Question

Statistics question?

The mean life of a battery is 50 hrs with a standard deviation of 6 hrs in a normal distribution. The producer wants to advertise that they will replace the batteries with the lowest 1% of battery life. Below how many hours of battery life should the producer advertise that they will replace?

Thanks in advance

Answer 1

Any battery with a life of less than 35 hours should be replaced.

Explanation:

This is a simplified application of statistical principles. The key things to note are the standard deviation and the percentage. The percentage (`1%`) tells us that we want only that part of the population that is less probable than `3sigma`, or 3 standard deviations less than the mean (this is actually at 99.7%).

So, with a standard deviation of 6 hours, the difference from the mean for the desired lifetime lower limit is:
`50 - 3xx6 = 50 - 18 = 32`hours

That means that any battery with less than 32 hours of life will be replaced.

What the statistics say is that the RANGE of 32 to 68 hours will include 99.7% of ALL batteries produced. For example, on the 'high' end, it means that only 0.3% of all batteries have a lifetime of 68 hours or more.

OK, the rigorous solution is to use the normal distribution curve and its Z-values to find the exact `sigma` value. `99`% corresponds to `2.57sigma` (one-tailed). Thus, the EXACT value to reject batteries would be:
`50 - 2.57xx6 = 50 - 15.42 = 34.6`hours

Answer 2

36hrs or less will be replaced

Explanation:

Wow, the producer of that battery company has a very high variance product you would be taking a huge risk when buying from them as you have no idea what you are getting.

We know the formula for z-score (which tells you what multiple of standard deviation away the value x is from the mean) is:
`z = \frac{x - \mu}{\sigma}`

From the 3 sigma rule of thumb (68.3% - 95.4% - 99.7% rule) we know that our answer will be somewhere between 2 to 3 standard deviations away from the mean in the negative direction.

Using a Ti-83 graphing calculator or a z-score table, find the value of z that corresponds to a cumulative probability from `(-infty, x]` of 1% :

`z =` invnorm(0.01) # = -2.32634787...
(just as expected it's between -2 and -3)

Solve for x:
`-2.32634787 = \frac{x - 50}{6}`

`-13.95808726 = x - 50`

`x = 36.04191274 ... \approx 36`

Thus, the batteries with a life of 36hrs or less will be replaced.