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Find P(x≤6) where f(x)=1/8 e^(-x/8) and e=2.71828 Please show each step, I can't seem to figure this out. Thank you

Ann
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    Note that if $b\gt 0$ then $\Pr(X\le b)=\int_0^b \frac{1}{8}e^{-t/8},dt$. – André Nicolas May 22 '14 at 19:06
  • I still don't understand. I'm sorry. The answer given in my text is .5276 – Ann May 22 '14 at 19:18
  • Do you know how to integrate the function $\frac{1}{8}e^{-t/8}$? – André Nicolas May 22 '14 at 19:21
  • No, I guess not because I'm not sure what you mean. Why do you have a t, when the original equation uses x? I know x is greater than or equal to 0 if that's what you mean. – Ann May 22 '14 at 19:24
  • The dummy variable of integration doesn't matter. An antiderivative of our function is $-e^{-t/8}$. Plug in $b$, take away the result of plugging in $0$. We get $1-e^{-b/8}$. For your problem, set $b=6$. – André Nicolas May 22 '14 at 19:27
  • Right so, 1-2.71828 to the -6/8 power right? I thought I knew how to compute negative fractional exponents, but I can't get this problem to equal .5276 which means I'm doing something wrong. – Ann May 22 '14 at 19:32
  • We want $1-e^{-6/8}$. Use your calculator. Using mine I get $e^{-6/8}\approx 0.472366552$. Take away this from $1$. To $4$ decimal places, you will get the answer in the book. I cannot tell you how to evaluate $e^{-6/8}$ on your calculator, they are not all the same. – André Nicolas May 22 '14 at 19:35
  • Thank you. I actually didn't realize my calculator could do that. I've always had to figure negative fractional exponents by hand. Wish I could repay you somehow. – Ann May 22 '14 at 19:40
  • Sorry, wait, I do have one more question. Why do you subtract .4724 from 1 rather than multiplying it by 1/8 or .125? – Ann May 22 '14 at 19:44
  • As I wrote before, an antiderivative (indefinite integral) of $\frac{1}{8}e^{-t/8}$ is $-e^{-t/8}$. Or if you prefer, and antiderivative of $\frac{1}{8}e^{-x/8}$ is $-e^{-x/8}$. To evaluate the definite integral, plug in $6$, take away the result of plugging in $0$. After minor simplification, we get $1-e^{-6/8}$. It looks as if you have trouble with integration. Did you take a calculus course? – André Nicolas May 22 '14 at 20:09
  • No, this is a statistics course and there was no pre-req. I've taken other stats courses and aced them. The equations are different here and the text gives very little information in some places but it was written by my prof, so... Anyway, thank you for the help. I don't completely understand it yet, but I know how to solve the equations now it seems. I really appreciate the help. – Ann May 22 '14 at 21:11
  • Then you may have been told the following formula. If the density function of $X$ is $\lambda e^{-\lambda x}$, then $\Pr(X\le x)=1-e^{-\lambda x}$, or equivalently that $\Pr(X\gt x)=e^{-\lambda x}$. No integration, just a formula to remember and use. – André Nicolas May 22 '14 at 22:06

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