# Normal distribution finding P area

Q: let X be a continuous random variable with NORMAL DENSITY

$$f(μ,σ(x)) = \frac{1}{\sqrt{2}π}*σ *e^{−(x−μ)^2/ 2σ^2}$$

We know that $μ = 70$ and $σ = 2$.

Find $P(68 \leq X \leq 74)$ and $P(X \geq 73)$:

my approach is ...
Since above is normal distribution..

$$P\left(\dfrac{a-μ}{σ} \leq Z \leq \dfrac{b-μ}{σ}\right) = P(1 \leq Z \leq 2) = P(Z\leq2) - P(Z\leq1)$$ so, by the table, we have $0.4772 - 0.3413$ .

Am I wrong ?

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Yes, there is a little mistake. You want $\Pr(Z\le 2)-\Pr(Z\lt -1)$.
This is because $\frac{68-70}{2}=-1$ ($68$ is to the left of the mean).
Your normal table (if you are using tables) may not tell you directly what $\Pr(Z\lt -1)$ is. But by symmetry it is $\Pr(Z\gt 1)$, which is $1-\Pr(Z\lt 1)$.
Remark: There is another error in your calculation. The probability that $Z\le 2$ is approximately $0.9772$, not $0.4772$. You seem to have used a table that gives the area to the right of the mean.
If you are going to use that table, you can calculate your ultimate answer by adding to $0.4772$ the probability given in your table for $1$. If you make a sketch you will see why this is true.