Question

Problem 4: (2 points) A life test is performed on 10 components. The hours to failure are as follows: 200 370 500 620 730 840 950 1050 1160 400 Assume this data follows a 2-parameter Weibull distribution. If an identical component is put on test and survives for 100 hours, what is the probability that it will survive for 200 additional hours?

Answer 1

Failure probability function estimation :

In the given problem we first have to set the weibull probability density function for the given data set. It is quite cumbersome to calculate, hence the matlab code to do this is given below. Use the code to get the weibull probability function parameters as follows,

T = [ 200 ; 370 ; 500 ; 620 ; 730 ; 840 ; 950 ; 1050 ; 1160 ; 1400 ] ;

pd = fitdist( T , 'Weibull' )

The above simple code would return two parameters as,

Scale parameter, a = 882.818

Shape parameter, b = 2.39029

Hence the probability of failure would be given by,

(t)ba-erp Ca

on putting, values of a and b we get,

2.3029 f(t) 2.172 10-7l. 3029erjp t 882.818 erp

Now the cumulative function of the above failure probability would be,

..... equation (1)

t2.3%029 F(t) = | f(t)dt = 1-exp |- 882.818

Component Testing prediction :

Now since the component under testing is identical and survived 100, and has been asked to survive 200 hrs more i.e., upto 300 hrs. This means that the component must not fail between 100 hrs. to 300 hrs. For that we can write, using simple probability algebra,

P(survival) 1- [F (300) - F(100)]

Hence form the equation (1), we get,

300 2.39029 F(300) 1 - 882.818 -0.072979

100 2.39029 F(100) = 1-exp |-(882.818/ 0.0054689

Hence we finally get the probability of the component to survive as,

P(survival) 1-10.072979-0.0054689] = 0.9325

Thus there are 93.25 % chances that the component would survive 200 hrs more !!

NOTE : Feel free to ask any further queries in the comment section, down below.