Question #116013
Find the value of Z if the normal curve area between 0 and Z (positive) is 0.4332.
1
Expert's answer
2020-05-18T19:57:30-0400

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)=1σ2πe(xa)22σ2.f(x)=\frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x-a)^2}{2\sigma^2}}.

So, we have

0.4332=1σ2π0zf(x)dx.0.4332 = \frac{1}{\sigma\sqrt{2\pi}}\intop_0^zf(x)dx.

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

0.4332=1σ2π0zf(x)dx=1σ2π0ze(aσza)22σ2d(aσz)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=12π0zez22dz,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



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