Introduction to Probability and Statistics for Engineers and Scientists

188 Chapter 5: Special Random Variables

n = 1

n = 3

n = 10

FIGURE 5.13 The chi-square density function with n degrees of freedom.

conclude that the density ofXis given by

f(x)=

1

2

e−x/2

(x

2

)(n/2)− 1

(n

2

) , x> 0

The chi-square density functions having 1, 3, and 10 degrees of freedom, respectively,
are plotted in Figure 5.13.
Let us reconsider Example 5.8c, this time supposing that the target is located in the
two-dimensional plane.

EXAMPLE 5.8d When we attempt to locate a target in two-dimensional space, suppose that
the coordinate errors are independent normal random variables with mean 0 and standard
deviation 2. Find the probability that the distance between the point chosen and the target
exceeds 3.

SOLUTION IfDis the distance andXi,i=1, 2 are the coordinate errors, then

D^2 =X 12 +X 22

SinceZi=Xi/2,i=1, 2, are standard normal random variables, we obtain

P{D^2 > 9 }=P{Z 12 +Z 22 >9/4}=P{χ 22 >9/4}=e−9/8≈.3247

where the preceding calculation used the fact that the chi-square distribution with
2 degrees of freedom is the same as the exponential distribution with parameter 1/2. ■

Since the chi-square distribution withndegrees of freedom is identical to the gamma
distribution with parametersα=n/2 andλ=1/2, it follows from Equations 5.7.3