If the mean of variate X in distribution 2.6, then the value of k is
Variate X 1 2 3 4 5
Frequency of X 4 5 k 1 2

8

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10

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12

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18

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Solution

The correct option is A $8$
$\frac{1 \times 4 + 2 \times 5 + 3 \times k + 4 \times 1 + 5 \times 2}{4 + 5 + k + 1 + 2} = 2.6$
$\Rightarrow 0.4 k = 3.2 \Rightarrow k = 8$

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Variate x	1	2	3	4	5
Frequency of x	4	5	k	1	2

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Frequency	4	5	y	1	2

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Frequency of X	4	5	k	1	2

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\begin{matrix} V a r i a t e & 6 & 7 & 1.9 & 9 & 10 F r e q u e n c y & 1 & 2 & k & 3 & 4 \end{matrix}

3.6

k

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