1. Data set and mean

The five observations are $3,\;7,\;12,\;a,\;43-a.$

Their sum is $3+7+12+a+(43-a)=65.$ So the mean is constant: $\bar{x}=\frac{65}{5}=13.$

2. Variance formula

Using variance of a data set of size $5$: $\sigma^2=\frac{1}{5}\sum (x_i-\bar{x})^2.$

Now compute each deviation from the mean $13$:

• $3-13=-10 \Rightarrow (-10)^2=100$
• $7-13=-6 \Rightarrow (-6)^2=36$
• $12-13=-1 \Rightarrow (-1)^2=1$
• $a-13 \Rightarrow (a-13)^2$
• $(43-a)-13=30-a \Rightarrow (30-a)^2$

Hence $\sigma^2=\frac{1}{5}\left(100+36+1+(a-13)^2+(30-a)^2\right).$

So, $\sigma^2=\frac{1}{5}\left(137+(a-13)^2+(30-a)^2\right).$

3. Simplify

Expand: $(a-13)^2=a^2-26a+169$ and $(30-a)^2=a^2-60a+900.$

Therefore, $\sigma^2=\frac{1}{5}\left(137+a^2-26a+169+a^2-60a+900\right).$

$\sigma^2=\frac{1}{5}\left(2a^2-86a+1206\right).$

$\sigma^2=\frac{2}{5}(a^2-43a+603).$

4. Condition for variance to be a natural number

We need $\sigma^2=\frac{2}{5}(a^2-43a+603)\in \mathbb{N}.$ Since $2$ is coprime to $5$, this requires $a^2-43a+603 \equiv 0 \pmod{5}.$

Reduce modulo $5$: $a^2-43a+603 \equiv a^2-3a+3 \pmod{5}.$ So we need $a^2-3a+3 \equiv 0 \pmod{5}.$

Check all residue classes modulo $5$:

• $a\equiv 0$: $0-0+3=3 \not\equiv 0$
• $a\equiv 1$: $1-3+3=1 \not\equiv 0$
• $a\equiv 2$: $4-6+3=1 \not\equiv 0$
• $a\equiv 3$: $9-9+3=3 \not\equiv 0$
• $a\equiv 4$: $16-12+3=7\equiv 2 \not\equiv 0$

No residue works.

Thus there is no natural number $a$ for which the variance is a natural number.

5. Conclusion

Number of such values of $a$ is $0.$ So the correct option is A.