Introduction to Probability and Statistics for Engineers and Scientists

152 Chapter 5: Special Random Variables

from past experience that, on the average, 3.2 suchα-particles are given off, what is a good
approximation to the probability that no more than 2α-particles will appear?

SOLUTION If we think of the gram of radioactive material as consisting of a large numbern
of atoms each of which has probability 3.2/nof disintegrating and sending off anα-particle
during the second considered, then we see that, to a very close approximation, the number
ofα-particles given off will be a Poisson random variable with parameterλ=3.2. Hence
the desired probability is

P{X≤ 2 }=e−3.2+3.2e−3.2+

(3.2)^2

2

e−3.2

=.382 ■

EXAMPLE 5.2d If the average number of claims handled daily by an insurance company is
5, what proportion of days have less than 3 claims? What is the probability that there will
be 4 claims in exactly 3 of the next 5 days? Assume that the number of claims on different
days is independent.

SOLUTION Because the company probably insures a large number of clients, each having a
small probability of making a claim on any given day, it is reasonable to suppose that the
number of claims handled daily, call itX, is a Poisson random variable. SinceE(X)=5,
the probability that there will be fewer than 3 claims on any given day is

P{X≤ 3 }=P{X= 0 }+P{X= 1 }+P{X= 2 }

=e−^5 +e−^5

51

1!

+e−^5

52

2!

=

37

2

e−^5

≈.1247

Since any given day will have fewer than 3 claims with probability .125, it follows, from
the law of large numbers, that over the long run 12.5 percent of days will have fewer than
3 claims.
It follows from the assumed independence of the number of claims over successive days
that the number of days in a 5-day span that has exactly 4 claims is a binomial random
variable with parameters 5 andP{X= 4 }. Because

P{X= 4 }=e−^5

54

4!

≈.1755

it follows that the probability that 3 of the next 5 days will have 4 claims is equal to
(
5
3

)

(.1755)^3 (.8245)^2 ≈.0367 ■