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Statistics Part 1

Section: Mathematics

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Question : 32 of 100

Marks: +1, -0

If the mean and variance of the following data:

6,10,7,13,a,12,b,12

\frac{37}{4}

respectively, then

(a-b)^2

[27 Jul 2021 Shift 1]

24
12
32
16

Solution:

Mean

=\frac{6+10+7+13+a+12+b+12}{8}=9

60+a+b=72

a+b=12

.......(1)

variance

=\frac{\sum\limits x_i^2}{n}-\left(\frac{\sum\limits x_i}{n}\right)^2=\frac{37}{4}

\sum\limits x_i^2=6^2+10^2+7^2

+13^2+a^2+b^2+12^2+12^2

=a^2+b^2+642

=\frac{a^2+b^2+642}{8}-(9)^2=\frac{37}{4}

=\frac{a^2+b^2}{8}+\frac{321}{4}-81=\frac{37}{4}

=\frac{a^2+b^2}{8}=81+\frac{37}{4}-\frac{321}{4}

=\frac{a^2+b^2}{8}=81-71

\therefore a^2+b^2=80

.......(2)

From

(1) a^2+b^2+2ab=144

80+2ab=144 \; \; \therefore 2ab=64

(a-b)^2=a^2+b^2-2ab

=80-64=16

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