$d_{i}$ is the deviation of a class mark $y_{i}$ from ‘a’ the assumed mean and $f_{i}$ is the frequency, if $M_{g} = x + \frac{1}{\sum f_{i}} (\sum f_{i} d_{i})$ , then x is

Lower limit

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Assumed mean

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Number of observations

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Class size

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Solution

The correct option is B Assumed mean
$M_{g} = a + \frac{1}{\sum f_{i}} (\sum f_{i} d_{i}) ∴ x = a i . e ., x = a s s u m e d m e a n .$

Suggest Corrections

Similar questions

¯ x = a + \frac{\sum f_{i} d_{i}}{\sum f_{i}}

d_{i}^{'} s

¯ ¯ ¯ x = A + \frac{\sum f_{i} d_{i}}{\sum f_{i}}, d_{i} =

(A) = 1100, \sum f_{i} d_{i} = 800

\sum f_{i} = 100

\bar{X} = a + \frac{Σ f_{i} d_{i}}{Σ f_{i}}

{d_{i}}^{'^{s}}

a

In the formula $¯ x = A + \frac{\sum f_{i} d_{i}}{\sum f_{i}}$ , the A stands for assumed mean.

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