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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