Problem 22 - Applied Statistics and Probability for Engineers

Question

The blade and the bearings are important parts of a lathe. The lathe can operate only when both of them work properly. The lifetime of the blade is exponentially distributed with the mean three years; the lifetime of the bearings is also exponentially distributed with the mean four years. Assume that each lifetime is independent. (a) What is the probability that the lathe will operate for at least five years? (b) The lifetime of the lathe exceeds what time with \(95 \%\) probability?

Step-by-Step Solution

Answer

(a) Probability is \( \approx 0.2636 \). (b) Lifetime exceeds \( \approx 5.14 \) years with 95% probability.

Step 1: Define the problem

We are given two components of a lathe: the blade and the bearings, with independent exponential lifetime distributions. The blade's mean lifetime is 3 years, and the bearings' is 4 years. We need to compute probabilities for different operational durations of the lathe.

Step 2: Determine parameters of exponential distributions

The exponential distribution is defined by its rate parameter, which is the reciprocal of the mean. For the blade, the rate \( \lambda_1 \) is \( \frac{1}{3} \); for the bearings, the rate \( \lambda_2 \) is \( \frac{1}{4} \).

Step 3: Find combined lifetime distribution

The lathe operates while both components are functioning. The minimum of two independent exponential distributions is also exponential with a rate equal to the sum of the individual rates: \( \lambda = \lambda_1 + \lambda_2 = \frac{1}{3} + \frac{1}{4} = \frac{7}{12} \). Thus, the combined lifetime is exponential with mean \( \frac{12}{7} \) years.

Step 4: Solve part (a) - Probability of operation for at least 5 years

For an exponential distribution with rate \( \lambda = \frac{7}{12} \), the survival function is \( S(t) = e^{-\lambda t} \). For \( t = 5 \): \[ S(5) = e^{-\frac{7}{12} \times 5} = e^{-\frac{35}{12}} \] Calculating gives \( S(5) \approx 0.2636 \), which is the probability that the lathe will operate for at least 5 years.

Step 5: Solve part (b) - Lifetime with 95% probability

We want to find the time, \( t \), such that the cumulative distribution function (CDF) equals 0.95. The CDF for an exponential distribution is \( F(t) = 1 - e^{-\lambda t} \). Set \( F(t) = 0.95 \), thus: \[ 1 - e^{-\frac{7}{12}t} = 0.95 \rightarrow e^{-\frac{7}{12}t} = 0.05 \] Solving for \( t \) gives: \[ -\frac{7}{12}t = \ln(0.05) \rightarrow t \approx \frac{12}{7} \times 2.9957 \approx 5.14 \text{ years} \] Hence, the lifetime of the lathe exceeds about 5.14 years with 95% probability.

Key Concepts

Exponential Distribution

The exponential distribution is a continuous probability distribution that is commonly used to model the time until an event occurs. This distribution is particularly useful in situations where we are interested in the duration of time until a particular event happens. For instance, we might want to know how long a machine part will last before it fails. The probability density function (PDF) of an exponential distribution is defined by its rate parameter \( \lambda \), where \( \lambda = \frac{1}{\text{mean}} \). This rate parameter indicates how often the event occurs.

In our case with the lathe, both the blade and the bearings follow exponential distributions, with mean lifetimes of 3 years and 4 years, respectively. This implies that their rate parameters are \( \lambda_1 = \frac{1}{3} \) and \( \lambda_2 = \frac{1}{4} \). A noteworthy property of the exponential distribution is that the minimum time until events for two independent components is also exponentially distributed. Thus, in this situation, the rate of failure for the lathe is the sum of the individual rates, \( \lambda = \lambda_1 + \lambda_2 = \frac{7}{12} \).

Independent Lifetimes

In the exercise, another critical component in our analysis is that the lifetimes of the blade and the bearings in the lathe are independent. Independence in this context means that the failure or functioning of one component does not affect the other.

Understanding independent lifetimes helps to manage reliability and risk in various fields, such as engineering and operations research. For components like those in a lathe, this assumption simplifies the calculation of the overall lifetime distribution. Since the lifetimes are independently distributed, we can use their separate distributions to determine the behavior of the entire system, like we did by summing up the individual rates.

• The blade lasts independently of the bearings’ lifetime.
• The bearings' functioning does not influence the blade's longevity.

These assumptions help us predict the performance of complex systems more effectively.

Survival Analysis

Survival analysis is a branch of statistics focused on analyzing and predicting the time until an event occurs, often termed as "time-to-event" data. This method is widely used in fields such as medicine, biology, and engineering, where understanding the duration before an event is crucial.

In our scenario with the lathe, survival analysis allows us to compute the probability that the lathe will operate beyond a specific time, such as 5 years. This is done using the survival function \( S(t) = e^{-\lambda t} \), which represents the probability of the system surviving, or continuing to function, past time \( t \). For example, we found that the probability of the lathe operating for at least 5 years is given by \( S(5) = e^{-\frac{35}{12}} \), which approximates to 0.2636. Thus, there's about a 26.36% chance that the lathe will function for this additional duration.

Statistical Inference

Statistical inference involves using data from a sample to make generalizations or predictions about a population. In the context of exponential distributions and survival analysis, statistical inference helps us determine probabilities and time estimates for the whole population of interest, like the lathe components.

Given that the lathe's components have known exponential distributions, we can infer about the lathe's operational life. Suppose we want to determine with 95% confidence the time by which the lathe will have operated before failing, known as a quantile calculation. This is achieved by manipulating the cumulative distribution function (CDF), \( F(t) = 1 - e^{-\lambda t} \), and setting it equal to a desired confidence level.

• We solve for \( t \) in \( 1 - e^{-\frac{7}{12}t} = 0.95 \).
• Finding \( t \approx 5.14 \) years, indicating a 95% confidence that the lathe will not fail before this time.

These estimates are crucial for planning maintenance and predicting failures, promoting efficiency and safety in operations.