Q: let X be a continuous random variable with NORMAL DENSITY
$$f(x;\mu,\sigma) = \frac{1}{\sigma\sqrt{2\pi}}e^{−(x−\mu)^2/ 2\sigma^2}$$
We know that $\mu = 70$ and $\sigma = 2$.
Find $P(68 \leq X \leq 74)$ and $P(X \geq 73)$:
my approach is ...
Since above is normal distribution..
$$ P\left(\dfrac{a-μ}{σ} \leq Z \leq \dfrac{b-μ}{σ}\right) = P(1 \leq Z \leq 2) = P(Z\leq2) - P(Z\leq1) $$
but this was wrong because the density function is not standard distribution so I could not use the table. How can I solve this ?
You are on the right track, but I think you wanted $P(-1\le Z\le2)$ instead of $P(1\le Z\le2)$; so this would give $P(-1\le Z\le0)+P(0\le Z\le2)=P(0\le Z\le1)+P(0\le Z\le2)$.
For the second part, you want to find $P(Z\ge1.5)=1/2-P(0\le Z\le1.5)$.