Question

The mean of five observations is 4 and their variance is $5.2$. If three of these observations are $2, 4$ and $6$, then the other two observations are

• A. $3$ and $5$
• B. $2$ and $6$
• C. $4$ and $4$
• D. $8$ and $10$
• E. $1$ and $7$

Hint:

Always use both sum and sum of squares when mean and variance are given.

Answer 1

The Correct Option is A

Concept:
• Mean: $\displaystyle \bar{x} = \frac{\sum x_i}{n}$
• Variance: $\displaystyle \sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$

Step 1: Total sum using mean
\[ \bar{x} = 4,\quad n = 5 \] \[ \sum x_i = 5 \times 4 = 20 \] Let the unknown numbers be $a$ and $b$. \[ 2 + 4 + 6 + a + b = 20 \] \[ 12 + a + b = 20 \Rightarrow a + b = 8 \]

Step 2: Use variance formula
\[ \sigma^2 = 5.2 \] \[ 5.2 = \frac{\sum x_i^2}{5} - 4^2 \] \[ 5.2 = \frac{\sum x_i^2}{5} - 16 \] \[ \frac{\sum x_i^2}{5} = 21.2 \] \[ \sum x_i^2 = 106 \]

Step 3: Compute known squares
\[ 2^2 + 4^2 + 6^2 = 4 + 16 + 36 = 56 \] \[ 56 + a^2 + b^2 = 106 \] \[ a^2 + b^2 = 50 \]

Step 4: Use identity
\[ (a+b)^2 = a^2 + b^2 + 2ab \] \[ 8^2 = 50 + 2ab \] \[ 64 = 50 + 2ab \Rightarrow 2ab = 14 \Rightarrow ab = 7 \]

Step 5: Solve quadratic
\[ x^2 - (a+b)x + ab = 0 \] \[ x^2 - 8x + 7 = 0 \] \[ (x-1)(x-7)=0 \] \[ a,b = 1,7 \] But check variance consistency carefully: Try options: \[ (3,5):\quad a+b=8,\ a^2+b^2=9+25=34 \neq 50 \] \[ (1,7):\quad a^2+b^2=1+49=50 \quad \checkmark \] Hence correct pair: \[ \boxed{1 \text{ and } 7} \] Correct option is (E)