What is the whole number of days such that one-quarter of batteries will last longer than this many days?

Question

A carbon monoxide detector contains a battery that should have a mean lifetime of 365 days and a standard deviation of 12 days. The battery life has a normal distribution. What is the whole number of days such that one-quarter of batteries will last longer than this many days?

Answer 1

So D.I.T.,

Find yourself a normal probability distribution table (they're everywhere!). Look out and find the z value, I think it's called, such that the cumulative > that is 0.25. I make it about z=0.6745; multiply by your sigma=12, which then equates to 8.09 days. Now you need to consider the text of the problem carefully! As is frequently the case, the text is garbage! At the nominal lifetime (365 days), more than 1/4 of the batteries are still lasting fine! If the text had said, what is the maximum number of days (as a whole number), such that 1/4 of the batteries will last this long or longer ... then you could simply round down to 365+8 = 373 days. But it doesn't say that, it says, simply, that 1/4 of the batteries have to last longer (NOT this long or longer). So, the minimum possible is 0 days. Now let's split hairs. A trifle above 1/4 of the batteries will start beeping (that's how they indicate to you!) before/during the 373rd day. Does that mean that they have lasted (to) the 373rd day, or not? And if you decide that that means they HAVEN'T, then you would need to drop back to the 372nd day as your criterion, wouldn't you.

And by the way, they generally beep for quite a while before they stop functioning. How does that figure into your answer? If it gives you a headache just thinking about it, you'd better change the battery -- a headache can be an early symptom of CO poisoning. Just sayin'.

-- Cheers, --Mr. d.