532 18. PROBABILITY AND STATISTICSrepresents the integralí/2ð Jodt,which is the area under a standard normal (bell-shaped) curve above the mean, but
by at most one standard deviation, as given in many tables.18.4. Use Daniel Bernoulli's concept of utility to explain why only a person with
astronomical amounts of money should play a Petersburg paradox-type game. In
your explanation, take account of what the utility of the stakes must be for a
gambler versus the utility of the gain. Make an analogy between risk and work in
this regard. A laborer exchanges time and effort for money; a gambler exchanges
risk for potential gain. Remembering that all economic decisions are made "at the
margin," at what point does additional work (or risk) not bring enough additional
utility to be worth the exchange?18.5. Radium-228 is an unstable isotope. Each atom of Ra-228 has a probability
of 0.1145 (about 1 chance in 9, or about the probability of rolling a 5 with two
dice) of decaying to form an atom of actinium within any given year. This means
that the probability that the atom will survive the year as an atom of Ra-228 is
1 —0.1145 = 0.8855. Denote this "one-year survival" probability by p. Because any
sample of reasonable size contains a huge number of atoms, that survival probability
(0.8855) is the proportion of the weight of Ra-228 that we would expect to survive
a year.
If you had one gram of Ra-228 to begin with, after one year you would expect to
have ñ = 0.8855 grams. Each succeeding year, the weight of the Ra-228 left would be
multiplied by p, so that after two years you would expect to have ñ^2 = (0.8855)^2 =
0.7841 grams. In general, after t years, if you started with Wo grams, you would
expect to have W = Wop* grams. Now push these considerations a little further
and determine how strongly you can rely on this expectation. Recall Chebyshev's
inequality, which says that the probability of being more than k standard deviations
from the expected value is never larger than (1/fc)^2. What we need to know to
answer the question in this case is the standard deviation ó.
Our assumption is that each atom decays at random, independently of what
happens to any other atom. This independence allows us to think that observing
our sample for a year amounts to a large number of "independent trials," one for
each atom. We test each atom to see if it survived as an Ra-228 atom or decayed
into actinium. Let No be the number of atoms that we started with. Assuming
that we started with 1 gram of Ra-228, there will be No = 2.642 · 10^21 atoms of Ra-
228 in the original sample.^22 That is a very large number of atoms. The survival
probability is ñ = 0.8855. For this kind of independent trial, as mentioned the
standard deviation with JV 0 trials is
We write the standard deviation in this odd-looking way so that we can express it
as a fraction of the number No that we started with. Since weights are proportional
to the number of atoms, that same fraction will apply to the weights as well.
(^22) According to chemistry, the number of atoms in one gram of Ra-228 is the Avogadro number
6.023 • 1023 divided by 228.