Applying the central limit theorem

[oyth94]

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]

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

$\chi$ $\sigma$

 

[awkward]

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]

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 $\chi$ $\sigma$

 

[blue_raver22]

The central limit theorem states that as the sample size increases, the distribution of sample means will approach a normal distribution regardless of the underlying distribution of the population. In this scenario, we can use the central limit theorem to estimate the probability that 40 components will together be sufficient to last at least 6 years.

Since the time until a component fails follows a gamma distribution with alpha = 5 and theta = 1/10, we can calculate the mean and standard deviation of this distribution as follows:

Mean = alpha * theta = 5 * (1/10) = 0.5
Standard deviation = sqrt(alpha * theta^2) = sqrt(5 * (1/10)^2) = 0.1581

Now, let's consider the 40 components as a sample of the population. We can calculate the mean and standard deviation of this sample as follows:

Sample mean = 40 * 0.5 = 20
Sample standard deviation = sqrt(40) * 0.1581 = 1.5811

We can now use these values to estimate the probability that the 40 components will last at least 6 years, which is equivalent to 6 * 365.25 = 2191.5 days.

Using a normal distribution with a mean of 20 and a standard deviation of 1.5811, we can calculate the z-score for 2191.5 days as follows:

z = (2191.5 - 20) / 1.5811 = 1388.5 / 1.5811 = 878.8

Using a z-score table, we can find the probability corresponding to a z-score of 878.8, which is essentially 1. This means that the probability that the 40 components will last at least 6 years is very close to 1, or 100%.

In conclusion, by applying the central limit theorem, we can estimate with a high degree of confidence that 40 components will be sufficient to last at least 6 years. This highlights the usefulness of the central limit theorem in making predictions and estimates in real-world scenarios.