Q.

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

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a

4 and 9

b

3 and 5

c

2 and 6

d

4 and 6

answer is A.

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Detailed Solution

Σxi=nx¯=5×4.4=22So     1+2+6+x4+x5=22          x4+x5=13                           (1)   σ2=1n∣∑x12−1n∑xi2⇒∑xi2=nσ2+x¯2=5(8.24+19.36)=138.⇒ 1+4+36+x42+x52=138x42+x52=97So x42+13−x42=97⇒x42+169+x42−26x4=97⇒ 2x42−26x4+72=0⇒ x42−13x4+36=0 x4=13±169−1442=13±52=9,4So x5=13−x42=4,9Hence, the other two nembers are 4 and 9.
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