Question

The mean of five observations is 4 and their variance is 5.2. If three of these observations are 1, 2 and 6, then the other two are

• A. 2 and 9
• B. 3 and 8
• C. 4 and 7
• D. 5 and 6

Hint:

Save time by checking the options against your first derived relation ($a + b = 11$):
Option (A): $2 + 9 = 11$
Option (B): $3 + 8 = 11$
Option (C): $4 + 7 = 11$
Option (D): $5 + 6 = 11$
Since all options sum to 11, do a quick mental sum-of-squares check ($a^2 + b^2 = 65$):
For option (C): $4^2 + 7^2 = 16 + 49 = 65$. It matches instantly!

Answer 1

The Correct Option is C

Step 1: Set names for the unknowns.
There are five numbers. Three are 1, 2, 6. Call the other two $a$ and $b$. Mean is 4 and variance is 5.2.
Step 2: Use the mean.
Mean is the sum divided by 5: \[ \frac{1+2+6+a+b}{5} = 4 \;\Rightarrow\; 9 + a + b = 20 \;\Rightarrow\; a+b = 11 \quad (1) \]
Step 3: Write the variance formula.
\[ \sigma^2 = \frac{\sum x_i^2}{5} - (\text{mean})^2 \]
Step 4: Plug in the variance.
\[ 5.2 = \frac{1 + 4 + 36 + a^2 + b^2}{5} - 16 \] Add 16: $21.2 = \dfrac{41 + a^2 + b^2}{5}$. Multiply by 5: $106 = 41 + a^2 + b^2$, so \[ a^2 + b^2 = 65 \quad (2) \]
Step 5: Solve the two equations.
From (1), $b = 11 - a$. Put into (2): \[ a^2 + (11-a)^2 = 65 \;\Rightarrow\; 2a^2 - 22a + 121 = 65 \] \[ 2a^2 - 22a + 56 = 0 \;\Rightarrow\; a^2 - 11a + 28 = 0 \]
Step 6: Factor and read the answer.
\[ (a-4)(a-7) = 0 \;\Rightarrow\; a = 4 \text{ or } 7 \] If $a=4$ then $b=7$ (and the other way round). The two numbers are 4 and 7. \[ \boxed{4 \text{ and } 7} \]