Question 316654
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Presuming that the life of tires is an approximately normally distributed statistic, at least distributed such that the average and the median values of the data are very close to equal, you can say that for a tire with a 50,000 mile average life, 50% of the tires placed in service will wear out earlier than 50,000 miles and 50% will last longer than 50,000.  Hence, the probability of one tire lasting 50,000 miles is 0.5.


The probability of *[tex \Large k] successes in *[tex \Large n] trials where the probability of success of one trial is *[tex \Large p] is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_n(k)\ =\ \left(n\cr k\right\)p^k\left(1\,-\,p\right)^{n\,-\,k}]


where *[tex \LARGE \left(n\cr k\right\)] is the number of combinations of *[tex \Large n] things taken *[tex \Large k] at a time and is equal to:  *[tex \Large \frac{n!}{k!(n\,-\,k)!}]


You need 4 successes out of 4 trials, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_4(4)\ =\ \left(4\cr 4\right\)0.5^4\left(0.5\right)^{0}]


Since *[tex \LARGE\left(4\cr 4\right\)\ =\ 1] and *[tex \LARGE \left(0.5\right)^{0}\ =\ 1] we can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_4(4)\ =\ 0.5^4\ =\ \frac{1}{16}\ =\ 0.0625]


In other words, the manufacturer is going to be making a price adjustment on better than 93% of the <i>sets</i> of tires sold.


It is possibly a slightly worse situation than this, though it is impossible to quantify with only the given data.  The actual situation depends on whether the data taken to calculate the average life of this particular brand of tire includes premature failures resulting from road hazards.  The manufacturer's liability in this regard would depend on the wording in his warranty that goes along with the "...An immediate adjustment will be made on any tire that does not last 50,000 miles."


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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