# Example question about the exponential distribution

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

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?

• Hint: the mean of an exponential is $1/\lambda$, not $\lambda$. – Alex R. Mar 20 '16 at 19:44
• 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. – gung - Reinstate Monica Mar 20 '16 at 19:56
• @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. – ahorn Mar 20 '16 at 20:46

You used the wrong parameterization. When the mean of the Exponential distribution is $\beta = 2$,
$$f(x) = \dfrac{1}{\beta}e^{-x/\beta}.$$