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