Introduction to Probability and Statistics for Engineers and Scientists

7.2Maximum Likelihood Estimators 235

By equating to zero, we obtain that the maximum likelihood estimateˆλequals

λˆ=

∑n

1

xi

n

and so the maximum likelihood estimator is given by

d(X 1 ,...,Xn)=

∑n

i= 1

Xi

n

For example, suppose that the number of people that enter a certain retail establishment
in any day is a Poisson random variable having an unknown meanλ, which must be
estimated. If after 20 days a total of 857 people have entered the establishment, then
the maximum likelihood estimate ofλis 857/20=42.85. That is, we estimate that
on average, 42.85 customers will enter the establishment on a given day. ■

EXAMPLE 7.2d The number of traffic accidents in Berkeley, California, in 10 randomly
chosen nonrainy days in 1998 is as follows:

4, 0, 6, 5, 2, 1, 2, 0, 4, 3

Use these data to estimate the proportion of nonrainy days that had 2 or fewer accidents
that year.

SOLUTION Since there are a large number of drivers, each of whom has a small probability
of being involved in an accident in a given day, it seems reasonable to assume that the daily
number of traffic accidents is a Poisson random variable. Since

X=

1

10

∑^10

i= 1

Xi=2.7

it follows that the maximum likelihood estimate of the Poisson mean is 2.7. Since the
long-run proportion of nonrainy days that have 2 or fewer accidents is equal toP{X≤ 2 },
whereXis the random number of accidents in a day, it follows that the desired estimate is

e−2.7(1+2.7+(2.7)^2 /2)=.4936

That is, we estimate that a little less than half of the nonrainy days had 2 or fewer
accidents. ■