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Random variable

Uniman

New member
Oct 19, 2012
11
Screen Shot 2012-10-31 at 6.49.47 PM.png

Work done so far.....

Integrating from 0 to infinity and equating it to 1, we get

(c/2*10^-3) = 1

c= 2/1000

=0.002

Is it correct?

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Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
View attachment 432

Work done so far.....

Integrating from 0 to infinity and equating it to 1, we get

(c/2*10^-3) = 1

c= 2/1000

=0.002

Is it correct?

Report Abuse
Hi Uniman, :)

Yes the method you have used is correct.

\[\int_{0}^{\infty}C\,\mbox{exp}\left(-\frac{2.1x}{1000}\right)dx=1\]

\[\Rightarrow C\left[-\frac{1000}{2.1}\mbox{exp}\left(-\frac{2.1x}{1000}\right)\right]^{\infty}_{0}=1\]

\[\Rightarrow \frac{1000}{2.1}C=1\]

\[\therefore C=\frac{2.1}{1000}=0.0021\]

Kind Regards,
Sudharaka.