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NCERT Class XI Mathematics - Statistics - Solutions

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Question : 29 of 34
Marks: +1, -0
The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations.
Solution:  
SLet the remaining two observations be x and y. Then we are given that
2+4+10+12+14+x+y7\frac{2+4+10+12+14+x+y}{7} = 8
⇒ 42 + x + y = 56 ⇒ x + y = 14 ... (i)
Also ,
17(22+42+10+122+142+x2+y2)−(8)2\frac{1}{7}(2^2+4^2+10^{+}12^2+14^2+x^2+y^2)-(8)^2
= 16
⇒
17(4+16+100+144+196+x2+y2)−64\frac{1}{7}(4+16+100+144+196+x^2+y^2)-64
= 16
⇒ 460 + x2+y2x^2+y^2 = 560 ⇒ x2+y2x^2+y^2 = 100 ... (ii)
Now (x+y)2+(x−y)2(x + y)^2 + (x - y)^2 = 2(x2+y2)2(x^2 + y^2)
⇒ (14)2+(x−y)2(14)^2 + (x - y)^2 = 2 × 100 [Using (i) & (ii)]
⇒ (x−y)2(x - y)^2 = 200 – 196 ⇒ (x−y)2(x - y)^2 = 4 ⇒ x – y = ± 2
When x – y = 2 and x + y = 14, we get x = 8 and y = 6
When x – y = –2 and x + y = 14 we get x = 6 and y = 8.
So, the remaining two observations are 6 and 8.
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