PDE_S2
16
8/17/2019 PDE_S2 http://slidepdf.com/reader/full/pdes2 1/16 Prelucrarea Datelor Experimentale APLICATII Felicia Stan Departamentul IF [email protected] Facultatea de Inginerie
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Transcript of PDE_S2
8/17/2019 PDE_S2
http://slidepdf.com/reader/full/pdes2 1/16
Prelucrarea Datelor
Experimentale APLICATII
Felicia Stan Departamentul IF
[email protected] Facultatea de Inginerie
8/17/2019 PDE_S2
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P1. Fie variabilele aloatoare discrete:
−
9 / 23 / 13 / 19 / 1
2101 X
9 / 23 / 19 / 4
321Y
Sa se calculeze: M(X), M(Y); D2(X), D2(Y), M(X+Y).
8/17/2019 PDE_S2
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n
n
p p p p
x x x x
X L
L
321
321
( ) ∑=+++===
N
iiinn p x p x p x p x X M m
12211 KK
( ) ( )[ ] ( )[ ] ( )[ ] N N p X M x p X M x X M X M X D 2
1
2
1
22 −++−=−= K
Media
Disperia
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n
n
p p p p
x x x x
X L
L
321
321
−
9 / 23 / 13 / 19 / 1
2101
4321
4321
p p p p
x x x x
X
( )
3 / 2)9 / 2(2)3 / 1(1)3 / 1(0)9 / 1)(1(
44332211
=+++−=
=+++== p x p x p x p x X m
( ) ( )[ ] [ ] [ ]
[ ] [ ] 4
2
43
2
3
2
2
21
2
1
22
3 / 23 / 2
3 / 23 / 2
p x p x
p x p x X M X M X D
−+−+
−+−=−=
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Fie variabilela aloatoare discreta:
Sa se calculeze: M(X); D2(X), D(X).
30 / 118 / 112 / 19 / 136 / 536 / 136 / 59 / 112 / 118 / 136 / 1
12111098765432 X
7....
12
14
18
13
36
12)( =++++= X M
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P2. Variabila aleatoare continua X reprezinta curentul masurat
pentru un fir de cupru in mA.Densitatea de repartitie este
mA20x005.0)x(f ≤≤=
Care este probabilitatea ca valoarea masurata a curentuluisa fie < 10 mA?
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x
f(x)
0.05
10 20
mA20x005.0)x(f ≤≤=
Densitatea de repartitie
5.0)010(05.0|x05.0dx05.0
dx05.0dx)x(f )mA10x(P
10
0
10
0
10
0
10
0
=−==
===<
∫
∫∫
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P3. Functia de densitate a unei variabile aleatoare X =
marimea incarcarii dinamice a unui pod (in Newtons) este
≤≤+=
restin0
208
3
8
1
)( x x
x f
Sa se determine probabilitatea ca incarcarea dinamica sa fiemai mare decat 1 P(X>1).
( ) ( ) )1(1111 F X P X P −=≤−=>
Rezolvare:
∫∫∫∞−∞−
+==
x
0
0x
dy)y(f dy)y(f dy)y(f )x(F
Functia de repartitie
8/17/2019 PDE_S2
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>
≤≤+
<
=
21
2016
3
8
00)( 2
x
x x x x
xF
16
3
88
3
8
1
)()()()(
2
00
x x
dy ydy y f dy y f dy y f xF
x x x x
+∫ =
+=∫∫ +=∫=
∞−∞−
( )16
111
16
3
8
11)1(11 2
=
+−=−=> F X P
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P4. Functia de densitate a unei variabile aleatoare X este
( )
≤≤−
=
restin0
101)(
2
23 x x
x f
Sa se calculeze media aritmetica a variabilei X
)()()()()(1
1
0
0 x f xdx x f xdx x f xdx x f x X M ∫ ⋅+∫ ⋅+∫ ⋅=∫ ⋅=
∞+
∞−
∞+
∞−
( ) ( )8
3
422
3
2
31)(
1
0
421
0
32
231
0 =
−∫ =−=−∫ ⋅=
=
=
x
x
x xdx x xdx x x X M
=0
=0
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P5. La fabricarea compresoarelor, rezulta deseuri sub forma tubulara.
Fiecare deseu are lungimea reprezentata de functia de densitate f(x).Determinati probabilitatea ca lungimea unui tub selectat aleator sa fie:a) mai mica sau egala cu 4 inches;b) mai mare decat 6c) intre 5.5 si 7.5
≤≤=
rest in
x x f
0
10010
1
)(
Rezolvare (a)
∫+∫=∫=∫==≤∞−∞−∞−
4
0
04
)()()()()()4( dy y f dy y f dy y f dy y f xF xP
x
4.010
4
1010
1)4(
4
0
4
0
===∫=≤ ydy xP
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b)
( ) ( ) )(11 aF a X Pa X P −=≤−=>
( ) ( )∫=<<2
1
21
x
x
dy y f x X xP
c)
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P6. Urmatoarele functii nu pot fi considerate densitati de
repartitie. Explicati de ce.
≤≤
=rest in
x x x f
0
41)(
≤≤−
=rest in
x x f
0
211)(
≤≤−
=
rest in
x x
x f
0
303
2)(
2
=
=rest in
x x f
0
101)(
este densitate derepartiei daca:
( ) 1=∫∞+
∞−
dy y f
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P7. Determinati constantele astfel incat urmatoarele functii sa fiedensitati de probabilitate (repartitie).
≤≤=
rest in
xa
x
x f
0
40)(
≤≤=
rest in
xbx
x f 0
10
)(
2
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≤≤
=
rest in
xa x
x f
0
40)(
1)( =∫∞+
∞−
dx x f
8181
2
11
1)()()(1)(
4
0
24
0
4
0 4
0
=⇒===⇒
=++⇒=
∫
∫ ∫∫∫+∞
∞−
+∞
∞−
aa
x
adx x
a
dx x f dx x f dx x x f dx x f
Rezolvare
=0 =0
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Variabila aleatoare X are densitatea de repartitie
( ) R x
x x f ∈
+π=
1
1)(
2
1. Sa se scrie functia de repartitie corespunzatoare.2. Sa se calculeze probabilitatea ca variabila X sa ia o valoare cuprinsaintre -1 si 1
2
1arctg
1arctg
1
1
1)()( 2 +
π=
π=
+π== ∞−
∞−∞− ∫∫ xuu
duduu f xF
x x x
2
1
44
1)1()1()11( =
π+
π
π
=−−=<<− F F xP
