1. Given data

The six observations are: $7,\ 10,\ 11,\ 15,\ a,\ b$

Mean $=10$ and variance $=\dfrac{20}{3}$.

We need to find: $|a-b|$

2. Use the mean

For 6 observations, if the mean is 10, then the sum is: $6\times 10=60$

So, $7+10+11+15+a+b=60$ $43+a+b=60$ $a+b=17$

3. Use the variance

Variance of observations is: $\sigma^2=\frac{\sum x_i^2}{n}-\bar{x}^2$

Here, $\sigma^2=\frac{20}{3},\quad \bar{x}=10,\quad n=6$

So, $\frac{7^2+10^2+11^2+15^2+a^2+b^2}{6}-10^2=\frac{20}{3}$

First compute known squares: $7^2=49,\quad 10^2=100,\quad 11^2=121,\quad 15^2=225$

Their sum is: $49+100+121+225=495$

Thus, $\frac{495+a^2+b^2}{6}-100=\frac{20}{3}$

Add 100 to both sides: $\frac{495+a^2+b^2}{6}=100+\frac{20}{3}=\frac{320}{3}$

Multiply by 6: $495+a^2+b^2=640$ $a^2+b^2=145$

4. Use identity relating $a+b$ and $a^2+b^2$

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

Substitute values: $17^2=145+2ab$ $289=145+2ab$ $2ab=144$ $ab=72$

So $a$ and $b$ are roots of: $x^2-(a+b)x+ab=0$ $x^2-17x+72=0$

Factorizing: $(x-8)(x-9)=0$

Hence, $a,b=8,9$

Therefore, $|a-b|=|8-9|=1$

5. Check options

• A: $9$ ❌
• B: $11$ ❌
• C: $7$ ❌
• D: $1$ ✅

So the correct option is D.

6. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.