# Let $T$ be a normal random variable that describes the temperature…

Let $$T$$ be a normal random variable that describes the temperature in Rome on the 2nd of June. It is known that on this date the average temperature is equal to $$µ_T = 20$$ centigrade degrees and that $$P (T ≤ 25) = 0.8212$$.

How can I calculate the variance of $$T$$?

From $$P(T \leq 25)=0.8212$$, you can find the $$z$$-score of $$25$$ (reverse-lookup in a $$z$$-score table).
The $$z$$-score of $$25$$ is also given by $$z=\frac{25-\mu_T}{\sigma}$$.
Set these two expressions for the $$z$$-score equal to each other and solve for $$\sigma$$. Finally, square it to get $$\sigma^2$$.
• Yes. Since you found the z score, you know $0.92 = (25-\mu_T)/\sigma$. You already know $\mu_T$, and $\sigma$ is what you are trying to find. – bob.sacamento May 20 at 21:22
Consider the standard version of $$T$$, $$\Pr(Z \leq \alpha) = 0.8212$$ where we subtract by the mean and divide by standard deviation $$\sigma$$, as $$\Pr(T \leq 25) = \Pr(\underbrace{\frac{T - \mu_T}{\sigma}}_Z \leq \underbrace{\frac{25 - 20}{\sigma}}_\alpha) = 0.8212$$ Using the z table, you can easily find $$\alpha$$, which gives you $$\sigma$$.