An archaeologist has two old pieces of wood and shall decide which piece of wood is the oldest. Radioactivity from the pieces of wood are recorded by a counter. Number of registrations per unit time is Poisson distributed with rates 120 per hour for the oldest piece of wood and 150 per hour for the youngest piece of wood. The wood piece that provides one more count on the counter compared to the other, will be categorized as the youngest.
What is the probability that the archaeologist draws the wrong conclusion about which piece of wood is the oldest, when measured for 10 minutes?
- A = 20 rate per 10 minute
- B = 25 rate per 10 minute
The conclusion can be drawn wrong in two separate ways. This is a normal distribution. Either B can only let out 19 or less radioactive signals in the 10 minute period, or A can let out 26 or more radio active signals. Adding these two probabilities together I get 0.1151+0.0901 = 0.2052 or 20.52%. I am not sure if I have done this correctly, and confirmation is appreciated. I have used a different procedure than what is written in the answers, but the answer is roughly the same (off by a few decimals).
How long must the registration go on for the archaeologist to be 99% sure to draw right conclusion about which piece of wood is the oldest?
Even though I have the answer (and procedure) I do not understand this at all, and would very much like an explanation.
The answer is 1.59 hours.