The mean of 5 observations is 4 and their variance is 5.2. If three of those observations are 1, 2 and 6, the other two are:

Correct option is C

Given:
Number of observations n = 5
Mean = 4
Variance = 5.2
Three observations = 1, 2, 6
Formula used:
Mean = $$\frac{\sum x}{n}$$​​
​Variance = $$\frac{\sum x^{2}}{n} - (\text{mean})^{2}$$​​
Solution:
From mean:​
​$$\sum x = 5 \times 4 = 20$$​​
Sum of given three observations:
1 + 2 + 6 = 9
So, sum of remaining two observations:
= 20 − 9 = 11
From variance:
​$$5.2 = \frac{\sum x^{2}}{5} - 16\\\sum x^{2} = 5(5.2 + 16)\\= 5 \times 21.2= 106$$​​
Sum of squares of given observations:
​$$1^{2} + 2^{2} + 6^{2} = 1 + 4 + 36 = 41$$​​
So, sum of squares of remaining two observations:
= 106 − 41 = 65
Let the remaining observations be a and b.
Then:
a + b = 11
​$$a^{2} + b^{2} = 65$$​​
​$$(a + b)^{2} = a^{2} + b^{2} + 2ab\\121 = 65 + 2ab\\2ab = 56\\ab = 28$$​​
So,
​$$t^{2} − 11t + 28 = 0\\(t − 4)(t − 7) = 0$$​​
Hence, the two observations are 4 and 7.
The correct answer is (c) 4 and 7.