Exponential Distribution calculation

Question

I don't understand the following problem.

Data collected suggests that an exponential distribution with mean value 2.725 hours is a good model for rainfall duration.
a. What is the probability that the duration of a particular rainfall event at this location is at least 2 hours? At most 3 hours? Between 2 and 3 hours?
b. What is the probability that the rainfall duration exceeds the mean value by more than 2 standard deviations? What is the probability that it is less than the mean value by more than one standard deviation?

For part a I'm doing the following. According to the book however the answers are supposed to be .449, .699, and .148

$$mean = 2.725$$

a. P(X >= 2) = P(2 <= Y) = $1 - e^{-2/2.725} = .519$

$P(x <= 3) = P(0 < Y < 3) = 1 - e^{-3/2.725} = .667$

$P(2 < Y < 3) = P(Y > 2) - P(Y > 3) = e^{-2/2.725} - e^{-3/2.725} = .147 $

b. This part I have no clue what to do.

Answer 1

I think there is a 'mistake' in the solution of your book or perhaps you mistakenly saw the solution, I don't know!? Anyway, this is my approach:

Let $X$ be the random variable that denotes the rainfall duration and $X$ follows the exponential distribution with parameter $\lambda=2.725$. The CDF of exponential distribution is $$ F(x;\lambda)=\text{Pr}[X\le x]=1-e^{-\frac{x}{\lambda}}, $$ also $$ \text{Pr}[X\le x]=e^{-\frac{x}{\lambda}}\;,\;\text{E}[X]=\mu=\lambda\;,\text{ and }\;\sigma=\sqrt{\text{Var}[X]}=\lambda. $$

Therefore

• The probability that the duration of a particular rainfall event at this location is at least 2 hours is $$\text{Pr}[X\ge2]=\text{Pr}[X>2]=e^{-\frac{2}{2.725}}\approx0.480.$$
• The probability that the duration of a particular rainfall event at this location is at most 3 hours is $$\text{Pr}[X\le3]=1-e^{-\frac{3}{2.725}}\approx0.667.$$
• The probability that the duration of a particular rainfall event at this location is between 2 and 3 hours is $$\text{Pr}[X\le3]-\text{Pr}[X\le2]=e^{-\frac{2}{2.725}}-e^{-\frac{3}{2.725}}\approx0.148.$$
• The probability that the rainfall duration exceeds the mean value by more than 2 standard deviations is$$\text{Pr}[X>\mu+2\sigma]=\text{Pr}\left[X>3\lambda\right]=e^{-\frac{3\lambda}{\lambda}}=e^{-3}\approx0.0498.$$
• The probability that it is less than the mean value by more than one standard deviation is$$\text{Pr}[X<\mu+\sigma]=\text{Pr}[X\le\mu+\sigma]=\text{Pr}\left[X\le{2\lambda}\right]=1-e^{-\frac{2\lambda}{\lambda}}=1-e^{-2}\approx0.8647.$$

For no. b, you can also approximate the probabilities by using standard normal distribution approximation. Note that, $$ Z=\frac{X-\mu}{\sigma}\sim\mathcal{N}(0,1) $$ Hence, $$ \text{Pr}[X>\mu+2\sigma]=\text{Pr}\left[\frac{X-\mu}{\sigma}>2\right]=\text{Pr}[Z>2]=1-\Phi(2)\approx0.977 $$ and $$ \text{Pr}[X\le\mu+\sigma]=\text{Pr}\left[\frac{X-\mu}{\sigma}\le1\right]=\text{Pr}[Z\le1]=\Phi(1)\approx0.159. $$ Well, the results are quite bad. So, the approximation using standard normal distribution is not recommended.

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