For part (b),
Figure 3.9 shows pX (x) for the discrete portion and fX (x) for the continuous
portion of X. They are given by:
and
;
Note again that the area under fX (x) is no longer one but is
To obtain P(X 2) and P(2 X 6), both the discrete and continuous
portions come into play, and we have, for part (a),
3 x
pX(x)
fX(x)
3 x
—^1
2e
—^1
3
—^1
3e
—^1
6e
(a) (b)
Figure 3. 9 (a) Partial probability mass function, pX (x), and (b) partial probability
density function, fX (x), of X, as described in Example 3.4
48 Fundamentals of Probability and Statistics for Engineers
P 2 <X 6 FX 6 FX 2
1
e^2
2
1 e^2 =^3 e^2 =^3
e^2
2
:
pX
x
1
2e
; atx 3 ;
0 ; elsewhere;
8
<
:
fX
x
dFX
x
dx
0 ; forx< 0 ;
1
3
ex=^3 ; for 0x<
1
6
ex=^3 ; forx 3 :
8
>>
>>
<
>>>
>:
3
1 pX
3 1
1
2e
:
> <
P X> 2
Z 1
2
fX
xdxpX
3
1
3
Z 3
2
ex=^3 dx
1
6
Z 1
3
ex=^3 dx
1
2e
e^2 =^3