline{x}$ represents mean, $x$ represents the mid-interval, $f$ represents the frequency.
$\textbf{\textcolor{gray}{Variance}}$ To calculate the variance, we use the formula,
$\sigma^{2} = \frac{\Sigma x^{2}f}{\Sigma f} - \overline{x}^{2}$ Where $\sigma^{2}$ represents variance, $x$ represents the mid interval, $f$ represents the frequency, $\overline{x}$ represents mean.
$\textbf{\textcolor{gray}{Standard Deviation}}$ Standard deviation is the square root of variance. Therefore, the formula for standard deviation is,
$\sigma = \sqrt{\frac{\Sigma x^{2}f}{\Sigma f} - \overline{x}^{2}}$ Where $\sigma$ represents standard deviation, $x$ represents the mid interval, $f$ represents the frequency, $\overline{x}$ represents mean.

Let’s look at some past paper questions.

1. Helen measures the lengths of $150$ fish of a certain species in a large pond. These lengths, correct to the nearest centimetre, are summarised, are summarised in the following table. (9709/52/F/M/20 number 7)

Length (cm)	0 – 9	10 – 14	15 – 19	20 – 30
Frequency	15	48	66	21

(a) Draw a cumulative frequency graph to illustrate the data.

Find the cumulative frequency,

Length (cm)	0 – 9	10 – 14	15 – 19	20 – 30
Cumulative Frequency	15	63	129	150

You will notice that there are gaps between the classes. To remove those gaps we need to do continuity correction. Simply subtract $0.5$ from the lower bounds and add $0.5$ to the upper bounds,

Length (cm)	0 – 9.5	9.5 – 14.5	14.5 – 19.5	19.5 – 30.5
Cumulative Frequency	15	63	129	150

Note: We do not subtract $0.5$ from $0$ because we would end up with a negative value for length, which does not exist.

Now that there are no gaps, we can plot the upper bounds against the cumulative frequency. Label the $y$ -axis with cumulative frequency. Label the $x$ -axis with the class title.

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(b) $40\%$ of these fish have a length of $d$ cm or more. Use your graph to estimate the value of $d$ .

This means $60\%$ of fish have a length less than $d$ cm. Let’s find $60\%$ of $150$ ,
$\frac{60}{100} \times 150$ $\textcolor{red}{90}$ Draw construction lines at a cumulative frequency of $90$ and read off the length,
$d = 16.5 \textmd{ cm}$ Therefore, the final answer is,
$d = 16.5 \textmd{ cm}$ The mean length of these $150$ fish is $15.295$ cm.
(c) Calculate an estimate for the variance of the lengths of the fish.

The formula for variance is,
$\sigma^{2} = \frac{\Sigma x^{2}f}{\Sigma f} - \overline{x}^{2}$ We already have the mean, we need to find $\frac{\Sigma x^{2}f}{\Sigma f}$ . $x$ represents the mid intervals so let’s find $x$ ,

Mid Interval	4.75	12	17	25
Frequency	15	48	66	21

Note: Use the classes after continuity correction to find the mid interval.

Now that we have the mid interval, let’s find $\frac{\Sigma x^{2}f}{\Sigma f}$ ,
$\frac{\Sigma x^{2}f}{\Sigma f} = \frac{4.75^{2}(15) + 12^{2}(48) + 17^{2}(66) + 25^{2}(21)}{150}$ $\frac{\Sigma x^{2}f}{\Sigma f} = 262.99653$ Note: Remember that $f$ represents frequency NOT cumulative frequency.

Substitute into the formula for variance,
$\sigma^{2} = \frac{\Sigma x^{2}f}{\Sigma f} - \overline{x}^{2}$ $\sigma^{2} = 262.99653 - (15.295)^{2}$ $\sigma^{2} = 29.059225$ $\sigma^{2} = 29.1$ Therefore, the final answer is,
$\sigma^{2} = 29.1$ 2. The heights in cm of 160 sunflower plants were measured. The results are summarised on the following cumulative frequency curve. (9709/53/M/J/21 number 1)

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(a) Use the graph to estimate the number of plants with heights less than $100$ cm.

Draw construction lines at the height of $100$ cm and read off the respective cumulative frequency,
$\textcolor{red}{60}$ Therefore, the final answer is,
$60 \textmd{ plants}$ (b) Use the graph to estimate the $65$ th percentile of the distribution.

The formula to find the $n$ th percentile is,
$\frac{xn}{100}$ Substitute the value of $x$ and $n$ ,
$\frac{(65)(160)}{100}$ $104$ Draw construction lines at a cumulative frequency of $104$ and read off the respective height,
$\textcolor{#0f0}{136}$ Therefore, the final answer is,
$136$ (c) Use the graph to estimate the interquartile range of the heights of these plants.

The formula to find the interquartile range is,
$IQR = q_{3} - q_{1}$ To find the upper quartile, use the formula,
$q_{3} = \frac{3}{4}n$ $q_{3} = \frac{3}{4}(160)$ $q_{3} = 120$ Draw construction lines at a cumulative frequency of $120$ and read off the respective height,
$q_{3} = 150$ To find the lower quartile, use the formula,
$q_{1} = \frac{1}{4}n$ $q_{1} = \frac{1}{4}(160)$ $q_{1} = 40$ Draw construction lines at a cumulative frequency of $40$ and read off the respective height,
$q_{1} = 76$ Substitute into the formula for interquartile range,
$IQR = q_{3} - q_{1}$ $IQR = 150 - 76$ $IQR = 74$ Therefore, the final answer is,
$IQR = 74$