Find the mean of the following distribution using step-deviation method:
Class: 50-52, 53-55, 56-58, 59-61, 62-64
Frequency: 15, 110, 135, 115, 25
Step-by-Step Solution
Key Concept: Continuous classes: 49.5-52.5, 52.5-55.5, 55.5-58.5, 58.5-61.5, 61.5-64.5 ($h = 3$).<br>Class marks $x_i$: 51, 54, 57, 60, 63. Let $a = 57$.<br>$u_i = (x_i - 57)/3$: -2, -1, 0, 1, 2.<br>$f_i u_i$: -30, -110, 0, 115, 50. $\sum f_i u_i = 25$. $\sum f_i = 400$.<br>$\bar{x} = 57 + 3(25/400) = 57 + 3/16 = 57 + 0.1875 = 57.19$.
Class marks $x_i = [51, 54, 57, 60, 63]$. Let $a = 57, h = 3$. [1.0 Mark]
$u_i = [-2, -1, 0, 1, 2] \Rightarrow \sum f_i u_i = 25, \sum f_i = 400$. [1.0 Mark]
$\bar{x} = 57 + 3(25/400) = 57 + 0.1875 = 57.19$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Finding class marks & assumed mean $a = 57$: 1.0 Mark
Calculating $\sum f_i u_i = 25$: 1.0 Mark
Evaluating mean $= 57.19$: 1.0 Mark
Correct Answer: