Question

Suppose a normal distribution has a mean of 98 and a standard deviation of 6. What is P(x<=104)

Answer 1

$\mathbb P(X\le104)=\mathbb P\left(\dfrac{X-98}6\le\dfrac{104-98}6\right)=\mathbb P(Z\le1)$

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 $0.841345$.