1. Given data

The 10 observations are: $65,68,58,44,48,45,60,\alpha,\beta,60$

Mean $=56$ and variance $=66.2$.

We need to find: $\alpha^2+\beta^2$

2. Use the mean

For 10 observations, $\frac{65+68+58+44+48+45+60+\alpha+\beta+60}{10}=56$

So the total sum is: $10\times 56=560$

Now sum the known terms: $65+68+58+44+48+45+60+60=448$

Hence, $\alpha+\beta=560-448=112$

3. Use the variance

Variance of a data set is $\sigma^2=\frac{\sum x_i^2}{n}-\bar{x}^2$

Given: $\sigma^2=66.2,\quad \bar{x}=56,\quad n=10$

Thus, $66.2=\frac{\sum x_i^2}{10}-56^2$

Since $56^2=3136,$ we get $\frac{\sum x_i^2}{10}=66.2+3136=3202.2$

Therefore, $\sum x_i^2=32022$

4. Compute sum of squares of known terms

$65^2=4225$ $68^2=4624$ $58^2=3364$ $44^2=1936$ $48^2=2304$ $45^2=2025$ $60^2=3600$ Another $60^2=3600$

So, $4225+4624+3364+1936+2304+2025+3600+3600=25678$

Hence, $\alpha^2+\beta^2=32022-25678=6344$

5. Final answer

$\boxed{6344}$

Since this matches the stored correct answer, the stored answer is correct.