Find the mean of the following distribution using short-cut method:
C.I $35 - 40$ $40 - 45$ $45 - 50$ $50 - 55$ $55 - 60$
F $7$ $6$ $9$ $5$ $3$

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Solution

The average of each class $= 37.5, 42.5, 47.5, 52.5, 57.5$
Frequencies $= 7, 6, 9, 5, 3$
Mean of the distribution = $\frac{\sum (C . I \times F_{i})}{F_{i}}$
$M e a n = \frac{7 \times 37.5 + 6 \times 42.5 + 9 \times 47.5 + 5 \times 52.5 + 3 \times 57.5}{30} = 46$

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