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Worked Examples · Example 6
Q.

Using the same distribution as Example 3, calculate the mode:

Marks0-1010-2020-3030-4040-50
No. of students5812105
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The highest frequency is 12, in the class 20-30 — this is the modal class.

L=20L=20, f1=12f_1=12 (modal class), f0=8f_0=8 (preceding class, 10-20), f2=10f_2=10 (succeeding class, 30-40), h=10h=10:

Mode=L+(f1−f02f1−f0−f2)×h=20+(12−82(12)−8−10)×10=20+(46)×10=20+6.67=26.67\text{Mode} = L + \left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h = 20 + \left(\frac{12-8}{2(12)-8-10}\right)\times10 = 20+\left(\frac{4}{6}\right)\times10 = 20+6.67 = 26.67

Independent cross-check — empirical relationship: from Examples 3 and 4, Mean =25.5=25.5 and Median =25.83=25.83 for this same distribution.

Mode≈3 Median−2 Mean=3(25.83)−2(25.5)=77.5−51=26.5\text{Mode} \approx 3\,\text{Median} - 2\,\text{Mean} = 3(25.83) - 2(25.5) = 77.5-51 = 26.5 …

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