Problem 35
Question
The wait time before a 911 call is answered in the state of California has a probability density function \(f(t)=0.23 e^{-0.23 t},\) for \(0 \leq t<\infty,\) where \(t\) is in seconds. (Source: California Government Code.) a) The state standard is that \(90 \%\) of 911 calls are to be answered within 10 sec. Verify that this standard is met using the probability density function \(f\). b) What is the probability that a 911 call is answered within 15 to 25 sec after being made?
Step-by-Step Solution
Verified Answer
89.97% of calls are answered within 10 sec; 2.86% probability for 15-25 sec answer time.
1Step 1: Define the Cumulative Distribution Function
To verify if 90% of calls are answered within 10 seconds, we need the cumulative distribution function (CDF) \( F(t) \) of the given probability density function (PDF) \( f(t) = 0.23 e^{-0.23 t} \). The CDF is found by integrating the PDF from 0 to \( t \): \[F(t) = \int_0^t 0.23 e^{-0.23 x} \, dx.\]
2Step 2: Calculate the CDF
Evaluate the integral: \[F(t) = -e^{-0.23 x}\Big|_0^t = -(e^{-0.23 t} - e^0) = 1 - e^{-0.23 t}.\]The CDF is \( F(t) = 1 - e^{-0.23 t} \).
3Step 3: Verify 90% Standard
To check if 90% of calls are answered within 10 sec, plug \( t = 10 \) into the CDF: \[F(10) = 1 - e^{-0.23 \times 10} = 1 - e^{-2.3} \approx 1 - 0.1003 = 0.8997.\]This result is approximately 89.97%, which is close to but slightly under the 90% standard.
4Step 4: Define the Probability for 15 to 25 Seconds
To find the probability that a call is answered between 15 to 25 seconds, calculate \( F(25) - F(15) \), where \( F(t) = 1 - e^{-0.23 t} \).
5Step 5: Calculate the Probability for 15 to 25 Seconds
First, calculate \( F(25) \): \[F(25) = 1 - e^{-0.23 \times 25} = 1 - e^{-5.75} \approx 1 - 0.0032 = 0.9968.\]Next, calculate \( F(15) \): \[F(15) = 1 - e^{-0.23 \times 15} = 1 - e^{-3.45} \approx 1 - 0.0318 = 0.9682.\]Now, subtract these results: \[P(15 \leq t \leq 25) = F(25) - F(15) = 0.9968 - 0.9682 = 0.0286.\]
6Step 6: Conclusion
The 911 call response rate is approximately 89.97% within 10 seconds, just shy of the 90% target. The probability of a call being answered between 15 to 25 seconds is 0.0286.
Key Concepts
Cumulative Distribution FunctionExponential DistributionProbability Calculations
Cumulative Distribution Function
The cumulative distribution function (CDF) is a fundamental concept in probability theory and statistics. It provides a way to understand how probabilities build up to a certain point over a range of values. For a probability density function (PDF), the CDF, denoted as \(F(t)\), is found by integrating the PDF over the range from the start point up to \(t\).
This integral gives us the probability that a random variable is less than or equal to \(t\). For instance, if we have a PDF \(f(t) = 0.23 e^{-0.23 t}\), its CDF \(F(t)\) is calculated by integrating from 0 to \(t\). The resulting CDF formula \(F(t) = 1 - e^{-0.23 t}\) shows how probability adds up over time.
The CDF helps figure out likelihoods such as the probability that a 911 call is answered within a certain number of seconds.
This integral gives us the probability that a random variable is less than or equal to \(t\). For instance, if we have a PDF \(f(t) = 0.23 e^{-0.23 t}\), its CDF \(F(t)\) is calculated by integrating from 0 to \(t\). The resulting CDF formula \(F(t) = 1 - e^{-0.23 t}\) shows how probability adds up over time.
The CDF helps figure out likelihoods such as the probability that a 911 call is answered within a certain number of seconds.
Exponential Distribution
The exponential distribution is widely used to model the time between events in a Poisson process. It is characterized by its constant hazard rate, meaning the event's occurrence is independent of any elapsed time. This makes it a suitable model for wait times, like how long a caller waits for a 911 response.
An exponential distribution is defined by the PDF \(f(t) = 0e^{-0 t}\). In this setup, \(0\) is the rate parameter, showing how quickly events occur. Here, \(0 = 0.23\), indicating a specific rate of calls being answered. When integrated, this PDF gives us the CDF \(F(t) = 1 - e^{-0.23 t}\).
This distribution's memoryless property means each second is an independent chance for the event, which is why it's often used in queuing theory and reliability studies.
An exponential distribution is defined by the PDF \(f(t) = 0e^{-0 t}\). In this setup, \(0\) is the rate parameter, showing how quickly events occur. Here, \(0 = 0.23\), indicating a specific rate of calls being answered. When integrated, this PDF gives us the CDF \(F(t) = 1 - e^{-0.23 t}\).
This distribution's memoryless property means each second is an independent chance for the event, which is why it's often used in queuing theory and reliability studies.
Probability Calculations
Calculating probabilities using a CDF involves determining how likely an event is within a particular time frame. For the 911 call example, probabilities are derived by plugging specific times into the CDF function.
For instance, to check if 90% of calls are answered within 10 seconds, substitute \(t = 10\) into the CDF, resulting in \( F(10) \approx 0.8997\), indicating an 89.97% probability. Similarly, to find the probability of calls being answered between two times, subtract the CDF values \( F(t_1) \) and \( F(t_2) \) like \( F(25) - F(15) \approx 0.0286 \).
These calculations help assess system standards and performance, guiding improvements and confirming if goals, like the 90% rule, are met.
For instance, to check if 90% of calls are answered within 10 seconds, substitute \(t = 10\) into the CDF, resulting in \( F(10) \approx 0.8997\), indicating an 89.97% probability. Similarly, to find the probability of calls being answered between two times, subtract the CDF values \( F(t_1) \) and \( F(t_2) \) like \( F(25) - F(15) \approx 0.0286 \).
These calculations help assess system standards and performance, guiding improvements and confirming if goals, like the 90% rule, are met.
Other exercises in this chapter
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