Question

Assume that failure times were 77, 27, 59, 89, 96, 16, 108 and 78 hours for an operation stress.

• Estimate the values of the parameters for a two-parameter Weibull distribution.
• Determine the reliability of the units at a time of 70 hours.
• Find MTTF
• Find B1 life (99% life)
• Find median time to failure
• Find the design life for a reliability of 0.90

Answer 1

1. Let us denote the sample observations as

Here and denotes failure time in the ^th instant

Now a r.v if it's pdf is given by ,

where

scale parameter, or characteristic life

shape parameter (or slope)

The MLE estimate of   and are obtained by solving the equations:

and

Solving these two we get and .

2. MTTF is the average time that an item will function before it fails. It is the mean lifetime of the item.

So it is equal to Total hours of operation/Total number of units

So clearly it is here

3. Here we want to find a such that . Here based on the sample we will estimate it by computing sample 99^th percentile. For that we will do the following:

Step 1. Arrange the data in ascending order: 16, 27, 59, 77, 78, 89, 96, 108

Step 2. Compute the position of the p^th percentile (index i):

i = (p / 100) * n), where p = 99 and n = 8

i = (99 / 100) * 8 = 7.92

Step 3. The index i is not an integer, round up. (i = 8) ⇒ the 99^th percentile is the value in 8^th position, or 108

So the 99^th percentile is 108.

So, B1 life is 108

4. Median time to failure= Median of T_i'

Here sorted observations are 16,27,59,77,78,89,96,108

n=8=even

So,Median Time to Failure=(77+78)/2=77.5

5. Design life is the time to failure (tD) that corresponds to a specified reliability. Now it is requeired to find t_D corresponding to reliability=0.90

ie

i.e

i.e  

Now we know the estimated values of and . Plugging in these values and solving we get