Probability: Evaluation of normal distribution using a table

Question

How to evaluate values of a function $$\Phi(x)=\frac{1}{\sqrt {2\pi}}\int_{-\infty}^xe^{-t^2/2}dt$$ using the following table:

I am given the example $\Phi(1.56)=0.9406$

This example is easy because we can directly read the value (two decimal places). In the $x$ column, we find $1.5$, and than in the $x$ row we find $6$.

What would be the value of $\Phi(1.563),\Phi(1.567),...$

Also, how to evaluate $\Phi$ function of transcendental numbers using the table?

What would be the value of $\Phi(\pi),\Phi(e),...$

Answer 1

The usual practice is to interpolate the table values. Here is an example of a linear interpolation to find $\Phi(0.7535)$. We look up the adjacent entries on either side to $0.7535$, namely $\Phi(0.75) = 0.7734$ and $\Phi(0.76) = 0.7764$. Then we note that $0.7535$ is $35\%$ of the way along the interval from $0.75$ to $0.76$, so $\Phi(0.7535)$ is also the same proportion between $\Phi(0.75)$ and $\Phi(0.76)$. More concretely, we have $$\frac{0.7535 - 0.75}{0.76 - 0.75} = 0.35,$$ hence linear interpolation gives the corresponding relationship $$\frac{\Phi(0.7535) - \Phi(0.75)}{\Phi(0.76) - \Phi(0.75)} = 0.35.$$ Solving for $\Phi(0.7535)$ we get $$\begin{align*} \Phi(0.7535) &\approx (0.35)(\Phi(0.76) - \Phi(0.75)) + \Phi(0.75) \\ &= (0.35)(0.7764 - 0.7734) + 0.7734 \\ &= 0.77445. \end{align*}$$ The same concept applies to other values. In general, if we have a continuous function $f(x)$ that is reasonably linear on some interval $[a,b]$, then for any $c \in [a,b]$, we have $$\frac{f(c) - f(a)}{f(b) - f(a)} \approx \frac{c-a}{b-a},$$ or $$f(c) \approx \frac{c-a}{b-a} (f(b) - f(a)) + f(a).$$

Answer 2

Vertically find 1.5 and horizontally 6. This gives you 1.56 For $\pi$, $e$ and so on, use linear interpolation. Find $\Phi(3.14)$ and $\Phi(3.15)$ and interpolate. I hope you have the 2nd page if this table.

Answer 3

You're given $\Phi(1.56) = 0.9406$ and $\Phi(1.57)=0.9418.$

Those values are rounded, so that \begin{align} & 0.94055 \le \Phi(1.56)\le 0.94065 \\[10pt] \text{and } \quad & 0.94175\le\Phi(1.57)\le 0.94185. \end{align} On an interval as short as that from $1.56$ to $1.57$, the curve that is the graph of $\Phi$ is very nearly a straight line. So draw the straight line from $(1.56,\,\,\, 0.94055)$ to $(1.57,\,\,0.94175)$ and also the straight line from $(1.56,\,\,\, 0.94065)$ to $(1.57,\,\,0.94185)$ and conclude that the curve is between those.

The equation of the line from $(1.56,\,\,\, 0.94055)$ to $(1.57,\,\,0.94175)$ is $$ y = \frac{(0.94175)(x-1.56) + (0.94055)(1.57-x)}{1.57 - 1.56}. $$ Plug in $x=1.563$ and find $y$. Then do the same with the other line. Conclude that the correct value of $y$ is between those bounds.

This is called linear interpolation since you're approximating a curve with a straight line.

Answer 4

You know all the derivatives of the function, so it's easy to set up a Taylor expansion around the tabulated value. You an write $\Phi(1.563)$ as $\Phi(1.56 + h)$ with $h = 0.003$. a Taylor expansion then yields:

$$\Phi(1.56 +h) = 0.9406 + h f(1.56) + \frac{h^2}{2}f'(1.56) + \mathcal{O}(h^3)$$

where

$$f(x) = \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{x^2}{2}\right)$$

The derivative is:

$$f'(x) = -\frac{x}{\sqrt{2\pi}}\exp\left(-\frac{x^2}{2}\right)$$

What you then see is that the quadratic term can be ignored because the tabulated values aren't accurate enough to make this small correction relevant, you find that $\Phi(1.563)\approx 0.9410$.

If you only want to use the tabulated values, then linear interpolation using the table can be done in the way explained in heropup 's answer. In this case it doesn't make sense to go beyond linear interpolation. But in general, you can use Newton's method, which you can derive as follows. We define the difference operator $\Delta$ that acts on the tabulated function $\Phi(x)$ as $\Delta\Phi(x) = \Phi(x+r) - \Phi(x)$, where $r$ is the step size, in this case $r = 10^{-2}$. We can then write:

$$\Phi(x + n r) = (1+\Delta)^n \Psi(x) = \sum_{k=0}^{n}\binom{n}{k}\Delta^k\psi(x) = \Phi(x) + n \Delta\Phi(x) + \frac{n(n-1)}{2}\Delta^2\Phi(x)+\cdots$$

You can insert fractional values for $n$ to get interpolated values from the table.