Let \(X_i\) denote a data point taken from the distribution, where \(1\le i\le40\), and let \(Y\) denote the average.
You want to find
\(\mathbb P\left(\displaystyle\frac1{40}\sum_{i=1}^{40}X_i<490\right)=\mathbb P(Y<490)\)
First, let's recall a few things. The PDF of a normal distribution with mean \(\mu\) and variance \(\sigma^2\) is
\(f(x;\mu,\sigma^2)=\displaystyle\frac1{\sigma\sqrt{2\pi}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)\)
Each of the \(X_i\) are presumably independently selected, so they are i.i.d. random variables.
The MGF of a normal distribution is
\(M_X(t)=\mathbb E(e^{tX})\)
\(M_X(t)=\displaystyle\int_{-\infty}^\infty e^{tx}f_X(x)\,\mathrm dx\)
\(M_X(t)=\exp\left(\mu t+\dfrac12\sigma^2t^2\right)\)
The MGF of a linear combination of i.i.d. random variables is
\(M_{c_1X_1+\cdots+c_nX_n}=M_{X_1}(c_1t)\times\cdots\times M_{X_n}(c_nt)=\displaystyle\prod_{i=1}^nM_{X_i}(c_it)\)
In this case, each \(c_i=\dfrac1{40}\). This product of MGFs reduces to an MGF of a normal distribution because the \(X_i\) are i.i.d..
\(M_Y(t)=\displaystyle\prod_{i=1}^{40}M_{X_i}(t)=\exp\left(\mu\left(\dfrac t{40}\right)+\frac12\sigma^2\left(\dfrac t{40}\right)^2\right)\times\cdots\times\exp\left(\mu\left(\dfrac t{40}\right)+\frac12\sigma^2\left(\dfrac t{40}\right)^2\right)\)
\(M_Y(t)=\exp\left(\mu t+\dfrac12\left(\dfrac\sigma{\sqrt{40}}\right)^2t^2\right)\)
which is indeed the MGF of a normal distribution with mean \(\displaystyle\mu\sum_{i=1}^{40}\frac1{40}=\mu\) and variance \(\sigma^2\displaystyle\sum_{i=1}^{40}\left(\frac1{40}\right)^2=\frac{\sigma^2}{40}\)
So, the PDF of \(Y\), given that \(\mu=485\) and \(\sigma=11.6\), is
\(f_Y(y)=\displaystyle\frac1{\frac{11.6}{\sqrt{40}}\sqrt{2\pi}}\exp\left(-\frac{(y-485)^2}{2\left(\frac{11.6}{\sqrt{40}}\right)^2}\right)\)
Now,
\(\mathbb P(Y<490)=\displaystyle\int_{-\infty}^{490}f(y)\,\mathrm dy\)
or, using the CDF of \(Y\),
\(\mathbb P(Y<490)=F_Y(490)\approx0.9968\)
