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Some Important Continuous Distributions 193
f (x)
X
1
—
8
x
–3 0 5
Figure 7.2 Probability density function, f (x), of X, in Example 7.1
X
We are interested in the probability P( 3 X 0). As seen from Figure 7.2, it
is clear that this probability is equal to the ratio of the shaded area and the unit
total area. Hence,
1 3
P
3 X 0 3 :
8 8
It is also clear that, owing to uniformity in the distribution, the solution can
be found simply by taking the ratio of the length from 3 to 0 to the total length
of the distribution interval. Stated in general terms, if a random variable X is
uniformly distributed over an interval A, then the probability of X taking
values in a subinterval B is given by
length of B
P
X in B :
7:4
length of A
7.1.1 BIVARIATE UNIFORM DISTRIBUTION
Let random variable X be uniformly distributed over an interval (a 1 , b 1 ), and let
random variable Y be uniformly distributed over an interval (a 2 , b 2 ). Further-
more, let us assume that they are independent. Then, the joint probability
density function of X and Y is simply
8
1
; for a 1 x b 1 ; and a 2 y b 2 ;
<
f XY
x; y f
x f
y
b 1 a 1
b 2 a 2
X
Y
:
0; elsewhere:
7:5
TLFeBOOK