Assertion (A): For a frequency distribution with unequal class intervals, the histogram is drawn with the area of the rectangle proportional to the class frequency. Reason (R): When class widths are unequal, the frequency density must be used to construct the histogram. Choose the correct option:
A. A is false, R is true
B. Both A and R are true, and R is the correct explanation of A
C. Both A and R are true, but R is not the correct explanation of A
D. A is true, R is false
Answer: Option B
Solution (By JKSSB Mock Tests)
In a histogram, area (not height) represents frequency; if intervals are unequal, height = frequency / width (frequency density). So R explains A.
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