# Find the probability of a Normal Distribution random variable

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.

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