How to use weibull distribution in matlab ?

Hi,

I'm new to Matlab and inexperienced too. I need to solve this problem. I understood the question but unable to decide how to solve after 2nd part. If anyone could help, it would be highly appreciated.

You have been supplied with a set of measurements for the lifetime of a bearing in the file ass3q1data.csv. You should use this data to construct a model for the behaviour of the real lifetime (the lifetime of the population). It has been well-established that the Weibull distribution is the best model for the reliability of objects (Juvinall & Marshek 2011).

For quality control and defining warranties, models for population lifetimes are used to determine how many products will fail within a stipulated period. Using the model for the population, determine which samples are in the bottom 2% of the population. Also determine the fraction of samples which do not last the desired minimum of 950 hours and which fraction of the population would not last 950 hours.

Requirements

For this assessment item, you must produce MATLAB code which:

1. Loads the data file ass3q1data.csv and verifies that it has been loaded correctly (first 5 values)

2. Determines the estimated parameters for the Weibull distribution and reports the values to the Command Window.

3. Compares the sample pdf with the population pdf graphically.

4. Compares the mean and standard deviation computed from the samples with the mean and standard deviation computed from the Weibull distribution’s parameters. Discuss why the results are the same or different. This comparison and discussion is to be reported to the Command Window.

5. Verifies the mean and standard deviation calculations by comparing the results for the first 5 values with hand calculations. Be careful how you calculate the standard deviation.

6. Computes the fraction of the samples that are in the bottom 2% of the population and reports the result to the Command Window.

7. Computes the fraction of the samples and fraction of the population that do not satisfy the minimum of 950 hours and reports the result to the Command Window.

Thanks!