Question

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

Answer 1

Given:

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

To find:

The other two observations.

Solution:

Mean = 4.8

N = 5

Mean = ∑ x / N = 1 + 3 + 8 + x + y / 5

4.8 = 1 + 3 + 8 + x + y / 5

4.8 × 5 = 1 + 3 + 8 + x + y

24 = 12 + x + y

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

Variance = 6.56

$\dfrac{1}{n}\sum (x_i-\bar{x})^2 = 6.56$

1/5 ∑(xi - 4.8)² = 6.56

∑(xi - 4.8)² = 6.56 × 5

(1 - 4.8)² + (3 - 4.8)² + (8 - 4.8)² + (x - 4.8)² + (y - 4.8)² = 32.8

14.44 + 3.24 + 10.24 + x² + 23.04 - 9.6x + y² + 23.04 - 9.6y = 32.8

74 + x² + y² - 9.6 (x + y) = 32.8

From (1), we get,

74 + x² + y² - 9.6 (12) = 32.8

74 + x² + y² - 115.2 = 32.8

x² + y² = 74 ........(2)

Using (1), we have,

(x + y)² = 12²

x² + y² + 2xy = 144

74 + 2xy = 144

xy = (144 - 74) / 2

xy = 35

x = 35/y ..........(3)

Again using (1), we have,

x + y = 12

35/y + y = 12

35 + y² = 12y

y² - 12y + 35 = 0

solving the above quadratic equation, we get,

(y - 7) (y - 5) = 0

y = 5, 7

From (3), we get,

x = 35/y

when, y = 5, x = 35/5 ⇒ x = 7

when, y = 7, x = 35/7 ⇒ x = 5

Therefore, the remaining two observations are 5 and 7.