Let $Y_{n}$ be the amount paid to the $n$-th policyholder assuming
that the claim is made. Let $\mathbb{I}_{n}$ be $0$ when the $n$-th
claim is not made and $1$ otherwise. Then, the total benefits paid
is
$$
X=\sum_{n=1}^{32}\mathbb{I}_{n}Y_{n}.
$$
We need to calculate
$$
\mathbb{E}\left[X\right]=\mathbb{E}\left[\sum_{n=1}^{32}\mathbb{I}_{n}Y_{n}\right]=\sum_{n=1}^{32}\frac{1}{6}\int_{0}^{1}yf\left(y\right)dy.
$$
I think you can do the rest yourself.
You're dealing with multiple issues here - one deals with an expected value calculation given truncation (probability), and another deals with estimation of parameters using data (statistics). Given that you haven't specified a method for estimation of your parameters, my guess is that there's a lot of background that you're missing.
First of all, assuming that $\mu$ and $\sigma^2$ are already specified, if you have a random variable $X \sim \mathcal{N}(\mu, \sigma^2)$ and you wish to truncate $X$ so that its support is limited to the interval $(1.1, 6.6)$, if we let $Y = X \mid (1.1 < X < 6.6)$ (read as "$X$ given $X$ is greater than $1.1$ and less than $6.6$"), then you would have to compute the PDF of $Y$, which would be
$$f_{Y}(y) = \dfrac{f_{X}(y)}{\mathbb{P}(1.1 < X < 6.6)} = \dfrac{f_{X}(y)}{\Phi\left( \dfrac{6.6-\mu}{\sigma}\right) - \Phi\left( \dfrac{1.1-\mu}{\sigma}\right)}\text{ for } 1.1 < y < 6.6$$
where $f_{X}(y) = \dfrac{1}{\sigma \sqrt{2\pi}}e^{-(y-\mu)^2/(2\sigma^2)}$ and
$$\Phi(x) = \int_{-\infty}^{x}\dfrac{1}{\sqrt{2\pi}}e^{-t^2/2}\text{ d}t$$
is the $\mathcal{N}(0, 1)$ (standard normal) cumulative distribution function.
Thus,
$$\mathbb{E}[Y] = \int_{1.1}^{6.6}yf_{Y}(y)\text{ d}y = \int_{1.1}^{6.6}y \cdot \dfrac{\dfrac{1}{\sigma \sqrt{2\pi}}e^{-(y-\mu)^2/(2\sigma^2)}}{\Phi\left( \dfrac{6.6-\mu}{\sigma}\right) - \Phi\left( \dfrac{1.1-\mu}{\sigma}\right)}\text{ d}y\text{.}$$
This will likely have to be computed using numerical approximation.
Second of all, you have a data set and you seem to suggest estimating $\mu$ and $\sigma^2$ using this data set. There is no one "correct" way to estimate $\mu$ and $\sigma^2$ with a provided data set. Two estimation procedures include the method of moments (MOM) and the method of maximum likelihood (MLE). Depending on the method you choose, your values for $\mu$ and $\sigma$ may vary. However, in the case of both of these methods, for the non-truncated normal distribution, the estimators for $\mu$ and $\sigma^2$ are identical (see MOM, MLE).
Best Answer
Let's define, for ease of notation, $Z=\lfloor X \rfloor$.
To get the expectation, let's first find the distribution law of $Z$.
Let $n\in\mathbb{N}_0$. Then $P(Z=n)=P(n\leq X < n+1)=\int_{n}^{n+1}\frac{1}{2}e^{-x}dx=\frac{e-1}{2e^{n+1}}$.
Because of the symmetry of the function $\frac{1}{2}e^{-|x|}$ around $0$, we have that $P(Z=-n-1)=P(-n-1\leq X<-n)=P(n\leq X < n+1)=P(Z=n)$.
Now we can simply calculate the expectation: \begin{align} E(Z) &= \sum\limits_{k=-\infty}^{+\infty}kP(Z=k)\\ &=\sum\limits_{n=0}^{+\infty}nP(Z=n)+\sum\limits_{k=-\infty}^{-1}kP(Z=k)\\ &=\sum\limits_{n=0}^{+\infty}nP(Z=n)+\sum\limits_{n=0}^{+\infty}(-n-1)P(Z=-n-1)\\ &= \sum\limits_{n=0}^{+\infty}nP(Z=n)+\sum\limits_{n=0}^{+\infty}(-n-1)P(Z=n)\\ &= -\sum\limits_{n=0}^{+\infty}\frac{e-1}{2e^{n+1}} = \frac{1-e}{2e}\cdot\frac{1}{1-\frac{1}{e}}\\ &=-\frac{1}{2} \end{align}