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I was given the following question. My answer C was marked incorrect. Quiz 6.2.

My method of calculating it was to use the exponential distribution with the parameter $\lambda = 2$: $$\int_{0.75}^\infty 2e^{-2x} \text {d} x = \left. \lim_{n \to \infty} -e^{-2x}\right]_{0.75}^n = 0+ e^{-2 \times 0.75} \approx 0.22 $$

What did I do incorrectly?

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  • $\begingroup$ Hint: the mean of an exponential is $1/\lambda$, not $\lambda$. $\endgroup$ – Alex R. Mar 20 '16 at 19:44
  • $\begingroup$ Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. $\endgroup$ – gung - Reinstate Monica Mar 20 '16 at 19:56
  • $\begingroup$ @gung Thank you for the suggestion. I did search for a [homework] tag, but I didn't find one, and I wouldn't have expected [self-study] to include questions that come from a university course anyway. It seems that your comment is rather generic, as I have actually explained my working sufficiently. You come across patronizing because you are not clear about how I could have asked my question differently - either that, or you haven't actually read my question. Next time, have a look at a new user's credentials on other SE sites. In any case, I have realized where I went wrong. $\endgroup$ – ahorn Mar 20 '16 at 20:46
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You used the wrong parameterization. When the mean of the Exponential distribution is $\beta = 2$,

$$f(x) = \dfrac{1}{\beta}e^{-x/\beta}. $$

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  • $\begingroup$ Oh, it's not the first time I've made that mistake. I'm still pretty new to the exponential distribution. $\endgroup$ – ahorn Mar 20 '16 at 19:48

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