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The time between calls to a plumbing supply business is exponentially distributed with a mean time between calls of 15 minutes. (a) What is the probability that there are no calls within a 30-minute interval? (b) What is the probability that at least one call arrives within a 10 -minute interval? (c) What is the probability that the first call arrives within 5 and 10 minutes after opening? (d) Determine the length of an interval of time such that the probability of at least one call in the interval is 0.90

Short Answer

Expert verified
(a) 0.1353; (b) 0.4866; (c) 0.1804; (d) 34.54 minutes.

Step by step solution

01

Understanding the exponential distribution

The exponential distribution is characterized by its rate parameter \( \lambda \), which is the reciprocal of the mean. Given that the mean time between calls is 15 minutes, we have \( \lambda = \frac{1}{15} \). The probability density function of the exponential distribution is \( f(t) = \lambda e^{-\lambda t} \).
02

Calculating probability for no calls in 30 minutes (Part a)

We use the cumulative distribution function (CDF) of the exponential distribution, \( P(T > t) = e^{-\lambda t} \), to find the probability of having no events in a period of time. For no calls in 30 minutes, \( t = 30 \). Therefore, \( P(T > 30) = e^{-\frac{1}{15} \times 30} = e^{-2} \). Calculating \( e^{-2} \) gives approximately 0.1353.
03

Calculating probability for at least one call in 10 minutes (Part b)

The problem asks for the probability of at least one call in 10 minutes, which is the complementary probability of having zero calls. So, \( P(\text{at least one call in 10 minutes}) = 1 - P(T > 10) = 1 - e^{-\frac{1}{15} \times 10} = 1 - e^{- rac{2}{3}} \). Calculating this value gives approximately 0.4866.
04

Probability first call between 5 and 10 minutes (Part c)

To find the probability that the first call arrives between 5 and 10 minutes, we calculate \( P(5 < T < 10) = P(T > 5) - P(T > 10) = e^{-\frac{1}{15} \times 5} - e^{-\frac{1}{15} \times 10} = e^{- rac{1}{3}} - e^{- rac{2}{3}} \). Calculating this gives approximately 0.1804.
05

Determining time interval for probability 0.90 (Part d)

For a probability of at least one call being 0.90, we set up the equation \( 1 - e^{-\lambda t} = 0.90 \). Thus, \( e^{-\lambda t} = 0.10 \). Solving for \( t \) gives \( -\lambda t = \ln(0.10) \) or \( t = -\frac{\ln(0.10)}{\lambda} \). Substituting \( \lambda = \frac{1}{15} \), we get \( t = 15 \times \ln(10) \), which is approximately 34.54 minutes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Calculation
In this exercise, the time between calls follows an exponential distribution. Calculating probability using this distribution requires understanding its characteristics and formulas. A key aspect is the **rate parameter** \( \lambda \), which is the reciprocal of the mean time interval between events. Given a mean time of 15 minutes, the rate is \( \lambda = \frac{1}{15} \).

The exponential distribution's cumulative distribution function (CDF) then helps in determining the probability of events within a certain time frame. The probability of no events occurring in a period is given by \( P(T > t) = e^{-\lambda t} \). To find the probability of at least one event occurring, we use the complement: \( P(\text{at least one event}) = 1 - P(T > t) \).

In this problem, probabilities for varying intervals were determined using these formulas, demonstrating how central they are to solving such problems effectively.
Mean Time Interval
The mean time interval is a pivotal factor in the exponential distribution, helping to determine the rate parameter \( \lambda \). The mean specifies the average time expected between events—in this case, between calls. Given a mean of 15 minutes, \( \lambda = \frac{1}{15} \) translates that mean into the distribution's functionality.

This interval fundamentally impacts how probabilities are calculated for any given time period. It forms the basis from which the cumulative distribution function (CDF) is utilized to predict how likely events are to occur within specified durations. Thus, recognizing and applying the mean time effectively are crucial skills in working with the exponential distribution.
Cumulative Distribution Function
The cumulative distribution function (CDF) of an exponential distribution is a powerful tool for calculating probabilities. It helps to determine the likelihood of an event occurring within a certain time frame. For our exponential distribution, the CDF is expressed as \( P(T > t) = e^{-\lambda t} \), where \( t \) represents the time duration of interest and \( \lambda \) is the rate parameter.

Using this function, you can calculate:
  • the probability of no events occurring in a specified interval by evaluating \( P(T > t) \)
  • for at least one event, simply find the complement: \( 1 - P(T > t) \)
Thus, the CDF is an essential component in working through problems involving exponential distributions, guiding the user to determine probabilities efficiently with clear cut-off points between time intervals.
Call Arrival Time
Call arrival time in relation to an exponential distribution can be better understood through the computations presented. Here, we're investigating specific time intervals to assess the probability of call arrivals within them. These intervals are examined by both the probability of no calls and the complementary probability for at least one call.

For example, when looking to determine the probability of calls in a 30-minute interval, the analysis uses the procedures outlined to show that there's a low likelihood (approximately 13.53%) of there being zero calls, thereby implying a high chance of at least one call. Calculating such times helps predict activity patterns based on historical data and is essential for businesses needing to react proactively to call times. Understanding how to use these models makes it easier for anyone to plan effectively regarding customer interactions.

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