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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%.

    Code:
    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?


    *************************

    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.
 
 
 
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