Draw a histogram for grouped frequency distribution and find mode graphically

Question

Given the following grouped frequency distribution, draw a histogram and use it to find the mode graphically:

Class Interval	Frequency
10 – 20	5
20 – 30	12
30 – 40	20
40 – 50	15
50 – 60	8

Solution — Step by Step

A histogram is a bar graph where the bars touch each other, showing continuous data. Each bar’s width is the class width (here, 10 units) and its height is the frequency. We plot class intervals on the x-axis and frequency on the y-axis.

Mark the class intervals on the x-axis: 10, 20, 30, 40, 50, 60. On the y-axis, mark frequencies up to at least 20. Draw vertical bars with heights 5, 12, 20, 15, and 8 for each interval. The bars must be adjacent — no gaps between them.

The bar for 30–40 is the tallest (height = 20). This is the modal class.

The modal class is the class interval with the highest frequency. Here that is 30–40 (frequency = 20).

We will now locate the mode geometrically within this class.

To find mode graphically from the histogram:

Take the bar of the modal class (30–40).
Draw a line from the top-left corner of the modal bar to the top-right corner of the bar immediately to its left (20–30 bar).
Draw a line from the top-right corner of the modal bar to the top-left corner of the bar immediately to its right (40–50 bar).
The two lines intersect at a point. Drop a perpendicular from this intersection to the x-axis.
The x-value at this perpendicular is the mode.

Mathematically, this gives the same result as the formula:

\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h

Where $l = 30$ , $f_0 = 12$ , $f_1 = 20$ , $f_2 = 15$ , $h = 10$ .

\text{Mode} = 30 + \frac{20 - 12}{2(20) - 12 - 15} \times 10

= 30 + \frac{8}{40 - 27} \times 10

= 30 + \frac{8}{13} \times 10

= 30 + \frac{80}{13}

= 30 + 6.15 \approx 36.15

Mode ≈ 36.15

This matches the x-value you read from the graphical construction.

Why This Works

The graphical method is essentially a visual shortcut to the same formula. When you draw the diagonal lines from adjacent bars and drop a perpendicular, you’re geometrically interpolating within the modal class based on how much taller the modal bar is compared to its neighbors.

The formula weights the modal class more heavily when the difference $(f_1 - f_0)$ is large relative to the total “excess” $(2f_1 - f_0 - f_2)$ . If both neighboring bars are equal, the mode falls exactly at the midpoint of the modal class.

This is why mode is never simply the midpoint of the modal class — it shifts toward the side with the smaller neighbor.

Alternative Method

You can also use the empirical formula to estimate mode from mean and median:

\text{Mode} \approx 3 \times \text{Median} - 2 \times \text{Mean}

This is less precise than the graphical/formula method, but useful when you’ve already calculated mean and median and need a quick check.

Common Mistake

Many students identify the modal class correctly but then use the class midpoint (35 in this case) as the mode. That’s wrong. The modal class just tells you WHERE mode lies — the actual mode requires interpolation using the formula or the graphical construction. Always apply the formula.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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