1. Given data

The 5 observations are: $9, 25, a, b, c$ with $a,b,c\in \mathbb N$ and $a<b<c$

Also given:

• Mean $=18$
• Mean deviation about the mean $=4$
• Variance $=\dfrac{136}{5}$

We need to find: $2a+b-c$

2. Use the mean

For 5 observations, mean $18$ gives: $\frac{9+25+a+b+c}{5}=18$ $34+a+b+c=90$ $a+b+c=56 \qquad (1)$

3. Use the variance

Variance about the mean is $\frac{1}{5}\sum (x_i-18)^2=\frac{136}{5}$ So, $\sum (x_i-18)^2=136$

Now, $9-18=-9 \Rightarrow (-9)^2=81$ $25-18=7 \Rightarrow 7^2=49$ Thus, $81+49+(a-18)^2+(b-18)^2+(c-18)^2=136$ $130+(a-18)^2+(b-18)^2+(c-18)^2=136$ $(a-18)^2+(b-18)^2+(c-18)^2=6 \qquad (2)$

Since $a,b,c$ are integers, the only way three squares sum to $6$ is: $1^2+1^2+2^2=6$ So, $|a-18|, |b-18|, |c-18| = 1,1,2$

Hence $a,b,c$ must be chosen from: $16,17,19,20$ with deviations from 18 matching $2,1,1$.

4. Use the sum condition

From (1), $a+b+c=56$ Now test triples from $\{16,17,19,20\}$ having deviations $2,1,1$:

• $16+17+19=52$
• $16+19+20=55$
• $17+19+20=56$
• $16+17+20=53$

So the only possible values are: $a=17,\quad b=19,\quad c=20$ which also satisfy $a<b<c$.

5. Check mean deviation about the mean

Mean deviation about mean $18$ is $\frac{1}{5}\left(|9-18|+|25-18|+|17-18|+|19-18|+|20-18|\right)$ $=\frac{1}{5}(9+7+1+1+2)$ $=\frac{20}{5}=4$ This matches the given condition.

6. Compute required value

$2a+b-c=2(17)+19-20$ $=34+19-20=33$

7. Comparison with stored answer

Derived answer: $\boxed{33}$ Stored correct answer: $33$

They agree.