Introduction to Probability and Statistics for Engineers and Scientists

238 Chapter 7: Parameter Estimation

it follows that the logarithm of the length of a randomly chosen grain has a normal
distribution with mean approximately equal to .7504 and with standard deviation approxi-
mately equal to .4351. Hence, ifXis the length of the grain, then

P{ 2 <X< 3 }=P{log(2)<log(X)<log(3)}

=P

{

log(2)−.7504

.4351

<

log(X)−.7504

.4351

<

log(3)−.7504

.4351

}

=P

{

−.1316<

log(X)−.7504

.4351

<.8003

}

≈ (.8003)− (−.1316)

=.3405 ■

In all of the foregoing examples, the maximum likelihood estimator of the population
mean turned out to be the sample meanX. To show that this is not always the situation,
consider the following example.

EXAMPLE 7.2g Estimating the Mean of a Uniform DistributionSuppose X 1 ,...,Xn consti-
tute a sample from a uniform distribution on (0,θ), whereθis unknown. Their joint
density is thus

f(x 1 ,x 2 ,...,xn|θ)=







1

θn

0 <xi<θ, i=1,...,n

0 otherwise

This density is maximized by choosingθas small as possible. Sinceθmust be at least as
large as all of the observed valuesxi, it follows that the smallest possible choice ofθis equal
to max(x 1 ,x 2 ,...,xn). Hence, the maximum likelihood estimator ofθis

θˆ=max(X 1 ,X 2 ,...,Xn)

It easily follows from the foregoing that the maximum likelihood estimator ofθ/2, the
mean of the distribution, is max(X 1 ,X 2 ,...,Xn)/2. ■

*7.2.1 Estimating Life Distributions

LetXdenote the age at death of a randomly chosen child born today. That is,X =iif
the newborn dies in itsith year,i≥1. To estimate the probability mass function ofX,
letλidenote the probability that a newborn who has survived his or her firsti−1 years

* Optional section.