Zmienna losowa: odpowiedzi
Å„Å‚
ôÅ‚0 dla x 2
ôÅ‚
ôÅ‚
ôÅ‚
ôÅ‚0, 2 dla x " (2, 3
ôÅ‚
òÅ‚
xi 2 3 4 5
1. a) b) F (x) = 0, 6 dla x " (3, 4
pi 0, 2 0, 4 0, 3 0, 1 ôÅ‚
ôÅ‚0, 9 dla x " (4, 5
ôÅ‚
ôÅ‚
ôÅ‚
ôÅ‚
ół1 dla x > 5
c) P (X < 3, 5) = F (3, 5) = 0, 6 = P (X = 2) + P (X = 3) = 0, 2 + 0, 4
P (3 X < 4, 5) = F (4, 5) - F (3) = 0, 9 - 0, 2 = 0, 7 = P (X = 3) + P (X = 4) = 0, 4 + 0, 3
e) P (X = 3) = F (3+) - F (3) - wysokość  skoku dystrybuanty w trójce
f) E(X) = 2 · 0, 2 + 3 · 0, 4 + 4 · 0, 3 + 5 · 0, 1 = 3, 3
E(Y ) = E(X2 - 4X) = (22 - 4 · 2) · 0, 2 + (32 - 4 · 3) · 0, 4 + (42 - 4 · 4) · 0, 3 + (52 - 4 · 5) · 0, 1 = -1, 5 lub najpierw znalez
ć
yi -4 -3 0 5
rozkład zmiennej Y :
pi 0, 2 0, 4 0, 3 0, 1
Å„Å‚
ôÅ‚0 dla x 0
ôÅ‚
ôÅ‚
ôÅ‚0, 5 dla x " (0, 10
ôÅ‚
ôÅ‚
ôÅ‚
ôÅ‚
ôÅ‚0, 8 dla x " (10, 100
ôÅ‚
òÅ‚
xi 1000 500 200 100 10 0
2. a) b) F (x) = 0, 9 dla x " (100, 200
2 8 10 20 60
pi 200 200 200 200 200 100 ôÅ‚
200 ôÅ‚0, 95 dla x " (200, 500
ôÅ‚
ôÅ‚
ôÅ‚
ôÅ‚
ôÅ‚0, 99 dla x " (500, 1000
ôÅ‚
ôÅ‚
ôÅ‚
ół
1 dla x > 1000

c) E(X) = 53, D2(X) = 20221, Ã = D2(X) H" 141, 2
Å„Å‚
ôÅ‚0 dla x -2
ôÅ‚
ôÅ‚
ôÅ‚
1
ôÅ‚
dla x " (-2, -1
ôÅ‚
ôÅ‚ 8
ôÅ‚
òÅ‚
1
dla x " (-1, 0
12 11
4
3. a) P (X > -1) = , P (0 X < 5) = b) F (x) =
16 16 3
ôÅ‚
dla x " (0, 4
ôÅ‚
4
ôÅ‚
ôÅ‚
ôÅ‚ 15
ôÅ‚ dla x " (4, 6
ôÅ‚
16
ôÅ‚
ół
1 dla x > 6

3
c) E(X) = , D2(X) = 5, 3125, Ã = D2(X) H" 2, 3
4
17
d) E(Y ) = E(3X + 2) = 3E(X) + 2 =
4
D(Y ) = D2(3X + 2) = D2(3X) = 9D2(X) = 47, 8125 Ã H" 6, 915
1 1 1 3
4. niech P (X = x3) = p3, p3 = 1 - ( + + ), stÄ…d p3 =
2 4 16 16
1 3 1
E(X) = (-2) · + 0 + x3 · + 12 · , stÄ…d x3 = 8
2 16 16
15 15
yi - 0
15 1 15
4 4
5. np. P (Y = ) = P (X2 - = ) = P (X = 2) + P (X = -2) = 0, 1 + 0, 1
4 X2 4
pi 0, 6 0, 2 0, 2
E(Y ) = -1, 5, D2(Y ) = 9
"

c 3 2
6. a) 1 = = c, stÄ…d c =
3n 2 3
n=0
10 10

2 36-1
b) P (5 X 10) = P (X = n) = = ... = H" 0, 0041
3n+1 311
n=5 n=5
"

