/*! 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 72 Suppose that when a type of tran... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that when a type of transistor is subjected to an accelerated life test, the lifetime \(X\) (in weeks) has a gamma distribution with mean 24 weeks and standard deviation 12 weeks. a. What is the probability that a transistor will last between 12 and 24 weeks? b. What is the probability that a transistor will last at most 24 weeks? Is the median of the lifetime distribution less than 24 ? Why or why not? c. What is the 99 th percentile of the lifetime distribution? d. Suppose the test will actually be terminated after \(t\) weeks. What value of \(t\) is such that only \(5 \%\) of all transistors would still be operating at termination?

Short Answer

Expert verified
a) Calculate CDF at 12 and 24 weeks to find probability. b) Median < 24 if probability at 24 > 0.5. c) Use inverse CDF for 99th percentile. d) Find time \( t \) using 95% CDF inverse.

Step by step solution

01

Identify Parameters of the Gamma Distribution

The gamma distribution is characterized by two parameters, shape \( \theta \) and scale \( \beta \). Given the mean \( \mu = \theta \beta = 24 \) weeks and standard deviation \( \sigma = \sqrt{\theta} \beta = 12 \) weeks. By solving these equations simultaneously, we can find \( \theta \) and \( \beta \).
02

Solve for Shape and Scale Parameters

Since \( \sigma = \sqrt{\theta} \beta \), we have \( 12 = \sqrt{\theta} \beta \). Also, given \( \mu = \theta \beta = 24 \), we can set \( \beta = \frac{24}{\theta} \). Substituting for \( \beta \) in the standard deviation equation, \( 12 = \sqrt{\theta} \times \frac{24}{\theta} \), leading to \( \theta = 4 \) and \( \beta = 6 \).
03

Calculate the Probability for 12 < X < 24

To find \( P(12 < X < 24) \), we calculate \( P(X < 24) - P(X < 12) \) using the cumulative distribution function (CDF) of the gamma distribution with parameters \( \theta = 4 \) and \( \beta = 6 \).
04

Calculate the CDF for X = 24 and X = 12

Using a statistical software or tables for the gamma distribution, find \( P(X < 24) \) and \( P(X < 12) \). Compute the probability for \( P(12 < X < 24) = P(X < 24) - P(X < 12) \).
05

Calculate Probability for X ≤ 24

The probability that a transistor will last at most 24 weeks is \( P(X \leq 24) = P(X < 24) \), already calculated in Step 4.
06

Determine if Median is Less than 24

Because the median is not necessarily equal to the mean for non-symmetric distributions like the gamma, you calculate the median by finding \( X \) such that \( P(X < \text{median}) = 0.5 \). Compare this \( X \) to 24.
07

Calculate the 99th Percentile

Find \( X \) such that \( P(X < X_{99}) = 0.99 \). Use inverse gamma CDF with \( \theta = 4 \) and \( \beta = 6 \) to find this value.
08

Find Termination Time for 5% Survival

Determine the time \( t \) such that \( P(X > t) = 0.05 \), equivalently \( P(X < t) = 0.95 \). Use the inverse gamma CDF with \( \theta = 4 \) and \( \beta = 6 \) to find \( t \).

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Key Concepts

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

Cumulative Distribution Function
A Cumulative Distribution Function (CDF) is an essential tool in statistics, especially when dealing with continuous probability distributions like the gamma distribution. It gives us the probability that a random variable takes on a value less than or equal to a specific point.
For a gamma distribution characterized by parameters shape \( \theta \) and scale \( \beta \), the CDF can be used to find probabilities like \( P(X \leq x) \), which is the probability that the lifetime of transistors is less than or equal to \( x \) weeks. In our example, to find the probability that a transistor lasts less than 24 weeks, we calculate \( P(X < 24) \) using the gamma CDF.

Once we have the CDF values for specific time points, like 12 weeks or 24 weeks, we can easily compute the probability of the transistor lasting between a range, such as 12 to 24 weeks, by subtracting the CDF value at 12 weeks from that at 24 weeks. Statistical software or gamma distribution tables typically provide these CDF values.
Shape and Scale Parameters
In a gamma distribution, the behavior and characteristics of the distribution are largely determined by its shape \( \theta \) and scale \( \beta \) parameters.
To find these parameters, we rely on the relationship between the mean \( \mu \) and standard deviation \( \sigma \) of the distribution. Given that \( \mu = \theta \beta \) and \( \sigma = \sqrt{\theta} \beta \), we can solve these to find the values of \( \theta \) and \( \beta \) when the mean is 24 weeks and the standard deviation is 12 weeks.

In this specific exercise, by solving \( \theta \beta = 24 \) and \( \sqrt{\theta} \beta = 12 \) simultaneously, we determine the shape parameter \( \theta = 4 \) and the scale parameter \( \beta = 6 \). These parameters are critical as they allow us to describe the distribution fully and perform additional calculations, like finding the cumulative distribution function values or percentiles.
Percentile Calculation
Percentiles are particularly useful in understanding how a value compares within a distribution. The 99th percentile, for example, represents a value below which 99% of the data fall.
To calculate a percentile in a gamma distribution, we use the inverse of the cumulative distribution function (inverse CDF). Suppose we want the 99th percentile of the lifetime distribution of transistors. In that case, we find a value \( X_{99} \) such that \( P(X < X_{99}) = 0.99 \). This involves using statistical software or a calculator that supports the gamma distribution's inverse CDF.

Percentile calculations can help set performance benchmarks. For instance, if a transistor must perform better than 99% of all others, the 99th percentile value tells us the minimum performance criteria it needs to meet.
Median of Distribution
The median of a distribution represents the middle value, where half the observations are below this point, and half are above. Unlike symmetric distributions such as the normal distribution, where the median equals the mean, the gamma distribution's median can differ from the mean due to its skewness.
To find the median for the gamma distribution, we look for a value \( X \) such that \( P(X < \text{median}) = 0.5 \). This can be done using the gamma distribution's inverse CDF. In this exercise, following such a calculation allows us to determine if the median lifetime for the transistors is less than the mean, confirming whether the distribution is skewed.

Understanding the median is fundamentally important as it provides another way to assess the central tendency of data, especially in skewed distributions where the mean might not be a representative measure.

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Most popular questions from this chapter

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