quick distribution help please!

**mathaction** · November 15th 2007, 10:17 PM

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.

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)

=
(b)

=

some help...

**CaptainBlack** · November 15th 2007, 10:32 PM

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.

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)

=
(b)

=

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

**CaptainBlack** · November 15th 2007, 10:37 PM

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

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