A distribution consists of three components with frequencies $45, 40$ and $15$ having their means $2, 2.5, 2$ respectively. The mean of the combined distribution is

2.1

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2.2

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2.3

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Solution

The correct option is B $2.2$
Given, $f_{1} = 45$ , $f_{2} = 40$ , $f_{3} = 15$ , $x_{1} = 2$ , $x_{2} = 2.5$ and $x_{3} = 2$
Now, mean $=$ $\frac{x_{1} f_{1} + x_{2} f_{2} + x_{3} f_{3}}{f_{1} + f_{2} + f_{3}}$
$=$ $\frac{45 \times 2 + 40 \times 2.5 + 15 \times 2}{45 + 40 + 15}$
$=$ $\frac{90 + 100 + 30}{100}$
$=$ $\frac{220}{90} = 2.2$

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