The inclusive classes 5–14, 15–24, 25–34 are to be made continuous. The adjustment factor to be used is ____.

Class 10StatisticsStatistics and Probability

A1
B0.5
C0.25
D5

Step-by-Step Solution

The gap between one upper limit and the next lower limit is 15 − 14 = 1. Half of that gap, 0.5, is subtracted from every lower limit and added to every upper limit, giving 4.5–14.5, 14.5–24.5, 24.5–34.5.

Answer: B) 0.5

Key idea: Mode for grouped data

**Modal class**: class with HIGHEST frequency.

**Mode formula**:
**Mode =

l + h \times (f

₁ −

f

₀) /

(2 f

₁ −

f

₀ −

f

₂)**

Where:
-

l

= lower limit of modal class
-

h

= class width
-

f

₁ = frequency of modal class
-

f

₀ = frequency of class BEFORE modal class
-

f

₂ = frequency of class AFTER modal class

**Example**: Modal class 30–40 with

f

₁ = 12,

f

₀ = 5,

f

₂ = 8.

l = 30, h = 10

.
Mode =

30 + 10 \times (12 - 5) / (24 - 5 - 8) = 30 + \frac{70}{11}

≈ 36.36.

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