Question

The mean of 5 observations is \(4.4\) and their variance is \(8.24\). If three of the observations are \(1, 2\) and \(6\), then the other two observations are

• A. \(4, 9\)
• B. \(5, 8\)
• C. \(3, 10\)
• D. \(4, 10\)

Hint:

For MCQs involving mean and variance: \[ \sum x_i=n\bar{x} \] and \[ \sum x_i^2=n(\sigma^2+\bar{x}^2) \] These formulas quickly reduce the problem to solving simple equations.

Answer 1

The Correct Option is A

Step 1: Note what mean and variance give us.
The mean of 5 numbers is $4.4$, so their total is $5 \times 4.4 = 22$. The variance is $8.24$, which we will use in a moment. Three numbers are already known: $1, 2$ and $6$.

Step 2: Set up the sum of the two unknowns.
Call the missing numbers $a$ and $b$. Since everything adds up to $22$, we have $1 + 2 + 6 + a + b = 22$, which gives $a + b = 13$.

Step 3: Bring in the variance using the deviations.
Variance is the average of the squared distances from the mean. So $\sigma^2 = \frac{1}{5}\sum (x_i - 4.4)^2 = 8.24$, meaning $\sum (x_i - 4.4)^2 = 41.2$.

Step 4: Put in the three known deviations.
For $1$: $(1-4.4)^2 = 11.56$. For $2$: $(2-4.4)^2 = 5.76$. For $6$: $(6-4.4)^2 = 2.56$. Adding these gives $19.88$.

Step 5: Find the squared deviations of the unknowns.
So $(a-4.4)^2 + (b-4.4)^2 = 41.2 - 19.88 = 21.32$.

Step 6: Test the answer that fits both conditions.
Try $a = 4$ and $b = 9$ (their sum is $13$, good). Then $(4-4.4)^2 = 0.16$ and $(9-4.4)^2 = 21.16$, and $0.16 + 21.16 = 21.32$. Both conditions match perfectly.

So the other two observations are $4$ and $9$. \[ \boxed{4,\ 9} \]