Where inside the class?

The idea

Mode = l + ((f1 − f0)/(2f1 − f0 − f2)) × h, with l the lower limit and f1 the frequency of the modal class, f0, f2 the frequencies of the preceding and succeeding classes, h the class size.

Family sizes

20 households gave this table (NCERT Example 5). The modal class is 3–5, with 8 families. The class before it, 1–3, has 7 families and the class after it, 5–7, has only 2.

The mode is somewhere inside 3–5.

Where do you think the mode is inside the class 3–5?

• Nearer to 3
• Exactly in the middle, at 4
• Nearer to 5

Change the neighbours

The amber bar is the modal class. Move f₀ (the class before, blue) and f₂ (the class after, red). Two lines cross over the modal class: the blue one joins the top of the class before to the top of the modal bar, the red one joins the top of the modal bar to the top of the class after. The foot of the crossing is the mode. Make the mode lean left, sit in the middle, and lean right.

The mode formula

Mode = l + ((f₁ − f₀) / (2f₁ − f₀ − f₂)) × h

l = lower limit of the modal class, h = class size (all sizes equal), f₁ = frequency of the modal class, f₀ = frequency of the class before it, f₂ = frequency of the class after it.

The bigger neighbour pulls the mode toward its side. If f₀ = f₂ the mode is exactly at the middle of the class. Example 5: l = 3, h = 2, f₁ = 8, f₀ = 7, f₂ = 2, so Mode = 3 + (1/7) × 2 = 3.286.

Notes

Mode = l + ((f₁ − f₀) / (2f₁ − f₀ − f₂)) × h, with l the lower limit and f₁ the frequency of the modal class, f₀ and f₂ the frequencies of the classes before and after it, and h the class size.

Check yourself

NCERT Example 6: for the marks of 30 students the modal class is 40–55, with f₁ = 7, f₀ = 3, f₂ = 6 and h = 15. Find the mode.

Answer: 52

2f₁ − f₀ − f₂ = 14 − 3 − 6 = 5, so Mode = 40 + (4/5) × 15 = 40 + 12 = 52.

For the 53 students the modal class is 70–80, with f₁ = 9, f₀ = 7, f₂ = 7 and h = 10. Find the mode.

Answer: 75

Mode = 70 + ((9 − 7)/(18 − 14)) × 10 = 70 + 5 = 75. The neighbours are equal, so the mode is in the middle of 70–80.

Example 5: l = 3, h = 2, f₁ = 8, f₀ = 7, f₂ = 2. Find the mode correct to 3 decimal places.

Answer: 3.286

2f₁ − f₀ − f₂ = 16 − 7 − 2 = 7, so Mode = 3 + (1/7) × 2 = 3.2857…, which is 3.286.

The modal class is 20–30. The class before it and the class after it have the same frequency. Where is the mode?

• At the middle of the class, 25 — correct. Yes! If f₀ = f₂, the fraction (f₁ − f₀)/(2f₁ − f₀ − f₂) equals 1/2, so the mode is l + h/2.
• At 20, the lower limit. It would be at 20 only if f₁ = f₀.
• At 30, the upper limit. It would be at 30 only if f₁ = f₂.