Find the normal distribution given two values

Question

Let's suppose that $W$ is the temperature at noon tomorrow. Let a, b and c be such that

$\mathbb{P}(W\leq a)=0.5$

$\mathbb{P}(W\leq b)=0.25$

I'm gonna choose specific values for a, b and c, say $a=77$ and $b=68$ (so, for example, the probability that tomorrow's temperature at noon is $77°F$ is $50%$). Now I need to find the normal distribution that adjusts to the values that I assigned to $a$ and $b$, yet I'm not sure how to do it. I understand that $a$ and $b$ can be seen as quantiles of W, but given that I choose them (almost at random), it doesn't seem clear how to use them to find the normal distribution.

Viewed in the context of bayesian statistics, it seems that we're assuming that $W$ has a normal distribution and somehow we need to use our new information about $a$ and $b$ to find the parameters of that normal distribution.

Answer 1

You can standarize your temperature distribution. Let $W\sim N(\mu,\sigma^ 2)$ and $Z\sim N(0,1)$, then $\frac{W-\mu}{\sigma}\sim Z$. So $$P(W\leq a) = P\left(\frac{W-\mu}{\sigma}\leq \frac{a-\mu}{\sigma}\right) = P\left(Z\leq \frac{a-\mu}{\sigma}\right)=0.5.$$

Given that the median of the standard normal is the value $z_m=0$, then $$\frac{a-\mu}{\sigma}=0,$$ so $a=\mu$. (This was easy-ish since you want $a$ that leaves half the probability at each side, and the normal is symmetric, so it is the mean.)

For $b$, same reasoning: $$P(W\leq b) = P\left(\frac{W-\mu}{\sigma}\leq \frac{b-\mu}{\sigma}\right) = P\left(Z\leq \frac{b-\mu}{\sigma}\right)=0.25.$$

The first quartile for the normal is $z_{q_1}=-0.67$, so $$\frac{b-\mu}{\sigma}=-0.67,$$ therefore $b=\mu-0.67\sigma$.

Answer 2

A normal distribution $N(\mu,\sigma^2)$ has two parameters, the mean $\mu$ and the variance $\sigma^2$. Thus, if you are given the value of its CDF at two points, you have enough information to solve for $\mu$ and $\sigma^2$.

In this case, things are simpler since the probabilities are nice. $\mathbb P(W\leq a)=1/2$ implies that $a$ is at the center of the normal distribution, so by symmetry the mean must be $a$.

Now that $\mu=a$ is known, we just need to find $\sigma$. Consider $Z=(W-a)/\sigma$, which is a standard normal random variable. Therefore $\mathbb P(W\leq b)$ can be converted to a Z-score, since $\mathbb P(W\leq b)=\mathbb P(Z\leq \frac{b-a}{\sigma})=.25$. This allows you to look up the numerical value of $(b-a)/\sigma$ in a table of Z-scores, and since $a$ and $b$ are given we can solve for $\sigma$.