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The mean and variance of five observations are 4.4 and 8.24 respectively. If three of these are 1, 2 and 6, find the other two observations.

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So,

⇒ 41.2 = 19.88 + (x2 + 19.36 – 8.8x) + (y2 + 19.36 – 8.8y)

⇒ 41.2 – 19.88 = x2 + 19.36 – 8.8x + y2 + 19.36 – 8.8y 

⇒ 21.32 = x2 + y2 + 38.72 – 8.8(x + y) 

⇒ x2 + y2 + 38.72 – 8.8(13) – 21.32 = 0 [from (i)] 

⇒ x2 + y2 + 17.4 – 114.4 = 0 

⇒ x2 + y2 – 97 = 0 

⇒ x2 + y2 = 97 …(ii)

From eq. (i)

x + y = 17.4 

Squaring both the sides, we get 

(x + y)2 = (13)2 

⇒ x2 + y2 + 2xy = 169 

⇒ 97 + 2xy = 169 [from (ii)] 

⇒ 2xy = 169 – 97 

⇒ 2xy = 72 

⇒ xy = 36

⇒ X = \(\frac{36}{y}\) ....(iii)

Putting the value of x in eq. (i), we get 

x + y = 13

⇒   \(\frac{36}{y}+y\) = 13

⇒   \(\frac{36+y^2}{y}\) = 13

⇒ y2 + 36 = 13y

⇒ y2 – 13y + 36 = 0 

⇒ y2 – 4y – 9y + 36 = 0 

⇒ y(y – 4) – 9(y – 4)= 0 

⇒ (y – 4)(y – 9) = 0 

⇒ y – 4 = 0 and y – 9 = 0 

⇒ y = 4 and y = 9 

For y = 4

x = \(\frac{36}{y}\) = \(\frac{36}{4}\) = 9

Hence, x = 9, y = 4 are the remaining two observations 

For y = 9

x = \(\frac{36}{y}\) = \(\frac{36}{9}\) = 4

Hence, x = 4, y = 9 are the remaining two observations 

Thus, remaining two observations are 4 and 9.

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