# Finding $\Phi(x)$ in normal distribution that is not in the table - linear interpolation intuitive explanation

I need help understanding something regarding to using the $$\Phi(x)$$ table for normal distribution calculation.

The question: For $$P(x)$$ a regular normal distribution ($$\mu = 0$$ and $$\sigma = 1$$), find: $$P\{x<0.1357\}$$

Solution attempt:

I found online the following $$\Phi(x)$$ table: https://math.ucalgary.ca/files/math/normal_cdf.pdf

$$0.1357$$ is between $$0.13$$ and $$0.14$$ so $$\Phi(0.13)<\Phi(0.1357)<\Phi(0.14)$$

The solution I was given to this was to perform linear interpolation and get: $$\Phi(0.1357) = \Phi(0.13) + 0.57\times(\Phi(0.14)-\Phi(0.13))$$

My question is - is there a simple and intuitive explanation to this and why it is true?

• It's not exactly true, but it is approximately. – Minus One-Twelfth Apr 6 '19 at 10:58
• Why is it a good approximation? Is there a simple/intuitive explanation? – PhysicsPrincess Apr 6 '19 at 11:00
• You are effectively connecting two nearby points on the graph of $y=\Phi(x)$ by a straight line and reading off the line to approximate the graph. Intuitively, since the two points are so close, the line connecting them is quite close to the actual curve $y=\Phi(x)$ (draw a picture to help see this). (In fact, you can find bounds on the error in this approximation too, see e.g. here, but this is more for formal stuff rather than simple intuitions). – Minus One-Twelfth Apr 6 '19 at 11:03
• It should be exact if function $\Phi$ would be linear on the interval. That is not the case but very close to it. – drhab Apr 6 '19 at 11:03
• Visually, the line segment connecting $(.13, \Phi(.13))$ and $(.14, \Phi(.14))$ is a good approximation to the graph of $\Phi$ between those two points. – littleO Apr 6 '19 at 11:07

As for why we obtain the formula: we draw a diagram like so and then take the ratio of distances on the top and the bottom like so - let's set $$x:=\Phi(0.1357)$$, then $$\frac{x-\Phi(0.13)}{\Phi(0.14)-\Phi(0.13)}=\frac{0.1357-0.13}{0.14-0.03},$$ and then solve for $$x$$.
You can also use calculus. $$\Phi(x_0 -dx) = \Phi(x_0) - \frac{d\Phi}{dx}\big|_{x=x_0} dx$$. You can find $$\frac{d\Phi}{dx}$$ by using the fundamental theorem of calculus.
\begin{align} \frac{d\Phi}{dx} &= \frac{2}{\sqrt\pi}\frac{d}{dx}\int_0^x e^{-t^2} dt \\ &= \frac{2}{\sqrt\pi} e^{-x^2} \end{align}
By plugging in $$x_0 = 0.14$$ and $$dx = 0.0043$$ we get
$$\Phi(0.1357) \approx \Phi(0.14) - 2/\sqrt\pi e^{-0.14^2} 0.0043 = \Phi(0.14) - 0.0048$$