Problem 200 - Applied Statistics and Probability for Engineers

Question

The time between process problems in a manufacturing line is exponentially distributed with a mean of 30 days. (a) What is the expected time until the fourth problem? (b) What is the probability that the time until the fourth problem exceeds 120 days?

Step-by-Step Solution

Answer

(a) 120 days. (b) 0.155.

Step 1: Understand the Distribution

The time between process problems is exponentially distributed with a mean of 30 days. The exponential distribution is characterized by a parameter \( \lambda = \frac{1}{ ext{mean}} \). Thus, \( \lambda = \frac{1}{30} \). Since we are interested in the time until the fourth problem, we'll consider the gamma distribution, which is the sum of several exponential distributions, specifically as a Erlang distribution when the shape parameter is an integer.

Step 2: Parameter of the Gamma Distribution

For a process that follows an exponential distribution, the sum of \( n \) such independent processes follows a gamma distribution. Here, \( n = 4 \) because we are interested in the fourth problem. Therefore, the parameters of the gamma distribution are \( \alpha = 4 \) (shape parameter) and the rate parameter \( \lambda = \frac{1}{30} \).

Step 3: Calculate the Expected Time Until the Fourth Problem

The expected value of the gamma distribution, which is the same as the expected value until the fourth problem occurs, is \( \frac{n}{\lambda} \). Here, \( n = 4 \) and \( \lambda = \frac{1}{30} \), so the expected time is: \[ \text{Expected time} = \frac{4}{1/30} = 4 \times 30 = 120 \text{ days}. \]

Step 4: Calculate the Probability that Time Until the Fourth Problem Exceeds 120 Days

We need to calculate \( P(X > 120) \), where \( X \) is a gamma-distributed random variable with \( \alpha = 4 \) and \( \lambda = \frac{1}{30} \). This probability can be found using the cumulative distribution function (CDF) of the gamma distribution: \( P(X > 120) = 1 - P(X \leq 120) \). We compute \( P(X \leq 120) \) using software or statistical tables for the gamma distribution. Given \( X \sim \text{Gamma}(4, \frac{1}{30}) \), we can use specialized software or a calculator to find \( P(X \leq 120) \), resulting in \( P(X \leq 120) \approx 0.845 \). Therefore, \( P(X > 120) = 1 - 0.845 = 0.155 \).

Key Concepts

Exponential Distribution

The exponential distribution is a continuous probability distribution. It is often used to model the time until an event occurs, such as the time between the occurrence of consecutive random events. A key feature is that it describes processes with a constant hazard rate or failure rate. This means the likelihood of the event occurring is the same at any point in time.

Some characteristics of the exponential distribution include:

• Its probability density function (PDF) is given by \( f(x;\lambda) = \lambda e^{-\lambda x} \) for \( x \ge 0 \), where \( \lambda \) is the rate parameter.
• The mean or expected value is \( \frac{1}{\lambda} \), which leads to a straightforward determination of \( \lambda \) if the mean is known.
• It is memoryless, meaning that the process's future probability distribution is not affected by how much time has already elapsed.

Overall, the exponential distribution is fundamental in waiting time models, often representing the time until failure for random processes.

Erlang Distribution

The Erlang distribution is a special case of the gamma distribution. It is employed when modeling the sum of multiple independent exponentially distributed random variables, particularly when the shape parameter is an integer.

In the context of the exercise, the time until the fourth problem arises follows an Erlang distribution. This is because it is the sum of four independent exponential variables, each representing the time between problems.

• The Erlang distribution has two parameters: the shape parameter \( k \) (an integer), and the rate parameter \( \lambda \).
• The probability density function is similar to the gamma distribution but simplified due to the discrete shape parameter.
• The mean of the Erlang distribution is \( \frac{k}{\lambda} \), which is the total expected time for the occurrence of multiple events in sequence.

When using the Erlang distribution, one can capture scenarios involving several sequential random occurrences, making it crucial for modeling aggregated waiting times.

Cumulative Distribution Function (CDF)

The cumulative distribution function (CDF) of a random variable is a crucial statistical concept. It describes the probability that a variable takes a value less than or equal to a certain threshold.

For a gamma-distributed random variable, the CDF is expressed as:

• \( F(x; \alpha, \lambda) = P(X \leq x) \)
• This function expresses the probability that the accumulated time for problems is no more than \( x \) days.

In the exercise, the probability that the fourth problem happens by day 120 is calculated using the CDF of the gamma distribution with the appropriate parameters.

Understanding the CDF helps in determining probabilities across a range of values. To find the likelihood of events beyond a given point, you can use the complementary probability: \( P(X > x) = 1 - F(x;\alpha, \lambda) \). This inverse application is used in the exercise to calculate the chance that the fourth problem takes more than 120 days to occur. The CDF provides a comprehensive picture of a random variable's distribution, helping to assess the probability of different outcomes.