5 The cumulative distribution function

We know that the normal probability density function $f\left(x\right)$ is given by the formula

$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{\text{e}}^{-{\left(x-\mu \right)}^2∕2{\sigma }^2}$

and so the cumulative distribution function $F\left(x\right)$ is given by the formula

$F\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{\int \; <!--nolimits\; -->}_{-\infty }^x{\text{e}}^{-{\left(u-\mu \right)}^2∕2{\sigma }^2}du$

In the case of the cumulative distribution for the standard normal curve, we use the special notation $\Phi \left(z\right)$ and, substituting 0 and 1 for $\mu$ and ${\sigma }^2$ , we obtain

$\Phi \left(z\right)=\frac{1}{\sqrt{2\pi }}{\int \; <!--nolimits\; -->}_{-\infty }^z{\text{e}}^{-u^2∕2}du$

The shape of the curve is essentially ‘ $S$ ’ -shaped as shown in Figure 9. Note that the curve runs from $-\infty$ to $+\infty$ . As you can see, the curve approaches the value 1 asymptotically.

Figure 9

Comparing the integrals

$F\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{\int \; <!--nolimits\; -->}_{-\infty }^x{\text{e}}^{-{\left(u-\mu \right)}^2∕2{\sigma }^2}du\text{and}\Phi \left(z\right)=\frac{1}{\sqrt{2\pi }}{\int \; <!--nolimits\; -->}_{-\infty }^z{\text{e}}^{-v^2∕2}dv$

shows that

$v=\frac{u-\mu }{\sigma }\text{and so}dv=\frac{du}{\sigma }$

and $F\left(x\right)$ may be written as

$F\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{\int \; <!--nolimits\; -->}_{-\infty }^{\left(x-\mu \right)∕\sigma }{\text{e}}^{-v^2∕2}\sigma dv$

$=\frac{1}{\sqrt{2\pi }}{\int \; <!--nolimits\; -->}_{-\infty }^{\left(x-\mu \right)∕\sigma }{\text{e}}^{-v^2∕2}dv=\Phi \left(\frac{x-\mu }{\sigma }\right)$

We already know, from the basic definition of a cumulative distribution function, that

$\text{P}\left(a<X<b\right)=F\left(b\right)-F\left(a\right)$

so that we may write the probability statement above in terms of $\Phi \left(z\right)$ as

$\text{P}\left(a<X<b\right)=F\left(b\right)-F\left(a\right)=\Phi \left(\frac{b-\mu }{\sigma }\right)-\Phi \left(\frac{a-\mu }{\sigma }\right).$

The value of $\Phi \left(z\right)$ is measured from $z=-\infty$ to any ordinate $z=z_1$ and represents the probability $\text{P}\left(Z<z_1\right)$ .

The values of $\Phi \left(z\right)$ start as shown below:

You should compare the values given here with the values given for the normal probability integral (Table 1 at the end of the Workbook). Simply adding 0.5 to the values in the latter table gives the values of $\Phi \left(z\right)$ . You should also note that the diagrams shown at the top of each set of tabulated values tells you whether you are looking at the values of $\Phi \left(z\right)$ or the values of the normal probability integral.

Exercises

1. If a random variable $X$ has a standard normal distribution find the probability that it assumes a value:
   1. less than 2.00
   2. greater than 2.58
   3. between 0 and 1.00
   4. between $-1.65$ and $-0.84$
2. If $X$ has a standard normal distribution find $k$ in each of the following cases:
   1. $\text{P}\left(X<k\right)=0.4$
   2. $\text{P}\left(X<k\right)=0.95$
   3. $\text{P}\left(0<X<k\right)=0.1$

Answer