Question

The commute time for people in a city has an exponential distribution with an average of 0.66 hours. What is the probability that a randomly selected person in this city will have a commute time between 0.55 and 1.1 hours? Answer: (round to 3 decimal places)

Answer 1

Answer:

$P(0.55 <X<1.1)= F(1.1) -F(0.55)$

And replacing we got:

$P(0.55 <X<1.1)= (1-e^{-\frac{1}{0.66} *1.1}) -(1-e^{-\frac{1}{0.66} *0.55})$

$P(0.55 <X<1.1)=e^{-\frac{1}{0.66} *0.55}- e^{-\frac{1}{0.66} *1.1}=0.2457$

And rounded the answer would be 0.246

Step-by-step explanation:

For this case we can define the random variable X as "The commute time for people in a city" and for this case the distribution of X is given by:

$X \sim exp (\lambda = \frac{1}{0.66}= 1.515)$

And for this case we want to find the following probability:

$P(0.55 <X<1.1)$

And we can use the cumulative distribution function given by:

$F(x) =1- e^{-\lambda x}$

And using this formula we got:

$P(0.55 <X<1.1)= F(1.1) -F(0.55)$

And replacing we got:

$P(0.55 <X<1.1)= (1-e^{-\frac{1}{0.66} *1.1}) -(1-e^{-\frac{1}{0.66} *0.55})$

$P(0.55 <X<1.1)=e^{-\frac{1}{0.66} *0.55}- e^{-\frac{1}{0.66} *1.1}=0.2457$

And rounded the answer would be 0.246