I have a set where its values follow a normal distribution, but I only have the sum of all of them, and the sum of its squares.
Having: $\sum_{i=0}^n X_i$ and $\sum_{i=0}^n X_i^2$.
I have the following "table":
((+ (fever 245 5975))
(- (fever 176 4540)))
There are two types of classes: Positive (+) and Negative(-). For both, I have the sum (245 and 176), and the sum of their squares (5975 and 4540).
How can I calculate P(Positive class given 45)?
P(Positive class given 45) or $P(Positive|45)$ means that given a fever of 45, it will be positive case and the patient will have measles. This is an invented example.
I think I have to use the mean and the variance, but I don't know how.
I'm going to hazard a guess as to what you mean, in ordinary terms. I'm going to guess that you have two normal distributions, with estimated means $\mu_c$ (class $c = 1, 2$) and standard deviations $\sigma_c$. These can be derived from the sum and sum of squares, provided that you know $n$ (for each class, if they're different).
The estimated mean is
$$ \mu = \frac1n \sum_{i=1}^n X_i $$
and the estimated standard deviation is
$$ \sigma = \sqrt{\frac{\left(\sum_{i=1}^n X_i^2\right) - n\mu^2}{n-1}} $$
We can then use the PDF of the normal distribution at a given value $x$ to compute the relative likelihoods of being positive or negative for the condition:
$$ f_c(x) = \frac{1}{\sqrt{2\pi\sigma_c^2}}\exp\left[-\frac{(x-\mu_c)^2}{2\sigma_c^2}\right] $$
Then I think the probability you want is
$$ P(\text{class $c$} \mid x) = \frac{p_cf_c(x)}{p_1f_1(x)+p_2f_2(x)} \qquad c = 1, 2 $$
where $p_c$ is the a priori probability of being in class $c$.