Problem 204 - Applied Statistics and Probability for Engineers

Question

The time between calls is exponentially distributed with a mean time between calls of 10 minutes. (a) What is the probability that the time until the first call is less than 5 minutes? (b) What is the probability that the time until the first call is between 5 and 15 minutes? (c) Determine the length of an interval of time such that the probability of at least one call in the interval is \(0.90 .\) (d) If there has not been a call in 10 minutes, what is the probability that the time until the next call is less than 5 minutes? (e) What is the probability that there are no calls in the intervals from 10: 00 to \(10: 05,\) from 11: 30 to \(11: 35,\) and from 2: 00 to \(2: 05 ?\) (f) What is the probability that the time until the third call is greater than 30 minutes? (g) What is the mean time until the fifth call?

Step-by-Step Solution

Answer

(a) 0.3935, (b) 0.3834, (c) 23.02 minutes, (d) 0.3935, (e) 0.2230, (f) 0.3233, (g) 50 minutes.

Step 1: Understanding Exponential Distribution

An exponential distribution has the probability density function (PDF) described by \( f(t) = \lambda e^{-\lambda t} \). The mean of an exponential distribution is \( \frac{1}{\lambda} \). In this problem, the mean time between calls is 10 minutes, hence \( \lambda = \frac{1}{10} \).

Step 2: Calculate Probability for Part (a)

Calculate the probability that the time until the first call is less than 5 minutes: \[ P(T < 5) = 1 - e^{-\lambda t} = 1 - e^{-\frac{1}{10} \times 5} \]. Substitute \( \lambda = \frac{1}{10} \).

Step 3: Result for Part (a)

Perform the calculation: \[ P(T < 5) = 1 - e^{-0.5} \approx 1 - 0.6065 = 0.3935 \].

Step 4: Calculate Probability for Part (b)

Probability that the time until the first call is between 5 and 15 minutes: \[ P(5 < T < 15) = P(T < 15) - P(T < 5) = \left(1 - e^{-1.5}\right) - \left(1 - e^{-0.5}\right) \].

Step 5: Result for Part (b)

Perform the calculations: \[ P(5 < T < 15) = (1 - e^{-1.5}) - (1 - e^{-0.5}) = 0.7769 - 0.3935 = 0.3834 \].

Step 6: Calculate Time Interval for Part (c)

Solve for \( t \) such that the probability of at least one call is 0.90: \[ 1 - e^{-\lambda t} = 0.90 \Rightarrow e^{-\lambda t} = 0.10 \Rightarrow \lambda t = -\ln(0.10) \].

Step 7: Result for Part (c)

Calculate \( t \): \[ t = \frac{-\ln(0.10)}{\lambda} = 10 \times 2.302 = 23.02 \text{ minutes} \].

Step 8: Memoryless Property for Part (d)

Due to the memoryless property of the exponential distribution, \( P(T < 5 \mid T > 10) = P(T < 5) \). Thus, calculate \( P(T < 5) \) again.

Step 9: Result for Part (d)

The probability remains \( 0.3935 \), as calculated in Step 3.

Step 10: Calculate Probability for Part (e)

There are three "no call" intervals of 5 minutes each. Probability for no calls in any 5-minute interval: \[ P(T > 5) = e^{-0.5} \approx 0.6065 \]. Use the multiplication rule for independent events.

Step 11: Result for Part (e)

For three intervals: \[ 0.6065^3 \approx 0.2230 \].

Step 12: Calculate for Part (f) with Gamma Distribution

The time until the third call follows a gamma distribution with parameters \( n = 3 \) and \( \lambda = \frac{1}{10} \). Calculate \( P(T > 30) \) for \( n = 3 \).

Step 13: Result for Part (f)

Using the cumulative distribution function (CDF) of the gamma distribution, the probability is approximately \( 0.3233 \).

Step 14: Mean for Part (g)

Mean time until the fifth call for an exponential distribution is \( \frac{n}{\lambda} = 5 \times 10 = 50 \, \text{minutes} \).

Step 15: Final Result Compilation

Now that we have calculated each part, summarize the results: (a) 0.3935, (b) 0.3834, (c) 23.02 minutes, (d) 0.3935, (e) 0.2230, (f) 0.3233, (g) 50 minutes.

Key Concepts

Mean Time Between Events

In any stochastic process like an exponential distribution, understanding the 'Mean Time Between Events' is crucial. This is the average time interval between two successive events in a process. For an exponential distribution, this mean is represented as the reciprocal of the rate parameter \( \lambda \), i.e., \( \frac{1}{\lambda} \). In scenarios such as the time between calls, the larger the mean time, the less frequent the events. The given problem specifies a mean time of 10 minutes between calls, allowing us to calculate \( \lambda = \frac{1}{10} \). This relationship helps in using the exponential distribution to predict various probabilistic outcomes.

Probability Density Function

The Probability Density Function (PDF) of an exponential distribution describes the likelihood of a specific outcome within a certain time frame. It is given by the formula: \( f(t) = \lambda e^{-\lambda t} \). Here:

• \( \lambda \) is the rate parameter.
• \( t \) represents the time.

The PDF is a crucial tool because it provides insight into how probabilities distribute over time. For instance, it allows us to calculate the probability of an event occurring before a given time, which is precisely what we do when determining if a call happens before 5 minutes. The PDF gives us a continuous probability distribution, which means it defines the probability of getting any specific time as zero, emphasizing probabilities between intervals.

Gamma Distribution

In stochastic processes, when considering the sum of multiple exponential random variables, we look at the Gamma Distribution. If you repeatedly measure time between events such as calls, the sum of these times adheres to a gamma distribution. This distribution generalizes the exponential distribution to accommodate multiple events. Its PDF is defined for a parameter \( n \), where \( n \) is the number of events, and the rate parameter \( \lambda \). For example, in determining the probability that the time until the third call exceeds 30 minutes, we consult the gamma distribution because the question involves the accumulation of multiple events (3 calls in this scenario). Thus, the time until the third call is a gamma distribution with parameters \( n = 3 \) and \( \lambda = \frac{1}{10} \). This is how it extends the framework of exponential distributions to cover scenarios involving multiple consecutive events.

Memoryless Property

One of the unique aspects of the exponential distribution is its memoryless property. This means that the probability of an event happening in the future is independent of the past. Mathematically, this is expressed as:\[ P(T < t + s \mid T > s) = P(T < t) \] where \( T \) is the time until the event. The property implies that the distribution "forgets" what has happened before time \( s \). In simple terms, whether an event (such as a phone call) hasn't occurred for a substantial time period, the probability of it occurring next remains unchanged. This principle is exemplified in the provided problem's part (d), where the probability of a call in the next 5 minutes is the same regardless of the 10 minute wait.