Question

The mean of 5 observation is 4.8 and the variance is 6.56. if three of the five observations are 1,3 and 8, find the other two observations

Answer 1

The mean of 5 observation is 4.8 and the variance is 6.56. three of the five observations are 1,3 and 8,

Given,

Let the two observations be a and b.

Mean = sum of observations / number of observations

M = ∑ xi / N

4.8 = ( 1 + 3 + 8 + a + b ) / 5

24 = 12 + a + b

a + b = 24 - 12

∴ a + b = 12 ..........(1)

Variance

$V = \dfrac{1}{n}\sum (x_i-\bar{x})^2$

6.56 = 1/5 [ (1 - 4.8)² + (3 - 4.8)² + (8 - 4.8)² + (a - 4.8)² + (b - 4.8)² ]

32.8 = 14.44 + 3.24 + 10.24 + (a - 4.8)² + (b - 4.8)²

(a - 4.8)² + (b - 4.8)² = 4.88

a² + 4.8² - 2 (a) (4.8) + b² + 4.8² - 2 (b) (4.8) = 4.88

a² + b² - 2 (4.8) (a + b)= 4.88 - 4.8² - 4.8²

using (1) we get,

a² + b² - 2 (4.8) (12) = -41.2

a² + b²  = -41.2 + 2 (4.8) (12) = 74

∴  a² + b²  = 74 ............(2)

consider equation (1),

(a + b)² = 12²

a² + b² + 2ab = 144

74 + 2ab = 144

ab = (144 - 74) / 2

ab = 35

a = 35/b  ...........(3)

substituting equation (3) in (1), we get,

35/b + b = 12

35 + b² = 12b

b² - 12b + 35 = 0

(b - 5) (b - 7) = 0

b = 5, 7

a = 35/b

a = 35/5 = 7  or a = 35/7 = 5

Therefore, when b = 7, a = 5  and when b = 5, a = 7

The other two observations are 5 and 7.