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I'm working on the following sample problem and there's no solution for it.

Let X be the amount of time a student needed to finish a midterm. Assume that X follows an exponential distribution with mean 30.

  • Compute V(X)
  • Compute P(X>=45)
  • What is the 60th percentile of X?

All three of these need to compute the integral of the exponential distribution right? I suppose that's where I'm stuck at. It doesn't seem trivial to compute this value.

Note that this isn't homework -- this is just a practice problem.

echo
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2 Answers2

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The exponential distribution has a straightforward variance and CDF. Your book will probably have the derivation, but you can also find them at e.g., this SO question.

The variance is just $\frac{1}{\lambda^2}$, where $\lambda^{-1}$ (also called $\beta$) is the parameter of the distribution, here 30. So $V$ is 900 min^2.

The CDF is $1 - \exp(-\lambda x)$, so $P(X \geq 45) = \exp(-\frac{45}{30}) = 0.23$.

For the 60th percentile, set the CDF to 0.6 and solve for x.

$$0.6 = 1 - \exp(-\frac{x}{30})$$

so $\exp(-\frac{x}{30}) = 0.4$ and x = 27.5

epursimuove
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If you wished to derive the formula for $V(X)$ and the PDF of an exponential distribution, then integration would be required. However, I think the problem assumes you are already given these. That is, it assumes you know that for an exponential distribution, $V(X)=\frac{1}{\lambda^{2}}$ and the PDF is $\mathbb{P}(X=k)=\lambda e^{-\lambda k}$, where $\lambda$ is the inverse of $\mathbb{E}(X)$ , in this case, $\frac{1}{30}$.

Andrew Gross
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