Test Index

NDA Math Statistics

Question : 77 of 100

Marks: +1, -0

Solution:

Consider the following the given data,we get

Mean

=E(x)

=\frac{\sum\limits_{i=1}^{10} x_i}{10}=4

=\operatorname{Var}(x)=E(x^2)-E(x)^2

=E(x^2)-4^2=2

\therefore E(x)^2=18

Hence the new mean will be,

E(x)'=\frac{2x_1+2x_2+2x_3\ldots+2x_{10}}{10}

=2\left(\frac{x_1+x_2+x_3\ldots x_{10}}{10}\right)

=2\left(\frac{\sum\limits_{i=1}^{10}x_i}{10}\right)

=8

Now the new mean of the square of the observation will be

E(x^2)'

E(x^2)'=\frac{(2x_1)^2+(2x_2)^2+(2x_3)^2\ldots+(2x_{10})^2}{10}

=\frac{4(x_1^2+x_2^2+\dots+x_{10}^2)}{10}

=4E(x^2)

=4(18)

=72

Hence the new variance will be equal to

\operatorname{Var}(x)'

={E(x^2)}'-{E(x)^2}'

72-8^2

72-64

=8

Therefore the mean and variance of the new set of data are 8 and 8 respectively.

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