Answer to Question #116013 in Statistics and Probability for Seifu

Question

Find the value of Z if the normal curve area between 0 and Z (positive) is 0.4332.

Answer 1

Initially, we have that the value of the integral of the normal distribution function is 0.4332. The normal distribution function can be expressed as $f(x)=\frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x-a)^2}{2\sigma^2}}.$

So, we have

$0.4332 = \frac{1}{\sigma\sqrt{2\pi}}\intop_0^zf(x)dx.$

Let's rewrite $x = a-\sigma z$, then

$0.4332 = \frac{1}{\sigma\sqrt{2\pi}}\intop_0^zf(x)dx = \frac{1}{\sigma\sqrt{2\pi}}\intop_0^ze^{(\frac{a - \sigma z - a)^2 }{2 \sigma^2}} d(a - \sigma z)$

We get that,

$0.4332 = \frac{1}{\sqrt{2\pi}}\intop_0^z e^{\frac{z^2 }{2}} dz,$

the last integral is laplace function and it is very nice, because all its values ​​are already calculated and you can see them, for example, here http://yunus.hacettepe.edu.tr/~ezgiyilmaz/files/tables.pdf.

As we can see the value of our integral reached when z is equal to 1.5.

Answer Z = 1.5