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# The mean and variance of seven observations are 8 and 16 , respectively. If six of the observations are 2, 4,10,12 and 14 , find the remaining two observations

**Solution:**

Let the remaining two observations be x and y

Therefore, the observations are 2, 4, 10, 12, 14, x, y

Mean,

x = (2 + 4 +10 + 12 + 14 + x + y)/7

8 = (42 + x + y)/7

42 + x + y = 56

x + y = 14 ....(1)

16 = 1/n ∑^{7}_{i = 1}(x_{i} - x)²

16 = 1/7 [(- 6)² + (- 4)² + (2)² + (4)² + (6)² + x² + y² - 2 x 8(x + y) + 2 x (8)²]

16 = 1/7 [36 + 16 + 4 + 16 + 36 + x² + y² - 16 (14) + 2 (64)]

16 = 1/7 [108 + x² + y² - 224 + 128]

16 = 1/7 [12 + x² + y²]

x^{2} + y^{2} = 100 ....(2)

From (1) , we obtain

x^{2} + y^{2} + 2xy = 196 ....(3)

From (2) and (3) , we obtain

2xy = 196 - 100

2xy = 96 ....(4)

Subtracting (4) from (2) , we obtain

x^{2} + y^{2} - 2xy = 100 - 96

(x - y)^{2} = 4

x - y = ± 2 ....(5)

Therefore, from (1) and (5) , we obtain

x = 8 and y = 6, when x - y = 2

x = 6 and y = 8, when x - y = - 2

Thus, the remaining observations are 6 and 8

NCERT Solutions Class 11 Maths Chapter 15 Exercise ME Question 2

## The mean and variance of seven observations are 8 and 16 , respectively. If six of the observations are 2, 4,10,12 and 14 , find the remaining two observations

**Summary:**

The mean and variance of seven observations are 8 and 16 , respectively. If six of the observations are 2, 4,10,12 and 14. Thus, the remaining two observations are 6 and 8

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