Applying the central limit theorem

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Discussion Overview

The discussion revolves around applying the central limit theorem to estimate the probability that 40 components, each with a gamma distribution for failure time, will last at least 6 years. The context includes theoretical exploration of statistical distributions and their applications in reliability engineering.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant calculates the mean time before failure for a single component as $\mu = \alpha\ \theta = \frac{1}{2}\ \text{day}$ and for 40 components as $40\ \mu = 20\ \text{days}$, expressing concern about achieving 6 years of functionality.
  • Another participant suggests using the central limit theorem to approximate the distribution of the sum of failure times, indicating that $Y = \sum_{i=0}^{40} X_i$ will be approximately normal.
  • A further reply provides the mean and variance calculations for the sum of the random variables, stating $\mu_{S} = 20$ and $\sigma^{2}_{S} = 2$, and presents the probability expression for $S > x$ days.
  • One participant notes that the calculated probability for $S$ being greater than 2191.5 days results in a very small number, suggesting it is numerically unvaluable.

Areas of Agreement / Disagreement

Participants express differing views on the feasibility of achieving the required functionality duration, with some calculations leading to concerns about the probability being extremely low. The discussion does not reach a consensus on the implications of these calculations.

Contextual Notes

There are assumptions regarding the distribution parameters and the interpretation of the central limit theorem's applicability to this scenario. The calculations depend on the accuracy of the gamma distribution parameters and the assumptions made about the number of components.

oyth94
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Suppose the time in days until a component fails has the gamma distribution with alpha = 5, and theta = 1/10. When a component fails, it is immediately replaced by a new component. Use the central limit theorem to estimate the probability that 40 components will together be sufficient to last at least 6 years. *Assume that a year has 365.25 days.
 
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Re: applying the central limit theorem

If the time is expressed in days, then the mean time before a failure for one component is $\mu = \alpha\ \theta = \frac{1}{2}\ \text{day}$ and for 40 components is $40\ \mu = 20\ \text{days}$... in this case to guarantee 6 x 365.25 = 2191.5 days of continuous functionality of the equipment seems a little problematic...Kind regards

$\chi$ $\sigma$
 
Re: applying the central limit theorem

Hint: We want the probability that $Y = \sum_{i=0}^{40} X_i$ is greater than 6 years, where each $X_i$ has a Gamma distribution with known parameters. By the Central Limit Theorem the distribution of $Y$ is approximately Normal. You can use the mean and variance of $X_i$ to find the mean and variance of $Y$. From there, it should be easy.
 
Re: applying the central limit theorem

Given n r.v. $X_{1}, X_{2},..., X_{n}$ with the same distribution, mean $\mu$ and variance $\sigma^{2}$, then for n 'large enough' the r.v. $S = X_{1} + X_{2} + ... + X_{n}$ is normal distributed with mean $\mu_{S} = n\ \mu$ and variance $\sigma^{2}_{S} = n\ \sigma^{2}$. In this case is...

$\displaystyle \mu= \alpha\ \theta = \frac {1}{2} \implies \mu_{S} = 40\ \mu = 20$

$\displaystyle \sigma^{2}= \alpha\ \theta^{2} = \frac{1}{20} \implies \sigma^{2}_{S}= 40\ \sigma^{2} = 2$

The probability that S is greater than x days is...

$\displaystyle P\{S > x\} = \frac{1}{2}\ \text{erfc}\ (\frac{x - \mu_{S}}{\sigma_{S}\ \sqrt{2}})\ (1)$

... and for x = 2191.5 we have...

$\displaystyle P\{S > x\} = \frac{1}{2}\ \text{erfc}\ (1085.75)\ (2)$

Of course the quantity (2) is numerically unvaluable and an approximate value is given in...

http://www.mathhelpboards.com/f52/unsolved-statistic-questions-other-sites-part-ii-1566/index4.html#post12076

In any case is a number 'very small'... exceeding our imagination (Wasntme)...Kind regards $\chi$ $\sigma$
 
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