Answer:
a) ![P(X \leq 100) = 1- e^{-0.01342*100} =0.7387](https://tex.z-dn.net/?f=%20P%28X%20%5Cleq%20100%29%20%3D%201-%20e%5E%7B-0.01342%2A100%7D%20%3D0.7387)
![P(X \leq 200) = 1- e^{-0.01342*200} =0.9317](https://tex.z-dn.net/?f=%20P%28X%20%5Cleq%20200%29%20%3D%201-%20e%5E%7B-0.01342%2A200%7D%20%3D0.9317)
![P(100\leq X \leq 200) = [1- e^{-0.01342*200}]-[1- e^{-0.01342*100}] =0.1930](https://tex.z-dn.net/?f=%20P%28100%5Cleq%20X%20%5Cleq%20200%29%20%3D%20%5B1-%20e%5E%7B-0.01342%2A200%7D%5D-%5B1-%20e%5E%7B-0.01342%2A100%7D%5D%20%3D0.1930)
b) ![P(X>223.547) = 1-P(X\leq 223.547) = 1-[1- e^{-0.01342*223.547}]=0.0498](https://tex.z-dn.net/?f=%20P%28X%3E223.547%29%20%3D%201-P%28X%5Cleq%20223.547%29%20%3D%201-%5B1-%20e%5E%7B-0.01342%2A223.547%7D%5D%3D0.0498)
c) ![m = \frac{ln(0.5)}{-0.01342}=51.65](https://tex.z-dn.net/?f=%20m%20%3D%20%5Cfrac%7Bln%280.5%29%7D%7B-0.01342%7D%3D51.65)
d) ![a = \frac{ln(0.05)}{-0.01342}=223.23](https://tex.z-dn.net/?f=%20a%20%3D%20%5Cfrac%7Bln%280.05%29%7D%7B-0.01342%7D%3D223.23)
Step-by-step explanation:
Previous concepts
The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:
Solution to the problem
For this case we have that X is represented by the following distribution:
![X\sim Exp (\lambda=0.01342)](https://tex.z-dn.net/?f=%20X%5Csim%20Exp%20%28%5Clambda%3D0.01342%29)
Is important to remember that th cumulative distribution for X is given by:
![F(X) =P(X \leq x) = 1-e^{-\lambda x}](https://tex.z-dn.net/?f=%20F%28X%29%20%3DP%28X%20%5Cleq%20x%29%20%3D%201-e%5E%7B-%5Clambda%20x%7D)
Part a
For this case we want this probability:
![P(X \leq 100)](https://tex.z-dn.net/?f=%20P%28X%20%5Cleq%20100%29)
And using the cumulative distribution function we have this:
![P(X \leq 100) = 1- e^{-0.01342*100} =0.7387](https://tex.z-dn.net/?f=%20P%28X%20%5Cleq%20100%29%20%3D%201-%20e%5E%7B-0.01342%2A100%7D%20%3D0.7387)
![P(X \leq 200) = 1- e^{-0.01342*200} =0.9317](https://tex.z-dn.net/?f=%20P%28X%20%5Cleq%20200%29%20%3D%201-%20e%5E%7B-0.01342%2A200%7D%20%3D0.9317)
![P(100\leq X \leq 200) = [1- e^{-0.01342*200}]-[1- e^{-0.01342*100}] =0.1930](https://tex.z-dn.net/?f=%20P%28100%5Cleq%20X%20%5Cleq%20200%29%20%3D%20%5B1-%20e%5E%7B-0.01342%2A200%7D%5D-%5B1-%20e%5E%7B-0.01342%2A100%7D%5D%20%3D0.1930)
Part b
Since we want the probability that the man exceeds the mean by more than 2 deviations
For this case the mean is given by:
![\mu = \frac{1}{\lambda}=\frac{1}{0.01342}= 74.516](https://tex.z-dn.net/?f=%20%5Cmu%20%3D%20%5Cfrac%7B1%7D%7B%5Clambda%7D%3D%5Cfrac%7B1%7D%7B0.01342%7D%3D%2074.516)
And by properties the deviation is the same value ![\sigma = 74.516](https://tex.z-dn.net/?f=%20%5Csigma%20%3D%2074.516)
So then 2 deviations correspond to 2*74.516=149.03
And the want this probability:
![P(X > 74.516+149.03) = P(X>223.547)](https://tex.z-dn.net/?f=%20P%28X%20%3E%2074.516%2B149.03%29%20%3D%20P%28X%3E223.547%29)
And we can find this probability using the complement rule:
![P(X>223.547) = 1-P(X\leq 223.547) = 1-[1- e^{-0.01342*223.547}]=0.0498](https://tex.z-dn.net/?f=%20P%28X%3E223.547%29%20%3D%201-P%28X%5Cleq%20223.547%29%20%3D%201-%5B1-%20e%5E%7B-0.01342%2A223.547%7D%5D%3D0.0498)
Part c
For the median we need to find a value of m such that:
![P(X \leq m) = 0.5](https://tex.z-dn.net/?f=%20P%28X%20%5Cleq%20m%29%20%3D%200.5)
If we use the cumulative distribution function we got:
![1-e^{-0.01342 m} =0.5](https://tex.z-dn.net/?f=%201-e%5E%7B-0.01342%20m%7D%20%3D0.5)
And if we solve for m we got this:
![0.5 = e^{-0.01342 m}](https://tex.z-dn.net/?f=%200.5%20%3D%20e%5E%7B-0.01342%20m%7D)
If we apply natural log on both sides we got:
![ln(0.5) = -0.01342 m](https://tex.z-dn.net/?f=%20ln%280.5%29%20%3D%20-0.01342%20m)
![m = \frac{ln(0.5)}{-0.01342}=51.65](https://tex.z-dn.net/?f=%20m%20%3D%20%5Cfrac%7Bln%280.5%29%7D%7B-0.01342%7D%3D51.65)
Part d
For this case we have this equation:
![P(X\leq a) = 0.95](https://tex.z-dn.net/?f=%20P%28X%5Cleq%20a%29%20%3D%200.95)
If we apply the cumulative distribution function we got:
![1-e^{-0.01342*a} =0.95](https://tex.z-dn.net/?f=%201-e%5E%7B-0.01342%2Aa%7D%20%3D0.95)
If w solve for a we can do this:
![0.05= e^{-0.01342 a}](https://tex.z-dn.net/?f=%200.05%3D%20e%5E%7B-0.01342%20a%7D)
Using natural log on btoh sides we got:
![ln(0.05) = -0.01342 a](https://tex.z-dn.net/?f=%20ln%280.05%29%20%3D%20-0.01342%20a)
![a = \frac{ln(0.05)}{-0.01342}=223.23](https://tex.z-dn.net/?f=%20a%20%3D%20%5Cfrac%7Bln%280.05%29%7D%7B-0.01342%7D%3D223.23)