MHB Applying the central limit theorem

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The discussion focuses on applying the central limit theorem (CLT) to estimate the probability that 40 components, each with a gamma distribution for failure time, will last at least 6 years. The mean failure time for one component is calculated as 0.5 days, leading to a total mean of 20 days for 40 components. The variance for one component is 0.05, resulting in a total variance of 2 for the sum of the components. Using the CLT, the probability that the total time exceeds 2191.5 days is expressed through the complementary error function, indicating that this probability is extremely small. The analysis concludes that achieving 6 years of functionality with the given parameters is highly unlikely.
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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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