# Applying the central limit theorem

#### oyth94

##### Member
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.

#### chisigma

##### Well-known member
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 continous functionality of the equipment seems a little problematic...

Kind regards

$\chi$ $\sigma$

#### awkward

##### Member
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.

#### chisigma

##### Well-known member
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/u...ther-sites-part-ii-1566/index4.html#post12076

In any case is a number 'very small'... exceeding our imagination ...

Kind regards

$\chi$ $\sigma$

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