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NDA Math Statistics

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Question : 13 of 100

Marks: +1, -0

The mean of 5 observations is 4.4 and variance is 8.24. If three of the five observations are 1, 2 and 6, then what are the other two observations?

9, 16
9, 4
81, 16
81, 4

Solution:

Let x and y be the remaining two observations

Then

\overline{X}

= mean

=4.4

\Rightarrow \frac{1+2+6+x+y}{5}=4.4

\Rightarrow 9+x+y=22

\Rightarrow x+y=22-9=13

....(i)

and variance

=8.24

\Rightarrow \frac{1}{5}(1^{2}+2^{2}+6^{2}+x^{2}+y^{2})-\text{Mean}^{2}=8.24

\Rightarrow \frac{1}{5}(1+4+36+x^{2}+y^{2})-(4.4)^{2}=8.24

\Rightarrow \frac{1}{5}(41+x^{2}+y^{2})-19.36=8.24

\Rightarrow \frac{1}{5}(41+x^{2}+y^{2})=8.24+19.36=27.60

\Rightarrow 41+x^{2}+y^{2}=27.60\times 5=138

\Rightarrow x^{2}+y^{2}=138-41=97

Now,

(x+y)^{2}+(x-y)^{2}=2(x^{2}+y^{2})

\Rightarrow 169+(x-y)^{2}=2(97)

\Rightarrow (x-y)^{2}=194-169=25

\Rightarrow x-y=\pm 5

....(ii)

from (i) and (ii), we get, if

x-y=5

Then,

2x=18\Rightarrow x=9

and

2y=8\Rightarrow y=4

If

x-y=-5\therefore 2x=8\Rightarrow x=4

and

2y=18\Rightarrow y=9

Hence, the remaining two observations are 9 and 4

or 4 and 9

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