• Nov 15th 2007, 11:17 PM
mathaction
The figure below shows the standard normal distribution from statistics, which is given by the following equation.
Statistics books often contain tables such as the following (below right), which show the area under the curve from 0 to b for various values of b.
http://www.webassign.net/hgmcalc/5-4-67.gif bArea1.34132.47723.49874.5000
Use the information given in the table and the symmetry of the curve about the y-axis to find the following.
(a) http://www.webassign.net/www26/symIm...0924648699.gif =
(b) http://www.webassign.net/www26/symIm...efff8af93c.gif =

some help...
• Nov 15th 2007, 11:32 PM
CaptainBlack
Quote:

Originally Posted by mathaction
The figure below shows the standard normal distribution from statistics, which is given by the following equation.
Statistics books often contain tables such as the following (below right), which show the area under the curve from 0 to b for various values of b.
http://www.webassign.net/hgmcalc/5-4-67.gif bArea1.34132.47723.49874.5000
Use the information given in the table and the symmetry of the curve about the y-axis to find the following.
(a) http://www.webassign.net/www26/symIm...0924648699.gif =
(b) http://www.webassign.net/www26/symIm...efff8af93c.gif =

some help...

a) $\frac{1}{\sqrt{2 \pi}} \int_1^2 e^{-~\frac{x^2}{2}} ~dx$ $
=~\frac{1}{\sqrt{2 \pi}} \int_0^2 e^{-~\frac{x^2}{2}} ~dx-\frac{1}{\sqrt{2 \pi}} \int_0^1 e^{-~\frac{x^2}{2}} ~dx
$

That is the required area is written as the difference between two areas both of which are of the form given in the table.

RonL
• Nov 15th 2007, 11:37 PM
CaptainBlack
b) Again write the area as the combination of two areas with $x=0$ as one end point:

$\frac{1}{\sqrt{2 \pi}} \int_{-3}^2 e^{-~\frac{x^2}{2}} ~dx$ $
=~\frac{1}{\sqrt{2 \pi}} \int_{-3}^0 e^{-~\frac{x^2}{2}} ~dx+\frac{1}{\sqrt{2 \pi}} \int_0^2 e^{-~\frac{x^2}{2}} ~dx$

Now use the symmetry about $x=0$ to change the first of the integrals abouve into the form given in the table:

$
\frac{1}{\sqrt{2 \pi}} \int_{-3}^0 e^{-~\frac{x^2}{2}} ~dx=
\frac{1}{\sqrt{2 \pi}} \int_{0}^3 e^{-~\frac{x^2}{2}} ~dx
$

So:

$\frac{1}{\sqrt{2 \pi}} \int_{-3}^2 e^{-~\frac{x^2}{2}} ~dx$ $
=~\frac{1}{\sqrt{2 \pi}} \int_{0}^3 e^{-~\frac{x^2}{2}} ~dx+\frac{1}{\sqrt{2 \pi}} \int_0^2 e^{-~\frac{x^2}{2}} ~dx$

RonL