2.5.6.3. Formula for Cumulative Frequency

2.5.6.3. Formula for cumulative (relative) frequency

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The cumulative frequency is calculated by summing all the previous frequencies and the frequency of the current category. This is where the use of the index numbers become useful.

Let’s convert again from text to formula notation. Cumulative Frequency for category i = sum of all absolute frequencies from the first category till category i.

Let’s use CF_i to indicate the Cumulative Frequency of category i, and again the summation sign for sum of. We then get:
CF_i = ∑ absolute frequencies from the first category till category i

Remember that the starting point for the index goes below, the summation sign, so that becomes a 1, while where the index should stop is indicated at the top, which is i. Since i is now used to indicate the category for which we want it to stop, we need another letter to indicate the index for the absolute frequencies, so let’s use j for that. Combining this now gives us the formula for the cumulative frequencies.

$CF_i=\sum_{j=1}^{i}F_j$

So if we wanted to know the cumulative frequency of the 3^rd category (i.e. i = 3) we would get:

$CF_3=\sum_{j=1}^{3}F_j$

When we discussed the cumulative frequency we also noticed we could use an alternative calculation method, by using the cumulative frequency of the previous category. In words this meant:
Cumulative Frequency of category i = Cumulative Frequency of the category prior to i, and add the absolute frequency of category i.

Note that ‘the category prior to i' would simply be category i – 1, and using our usual abbreviations we can now get the following alternative formula.

$CF_i=CF_{i-1}+F_i$

We need to add one small deviation on this formula since if we apply it for the first category we get $CF_1=CF_{1-1}+F_1=CF_{0}+F_1$ . We need to define CF₀ then as 0 so with this addition, we get:

$CF_i=CF_{i-1}+F_i\textup{, with }CF_0=0$

Proof of equivalence

We have seen two formulas for cumulative frequencies. These two formulas will lead to the same result as can be shown as follows. We need to proof that:

$\sum_{j=1}^{i}F_j=CF_{i-1}+F_i$

To proof this we start with the first formula: 〖

$CF_i=\sum_{j=1}^{i}F_j$

If we write the summation sign further out, we get:

$CF_i=\frac{F_1}{n}+\frac{F_2}{n}+...+\frac{F_{i-1}}{n}+\frac{F_i}{n}$

If we replace i with i – 1 in the formula, we get:

$CF_{i-1}=\sum_{j=1}^{i-1}F_j$

Which if we write it out in full becomes:

$CF_{i-1}=\frac{F_1}{n}+\frac{F_2}{n}+...+\frac{F_{i-1}}{n}$

Note that the only difference between the full version of CF_i and CF_i-1 is that in the first one there is one more term, the ‘ + F_i’ one. We can therefore also rewrite this into:

$\sum_{j=1}^{i}F_j=CF_{i-1}+F_i$

Which is actually the second formula.