/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 The following failure time obser... [FREE SOLUTION] | 91Ó°ÊÓ

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The following failure time observations (1000s of hours) resulted from accelerated life testing of 16 integrated circuit chips of a certain type: \(\begin{array}{rrrrrr}82.8 & 11.6 & 359.5 & 502.5 & 307.8 & 179.7 \\ 242.0 & 26.5 & 244.8 & 304.3 & 379.1 & 212.6 \\ 229.9 & 558.9 & 366.7 & 204.6 & & \end{array}\) Use the corresponding percentiles of the exponential distribution with \(\lambda=1\) to construct a probability plot. Then explain why the plot assesses the plausibility of the sample having been generated from any exponential distribution.

Short Answer

Expert verified
Organize the data, calculate expected values, plot data against expected order statistics, and assess line fit to check exponential distribution fit.

Step by step solution

01

Arrange the Data in Ascending Order

First, we need to arrange the given failure time observations in ascending order: 11.6, 26.5, 82.8, 179.7, 204.6, 212.6, 229.9, 242.0, 244.8, 304.3, 307.8, 359.5, 366.7, 379.1, 502.5, 558.9.
02

Calculate the Exponential Order Statistics

Using the formula for the exponential distribution order statistics, calculate the expected values for each data point. The expected value of the i-th order statistic for an exponential distribution with rate \( \lambda = 1 \) is given by:\[E(T_{(i)}) = \frac{i}{n+1}\]Where \( n \) is the total number of observations (16 in this case). Calculate for each i from 1 to 16.
03

Determine Percentiles for Comparison

Since we have used \( \lambda = 1 \), the percentiles can be directly compared to the values obtained because the exponential distribution simplifies the scale:\[ \text{Percentiles} = -\log\left(1 - \frac{i}{n+1}\right) \].
04

Plot the Data against the Expected Values

Create a probability plot by plotting each of the ordered failure times against the corresponding expected order statistics. Use a scatter plot format where the y-axis represents ordered failure times, and the x-axis represents the exponential order statistics.
05

Analyze the Probability Plot

The probability plot allows us to visually assess how closely the sample data follows an exponential distribution. If the data points fall approximately along a straight line, this indicates that the data may plausibly come from an exponential distribution.

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Key Concepts

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

Accelerated Life Testing
In the world of statistics and reliability analysis, accelerated life testing (ALT) is a method used to estimate the lifespan of a product by subjecting it to conditions like higher stress levels than usual. This process speeds up the occurrence of failures so that data can be acquired faster than under normal usage conditions. Here’s why ALT is important: - It helps in predicting the reliability and expected lifetime of the product in a shorter time frame. - By identifying potential failures early, manufacturers can improve product designs and ensure product quality. - It provides insights into the weak spots of design and material choices, enabling cost-effective improvements. Using ALT in the given exercise, the failure times of 16 integrated circuit chips are obtained, which helps in creating a probability plot to verify if the data fits the exponential distribution pattern.
Probability Plot
A probability plot is a graphical method for assessing if a dataset follows a given distribution. By plotting the observed data against the expected values based on a theoretical distribution, one can visually inspect the fit. In the context of the exercise, here's how a probability plot is useful:- **Data Organization**: First, organize the data in ascending order, as done with the failure times.- **Expected Values**: Calculate expected order statistics based on the exponential distribution. For our case, the rate \( \lambda = 1 \) is used.- **Comparing Values**: By plotting the actual ordered failure times versus the expected values, a visual inspection reveals how closely the empirical data follows the theoretical distribution.If the points on the plot fall approximately along a straight line, it suggests that the data is consistent with the exponential distribution, supporting the hypothesis.
Failure Time Analysis
Understanding failure time analysis is crucial in reliability engineering as it focuses on estimating the time until a product experiences a failure. It helps companies understand product longevity and make decisions to enhance durability and safety. In our exercise, the failure time data collected from accelerated life testing is used for: - **Predictive Analysis**: Understanding the time distribution before failure, which is essential for maintenance planning. - **Reliability Estimates**: Calculating key reliability metrics such as mean time to failure (MTTF). - **Design Improvements**: Identifying trends and outliers in failure times can lead to design tweaks for enhanced reliability. The objective is to use these insights to refine product quality, optimize performance, and ensure customer satisfaction.
Order Statistics
Order statistics deals with the properties and behavior of ordered data points in a sample. When applying it to exponential distributions, it helps derive expected values for plotting against observed values.In this exercise, here’s the step-by-step guide to utilize order statistics:- **Ordering Data**: Start by arranging your data, e.g., failure times, in ascending order.- **Calculate Expected Order Statistics**: For an exponential distribution with \( \lambda = 1 \), expected values are calculated using \[ E(T_{(i)}) = \frac{i}{n+1} \], providing a theoretical basis for comparison.- **Graphical Analysis**: These expected order statistics are instrumental in crafting a probability plot. Observed values are compared against these expected values to visually assess distribution alignment.Order statistics simplifies the process of understanding where a given sample stands in relation to a theoretical distribution, enhancing the accuracy of analyses like the one demonstrated in our exercise.

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Most popular questions from this chapter

The weight distribution of parcels sent in a certain manner is normal with mean value \(12 \mathrm{lb}\) and standard deviation \(3.5 \mathrm{lb}\). The parcel service wishes to establish a weight value \(c\) beyond which there will be a surcharge. What value of \(c\) is such that \(99 \%\) of all parcels are at least \(1 \mathrm{lb}\) under the surcharge weight?

Let \(X=\) the time (in \(10^{-1}\) weeks) from shipment of a defective product until the customer returns the product. Suppose that the minimum return time is \(\gamma=3.5\) and that the excess \(X-3.5\) over the minimum has a Weibull distribution with parameters \(\alpha=2\) and \(\beta=1.5\) (see "Practical Applications of the Weibull Distribution" Industrial Quality Control, Aug. 1964:71-78). a. What is the cdf of \(X\) ? b. What are the expected return time and variance of return time? [Hint: First obtain \(E(X-3.5)\) and \(V(X-3.5)\).] c. Compute \(P(X>5)\). d. Compute \(P(5 \leq X \leq 8)\).

Let \(Z\) be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. a. \(P(0 \leq Z \leq 2.17)\) b. \(P(0 \leq Z \leq 1)\) c. \(P(-2.50 \leq Z \leq 0)\) d. \(P(-2.50 \leq Z \leq 2.50)\) e. \(P(Z \leq 1.37)\) f. \(P(-1.75 \leq Z)\) g. \(P(-1.50 \leq Z \leq 2.00)\) h. \(P(1.37 \leq Z \leq 2.50)\) i. \(P(1.50 \leq Z)\) j. \(P(|Z| \leq 2.50)\)

The automatic opening device of a military cargo parachute has been designed to open when the parachute is \(200 \mathrm{~m}\) above the ground. Suppose opening altitude actually has a normal distribution with mean value \(200 \mathrm{~m}\) and standard deviation \(30 \mathrm{~m}\). Equipment damage will occur if the parachute opens at an altitude of less than \(100 \mathrm{~m}\). What is the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes?

The breakdown voltage of a randomly chosen diode of a certain type is known to be normally distributed with mean value \(40 \mathrm{~V}\) and standard deviation \(1.5 \mathrm{~V}\). a. What is the probability that the voltage of a single diode is between 39 and 42 ? b. What value is such that only \(15 \%\) of all diodes have voltages exceeding that value? c. If four diodes are independently selected, what is the probability that at least one has a voltage exceeding 42 ?

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