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Statistics — MATHS STD 9 — Question

Rajasthan BoardEnglish MediumSTD 9MATHSStatistics1 Mark
Question
Let m be the mid-point and I be the upper-class limit of a class in a continuous frequency distribution. The lower class limit of the class is:
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Answer

  1. 2m - I
Solution:
Let the lower limit = k
Midpoint = m
Upper limit = I
$\text{Mid-point}=\frac{\text{(Upper limit + lower limit)}}{2}$
$\text{m}=\frac{\text{(K + I)}}{2}$
$2\text{m}=\text{k + I}$
$\text{k}=\text{2m - I}$
Therefore, lower limit = 2m - I

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