You have k(a) Android devices and k(i) failed devices. k(i) divided by n(i) gives you 58%.

No, that's what failure rate is **supposed** to mean. However, what the numbers actually said are:

- iPhone 6 had the highest failure rate of 29%
- iOS devices as a whole had a failure rate of 58%

These two statements cannot both be true simultaneously by any proper definition of "failure rate". The iPhone 6 is a subset of all iOS devices. The claim is made that its failure rate was 29%. For the failure rate of all iOS devices to be 58%, that would mean that at least one iOS device must have a failure rate greater than 58% to pull the average up from 29% to 58%, which contradicts the statement that the iPhone 6 had the highest failure rate at 29%.

Q.E.D.

The only way you could even halfway make those numbers plausible would be if you erroneously divided the iPhone numbers by either the total number of iOS devices or worse, the total number of devices. Either of those approaches makes the numbers meaningless because you don't know the relationship between... to use your terminology... k(i) and n(i) at that point.

In your ramblings, you fail to consider that the vast majority of people who want to avoid expensive shipping charges will often bring their unit into a store... which eliminates many of the simpler problems.

The vast majority of people who want to avoid expensive shipping charges will Google the problem and find an answer themselves. People go to a store when that fails.