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?

  • $\begingroup$ 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

You used the wrong parameterization. When the mean of the Exponential distribution is $\beta = 2$,

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.