Let $X$ be a random variable that is normally distributed and $X_1,\ldots,X_n$ be (independet) copies of $X$, then we can estimate this probability by using a simple Monte-Carlo estimator: $p := P (X \ge 5) =E(\xi_{X\ge5}) \approx \frac{1}{100000} \sum_{i=1}^{100000} \xi_{X_i \ge 5}.$
But the problem is $P(X \ge 5)$ is a comparably rare event. So I am supposed to find a way to estimate $P(X \ge 5)$ by using a new measure $dP_s = e^{s S_n - n \gamma(s)} dP,$ where $S_n= X_1+\cdots+X_n$ and $\gamma(s) = \frac{s^2}{2}.$ Of course, $dP = \frac{1}{(2 \pi)^{\frac{n}{2}}} \exp\left(-\frac{x_1^2+\cdots+x_n^2}{2} \right) dx$ is the respective normal distribution
$s:=\gamma^*(5)= \frac{25}{2},$ where $\gamma^*$ is the Legendre-Fenchel transform of $\gamma$. (So you can regard $s$ as being fixed).
Does anybody know how the new estimator could look like( it is supposed to be the same estimator, but should somehow use this new measure)? I don't know how to transform this old event into this new frame with the new measure. Does anybody know this?
If anything is unclear, please let me know.
By definition, $P(X_1 \geq 5) = e^{\gamma(s)} E_{P_s}[e^{-sX_1}\mathbf{1}_{X_1 \geq 5}]$ and this latter expectation can be estimated by the Monte-Carlo method (the random variables $X_1,\dots,X_n$ are still independent): $$ \frac{1}{n}\sum_{k=1}^n e^{-sX_k}\mathbf{1}_{\{X_k \geq 5\}} \xrightarrow[n\to\infty]{P_s-a.s.} E_{P_s}[e^{-sX}\mathbf{1}_{X \geq 5}]. $$ Finally, you need to simulate the random variables $X_1,\dots,X_n$ under the measure $P_s$ but this is easy enough since they still have a normal distribution.
By the way, I think that you should take $s=5$ so that $E_{P_s}[X_1]=\gamma'(s)=5$.
Using some simple facts about normal distributions, you should be able to derive a somewhat better estimator: $$ \frac{1}{2}e^{-\frac{25}{2}}\frac{1}{n}\sum_{k=1}^n e^{-5|X_k|} \xrightarrow[n\to\infty]{P-a.s.} P(X \geq 5) $$ With $n=100000$, I obtain $P(X \geq 5) \approx 2.86\,10^{-7}$.