\(\mathbb P(X\le104)=\mathbb P\left(\dfrac{X-98}6\le\dfrac{104-98}6\right)=\mathbb P(Z\le1)\)
The empirical rule states that approximately 68% of a normal distribution falls within one standard deviation of the mean, or \(\mathbb P(-1\le Z\le1)\approx0.68\). This means, by symmetry of the distribution, that \(\mathbb P(0\le Z\le1)\approx0.5\times0.68=0.34\)
Now, since the distribution is symmetric about the mean, you have \(\mathbb P(Z\le0)=0.5\).
So,
\(\mathbb P(Z\le1)=\mathbb P(Z\le0)+\mathbb P(0\le Z\le 1)\approx0.5+0.34=0.84\)
More precisely, this probability is around \(0.841345\).
