1. Given data

The numbers are: $6,\;4,\;a,\;8,\;b,\;12,\;10,\;13$ There are $n=8$ observations.

Given:

• Mean $=9$
• Variance $=9.25$

We need to find: $a+b+ab$

2. Use the mean

Mean of 8 observations is 9, so total sum is: $8\times 9=72$

Now sum of the known terms: $6+4+8+12+10+13=53$

Hence, $a+b=72-53=19$

So we get: $a+b=19 \qquad (1)$

3. Use the variance

For ungrouped data, $\text{Variance} = \frac{\sum x_i^2}{n} - (\text{mean})^2$

Given variance $=9.25$ and mean $=9$, so: $9.25=\frac{\sum x_i^2}{8}-81$

Therefore, $\frac{\sum x_i^2}{8}=90.25$ $\sum x_i^2 = 8\times 90.25 = 722$

Now compute squares of known terms: $6^2+4^2+8^2+12^2+10^2+13^2$ $=36+16+64+144+100+169=529$

Thus, $a^2+b^2 = 722-529=193 \qquad (2)$

4. Find $ab$

We know: $(a+b)^2=a^2+b^2+2ab$

Using (1) and (2): $19^2=193+2ab$ $361=193+2ab$ $2ab=168$ $ab=84$

5. Compute required expression

$a+b+ab=19+84=103$

6. Check options

• A: $103$ ✅
• B: $106$
• C: $100$
• D: $105$

So the correct option is: $\boxed{103}$

7. Comparison with stored answer

Stored correct answer: A

Our derived answer is also A = 103, so it agrees.