I have a problem as follows.
Life of tyres normally distributed for a specific make. mean=24,000 km and sd= 2500 km.
Question is: As a result of improvements in manufacture, the length of life is still normally distributed, but the proportion of tyres failing before 20,000 km is reduced to 1.5%.
Here is how I incorrectly tacked the problem. How can I do this correctly.
In this example, 20000 - mean will be negative so to make the calculation simpler, we use Normal distribution symmetry property and instead use an x value above the mean. ie 24000 - 20000 = 4000. So just add 4000 to mean, ie 24000 + 4000 = 28000. We also reverse 1.5% as in 100% - 1.5% = 98.5%.
and we know phi(z) = 0.985 so reverse lookup z = 2.17Code:z = 28000 - mean ------------ 2500
2.17 = (28000 - mean) / 2500
28000 - mean = 2.17 x 2500 = 5425
mean = 28000 - 5425 = 22575
Therefore, new mean = 22,575 km
But this is incorrect. Obviously, the mean should be HIGHER now - with the improvement to the tyre. The answer should be 25425 km.
How can this be correctly calculated?
Its ok I worked it out.
All, I need to do is handle like this.
Get the z value as above. But convert it to negative. ie
-2.17 = (20000 - mean) / 2500
mean = 25425 as expected.
S2 Find new mean from Normal probability Watch
- Thread Starter
Last edited by acomber; 23-01-2015 at 16:05. Reason: worked it out on my own
- 23-01-2015 15:51