2 1
P (X 100) = =
3n+1 3100
n=100
" " " "

2n 2n
c) E(X) = nP (X = n) = nP (X = n) = . Niech S(x) = xn-1. Wtedy E(X) = S(1), zaÅ› S(x) =
3n+1 3n+1
n=0 n=1 n=1
n=1

" "

2 2 2x 18 1
(xn) = xn = ... = = . StÄ…d E(X) = S(1) =
3n+1 3n+1 9-3x (9-3x)2 2
n=1 n=1
k k 1

1-
2 2
3k+1
d) dla x " (k, k+1 , gdzie n = 0, 1, 2, ... mamy F (x) = P (X < x) = P (X = n) = P (X = n) = = =
1
3n+1 3
1-
3
n
0 dla x 0
1
1 - . Zatem F (x) =
3k+1 1
1 - dla x " (k, k + 1 oraz k = 0, 1, 2...
3k+1
7. P (-1 X < 2) = F (2) - F (-1) = 0, 6
xi -2 1 3
pi 0, 2 0, 6 0, 2
+"
1
A
8. a) 1 = f(x)dx = xdx = , stÄ…d A = 2
2
-" A
3
4

1 3 5
b) P ( < X ) = 2xdx = ... =
2 4 16
1
2
+"
1
2
c) E(X) = xf(x)dx = xdx =
3
-" 2
x

d) F (x) = P (X < x) = f(t)dt
-"
dla x 0:
x x

F (x) = f(t)dt = 0dt = 0
-" -"
dla x " (0, 1 :
x x x
0
F (x) = P (X < x) = f(t)dt = f(t)dt + f(t)dt = 0 + 2tdt = x2
-" -" 0 0
dla x > 1:
x x x
0 1 1
F (x) = P (X < x) = f(t)dt = f(t)dt + f(t)dt + f(t)dt = 0 + 2tdt + 0dt = 1
-" -" 0 1 0 1
+"

3
9. podobnie jak w zadaniu 8, wyznaczymy B z równości f(x)dx = 1. Otrzymamy B = .
26
-"
Å„Å‚
ôÅ‚
òÅ‚0 dla x 1
+"
3
x3-1 3 30
F (x) = dla x " (1, 3 E(X) = xf(x)dx = x3dx =
26 13
ôÅ‚
ół1 26 -" 1
dla x > 3
+" +" -1 +"

A A
10. podobnie jak w zadaniu 8, wyznaczymy A z równości f(x)dx = 1. Otrzymamy f(x)dx = dx - dx =
x4 x4
-" -" -" 1
2 3
... = A, stÄ…d A = .
3 2
Å„Å‚a
1
ôÅ‚- dla x -1
òÅ‚
2x3 -1 +"

1 A A
F (x) = dla x " (-1, 1 E(X) = dx - dx = 0
2 x3 x3
ôÅ‚
ół1 - 1 dla x > 1 -" 1
2x3

1
x2 dla x " (0, 3),
9
11. f(x) = F (x) =
0 dla pozostałych x.
Ä„ Ä„
12. lim F (x) = lim (A+B arc tg x) = A+B , lim F (x) = lim (A+B arc tg x) = A-B . Korzystamy z równości:
2 2
x+" x+" x-" x-"
Ä„ Ä„ 1 1 1
F (+") = 1, F (-") = 0, czyli A + B = 1, A - B = 0. StÄ…d A = , B = . Wtedy f(x) = F (x) = dla x " R
2 2 2 Ä„ Ä„(1+x2)
+"
2
x
"
13. E(X) = xf(x)dx = dx = ... = 0
Ä„ 4-x2
-" -2
+"
2
2
"x
D2(X) = (x - E(X))2f(x)dx = dx = ... = 2
Ä„ 4-x2
-" -2