What I would suggest is to look at square differences between your given $y_i$s and the value implied by a normal distribution at the corresponding $x_i$ value.
Lets call your pairs $(x_i,y_i)_{i=1}^N$, so for example $(x_1,y_1) = (1.380211242, 0.06984097)$. I'll also denote by $\Phi$ the cdf of a standard normal distribution, while $F_{\mu,\sigma}$ is the cdf of a normal distribution with parameters $\mu$ and $\sigma$. Then we have the relationship $F_{\mu,\sigma}(x) = \Phi(\frac{x - \mu}{\sigma})$.
Now I suggest to minimize the sum of square differences, i.e. define
\begin{align}
&D(\mu,\sigma) = \sum_{i=1}^N (y_i - F_{\mu,\sigma}(x_i))^2, \mbox{ and}\\
&\min_{\mu,\sigma} D(\mu,\sigma)
\end{align}
The minimization is carried out over the two parameters $\mu \in \mathbb{R}$ and $\sigma > 0$, and the optimal parameters (i.e the parameters that give the lowest value) will be your best (in the least square sense) estimates.
If you want to go there you can calculate the derivative of $D$ with respect to $\mu$ and $\sigma$, see https://stats.stackexchange.com/questions/154133/how-to-get-the-derivative-of-a-normal-distribution-w-r-t-its-parameters
Edit: running your values through the R-script below with your starting values of $2.8$ and $0.9$ gives me $\mu = 2.7750658, \sigma = 0.8981213$.
XX = c(
1.380211242,
1.62324929,
1.84509804,
2.133538908,
2.423245874,
2.73239376)
YY = c(
0.06984097,
0.09854223,
0.15019463,
0.227332755,
0.352771719,
0.481883449)
squareDiff <- function(param,dat){
x = dat\$x
y = dat\$y
m=param[1]
s=param[2]
D = sum((y - pnorm(x,mean=m,sd=s))^2)
return(D)
}
data = list(x=XX,y=YY)
result = optim(par=c(2.8,0.9), squareDiff, dat=data)
print(result)
