Find the probability of a Normal Distribution random variable

Question

The excercise is given as it follows:

Let the Temperature $T$ during a month of a year has a normal distribution with mean $68°$ and a standard deviation of $6°$. Find the probability $p$ that the temperature is between the $70°$ and $80°$.

Sol: Since $T$ is a random variable with Normal Distribution and parameters $\mu = 68 , \sigma = 6 \text{ (or is it } \sigma^2?)$, then its pdf is given by: $$f_{_T}(x) = \frac{1}{6\sqrt{2\pi}}e^{\frac{-(x-68)^2}{72}}$$ Thus $p = \mathbb{P}(70 \leq T\leq 80) = \int_{70}^{80}f_{_T}(x)dx$.

This integral seems pretty difficult, so I'm wondering is there another way to find $p$. Maybe using the Normal Distribution tables or something likely.

Answer 1

Yes you will need to use a normal distribution table or a computer. And you are correct that $\sigma=6$. In order to use the normal table (which is for mean zero and variance $1$ normal distribution), you should standardize your random variable. Specifically, if $T$ is normal with mean $\mu$ and variance $\sigma^2$, then $Z := (T-\mu) / \sigma$ is normal with mean $0$ and variance $1$. Then things like $P(a \le Z \le b)$ can be computed from a normal table.

Answer 2

$$\int\limits_{70}^{80} \frac{e^{-(x-68)^2/(2 \cdot 6^2)}}{\sqrt{2 \pi} 6}\ dx = \frac{1}{2} \left(\text{erf}\left(\sqrt{2}\right)-\text{erf}\left(\frac{1}{3 \sqrt{2}}\right)\right) = 0.346691$$