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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%.

z = 28000 - mean
    ------------
     2500

and we know phi(z) = 0.985 so reverse lookup z = 2.17

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?

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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.
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    $\begingroup$ This is under the assumption, that the standard deviation has not changed. It's a bad question because that was not stated. $\endgroup$
    – Bernhard
    Commented Nov 10, 2016 at 8:24

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