Find the height of a normal curve at a value

The value row, labelled $p$, shows $0.01613138163$. That is the height of the curve, so $f(115)\approx 0.01613$. On the curve picture, a dashed vertical line marks the height at $115$.

This is a density, not a probability or a percentage. For a continuous normal variable, $P(X=115)=0$, and a density can exceed $1$. To find the probability between two limits, use Normal CD instead: see find a normal probability.

To check the symmetry, tap the $x$ box, press AC once to empty it, then enter $85$. It is the same distance from the mean as $115$, so the screen shows the same height, $0.01613138163$.

• You see a value before entering the example: a fresh screen shows $0.3989422804$ for $x=0$, $\mu =0$ and $\sigma =1$. A returning screen can keep earlier values. Enter all three example values before reading the answer.
• The result becomes zero while you are typing: after entering $115$ for $x$, the result is $0$ if $\mu =0$ and $\sigma =1$ are still set. Finish entering $\mu =100$ and $\sigma =15$.
• The $\sigma$ box is dashed and the result shows Argument ERROR: the standard deviation must be positive. Tap the $\sigma$ box, press AC once to empty it, then enter $15$.