If the probability density function (p.d.f.) of a variable Y is
f(y) = ky( 2- y) ; 0 ≤ y < 2 then the value of k is
a) 2/3
b) 3/2
c) 4/3
d) 3/4
Answers
Answered by
1
Since f(x) is a p.d.f. ∫
−∞
∞
f(x)dx=1
⇒∫
0
1
k(1−x
2
)dx=1
⇒k[x−
3
x
3
]
0
1
=1
⇒k(1−
3
1
)=1
⇒
3
2k
=1
⇒k=
2
3
(ii) The distribution function F(x)=∫
−∞
x
f(t)dt
(a) When x∈(−∞,0]
F(x)=∫
−∞
x
f(t)dt=0
(b) When xε(0,1]
F(x)=∫
−∞
x
f(t)dt=∫
−∞
0
f(t)dt+∫
0
x
f(t)dt
=0+
2
3
∫
0
x
(1−t
2
)dt
F(x)=
2
3
(x−
3
x
3
)
(c) When x∈[1,∞)
F(x)=∫
−∞
x
f(t)dt
=∫
−∞
0
f(t)dt+∫
0
1
f(t)dt+∫
1
x
f(t)dt
=0+∫
0
1
2
3
(1−t
2
)dt+0
=
2
3
[t
3
t
3
]
0
1
=1
∴F(x)=
⎩
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎧
0
2
3
(x−
3
x
3
),
1
−∞<x≤0
0<x<1
1≤x∞
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