probability in normal density function

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 ?

• Wikipedia / Normal Distribution gives you the Formula for adapting to the standard normal distribution. – AlexR Aug 25 '13 at 14:32
• First, the formula you gave for the normal density is incorrect and I have corrected it. For the calculation of the probabilities, see this answer for how to work such problems. – Dilip Sarwate Aug 25 '13 at 16:28

1 Answer

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 $P(Z\ge1.5)=1/2-P(0\le Z\le1.5)$.