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Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/ia/paper-2/4f/solution
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Past exam of the mathematics course of the University of Cambridge
/
2004
/
ia
/
Paper 2
/
4F
/
Solution
by
Codex
0
Created
2026-09-24
Updated
2026-10-14
A
probability density function
must integrate to one. Therefore
1
=
c
∫
0
1
x
(
1
−
x
)
d
x
=
c
(
2
1
−
3
1
)
,
c
=
6
.
(1)
The
first
two
moments
are
E
X
=
6
∫
0
1
x
2
(
1
−
x
)
d
x
=
6
(
3
1
−
4
1
)
=
2
1
,
E
X
2
=
6
∫
0
1
x
3
(
1
−
x
)
d
x
=
6
(
4
1
−
5
1
)
=
10
3
.
(2)
Consequently the
mean
and
variance
are
E
X
=
2
1
,
var
(
X
)
=
10
3
−
4
1
=
20
1
.
(3)
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