Probability and Statistics for Engineering and Science, 9th Edition
Authors: Jay L. Devore
ISBN-13: 978-1305251809
See our solution for Question 59E from Chapter 4 from Devore's Probability and Statistics for Engineering and Science.
Problem 59E
Chapter:
Problem:
Let X =the time between two successive arrivals at the drive-up window of a local bank. If X has an exponential distribution with λ = 1(which is identical to a standard gamma distribution withα = 1 ), compute the following: a. The expected time between two successive arrivals b. The standard deviation of the time between successive arrivals c. P(X ≤ 4) d. P( 2 ≤ X ≤ 5 )
Step-by-Step Solution
Step 1
We are given with an exponential distribution with $\lambda =1$.
Step 2: (a)
Expected time between two successive arrivals is given by the expected value of X\[\begin{array}{l}E\left( X \right) = \dfrac{1}{\lambda }\\ = \dfrac{1}{1}\\ = 1\end{array}\]Therefore, \[E\left( X \right) = 1\]
Step 3: (b)
Standard deviation of exponential distribution\[\begin{array}{l}\sigma = \dfrac{1}{\lambda }\\ = \dfrac{1}{1}\\ = 1\end{array}\]Therefore, \[\sigma = 1\]
Step 4: (c)
Cumulative distribution function of exponential function is given by:\[F\left( x \right) = 1 - {e^{ - \lambda x}}\]To find the probability of time less than 4 is given by:\[\begin{array}{l}P\left( {X \le 4} \right) = F\left( 4 \right) = 1 - {e^{ - \lambda x}}\\ = 1 - {e^{ - \left( 1 \right)\left( 4 \right)}}\\ = 1 - {e^{ - 4}}\\ = 0.98168\\ \approx 0.982\end{array}\]Therefore, \[P\left( {X \le 4} \right) \approx 0.982\]
Step 5: (d)
\[\begin{array}{l}P\left( {2 \le X \le 5} \right) = F\left( 5 \right) - F\left( 2 \right)\\ = \left( {1 - {e^{ - \lambda {x_1}}}} \right) - \left( {1 - {e^{ - \lambda {x_2}}}} \right)\\ = \left( {1 - {e^{ - 1 \times 5}}} \right) - \left( {1 - {e^{ - 1 \times 2}}} \right)\\ = {e^{ - 2}} - {e^{ - 5}}\\ = 0.135335 - 0.006737\\ = 0.128597\\ = 0.129\end{array}\]Therefore, \[P\left( {2 \le X \le 5} \right) = 0.129\]
We are given with an exponential distribution with $\lambda =1$.
Step 2: (a)
Expected time between two successive arrivals is given by the expected value of X\[\begin{array}{l}E\left( X \right) = \dfrac{1}{\lambda }\\ = \dfrac{1}{1}\\ = 1\end{array}\]Therefore, \[E\left( X \right) = 1\]
Step 3: (b)
Standard deviation of exponential distribution\[\begin{array}{l}\sigma = \dfrac{1}{\lambda }\\ = \dfrac{1}{1}\\ = 1\end{array}\]Therefore, \[\sigma = 1\]
Step 4: (c)
Cumulative distribution function of exponential function is given by:\[F\left( x \right) = 1 - {e^{ - \lambda x}}\]To find the probability of time less than 4 is given by:\[\begin{array}{l}P\left( {X \le 4} \right) = F\left( 4 \right) = 1 - {e^{ - \lambda x}}\\ = 1 - {e^{ - \left( 1 \right)\left( 4 \right)}}\\ = 1 - {e^{ - 4}}\\ = 0.98168\\ \approx 0.982\end{array}\]Therefore, \[P\left( {X \le 4} \right) \approx 0.982\]
Step 5: (d)
\[\begin{array}{l}P\left( {2 \le X \le 5} \right) = F\left( 5 \right) - F\left( 2 \right)\\ = \left( {1 - {e^{ - \lambda {x_1}}}} \right) - \left( {1 - {e^{ - \lambda {x_2}}}} \right)\\ = \left( {1 - {e^{ - 1 \times 5}}} \right) - \left( {1 - {e^{ - 1 \times 2}}} \right)\\ = {e^{ - 2}} - {e^{ - 5}}\\ = 0.135335 - 0.006737\\ = 0.128597\\ = 0.129\end{array}\]Therefore, \[P\left( {2 \le X \le 5} \right) = 0.129\]