If {x} and {y} are the means of two distributions such that {x} [NDA II 2014]

Correct Answer is : x‾<z‾<y‾{x} < {z} < {y}x<z<y

Correct Answer is : x‾<z‾<y‾{x} < {z} < {y}x<z<y

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Question : 86 of 100

Marks: +1, -0

\overline{x}

\overline{y}

\overline{x} < \overline{y}

\overline{z}

Solution:

Let the first distribution has

N_1

terms and

the second distribution has

N_2

terms.

Hence,the combined means

\overline{z}=\frac{\overline{N_1x}+\overline{N_2y}}{N_1+N_2}

=\overline{x}+\frac{N_2}{N_1+N_2}(\overline{y}-\overline{x})

>\overline{x}(\because \overline{y}-\overline{x}>0)\dots(1)

Also

\overline{z}=\frac{\overline{N_1x}+\overline{N_2y}}{N_1+N_2}

=\overline{y}-\frac{N_1}{N_1+N_2}(\overline{y}-\overline{x})

< \overline{y}(\because \overline{y}-\overline{x}>0)

.....(2)

From 1 and 2 we get

\overline{x} < \overline{z} < \overline{y}

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