population variance vs sample mean variance

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Central Limit Theorem states that the relationship between population variance and sample mean variance is $var(\bar{x})=\sigma^2/n$.

I'm trying to verify this with dataset (0,0,0,1,2,9) which has $\mu=2$ and $\sigma^2=10.33$ by conducting experiment with sample size of 2 with replacement $(x_1, x_2)$, and calculate the mean variance from all possible outcomes.

the result I got was:

$$var(\bar{x})=\frac{1}{36}\Sigma (\bar{x_i}-\mu)^2=\frac{186}{36}=5.17$$

$$5.17 \ne \frac{10.33}{36}=0.29$$

Could anyone help point out whether I get anything wrong in my calculation?