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 $\sigma^2$.

  • $\begingroup$ Ok, so the z-score is 0,92? $\endgroup$ – user675557 May 20 at 20:49
  • $\begingroup$ @user675557 Yes, that's it. $\endgroup$ – bob.sacamento May 20 at 20:50
  • $\begingroup$ @bob.sacamento I have to put 0.92 equal to the z equation? $\endgroup$ – user675557 May 20 at 20:53
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    $\begingroup$ 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. $\endgroup$ – 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 $\sigma$.


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