ΛΥΣΕΙΣ ΘΕΜΑΤΩΝ ΣΤΑ ΜΑΘΗΜΑΤΙΚΑ ΚΑΙ ΣΤΟΙΧΕΙΑ ΣΤΑΤΙΣΤΙΚΗΣ

°ENIKH ¶AI¢EIA
ZHTHMA 1
A1 ÛÂÏ 152 A2 ÛÂÏ 142 A3 ÛÂÏ 65
A4  ·) @ §  ‚) @ §  Á) @ ™  ‰) @ §  Â) @ ™
Œ¯Ô˘Ì ٷ ‰Â‰Ô̤ӷ: 64 < N(ø) < 72 Î·È P(M) =
1
4
, P( A) = 4Ï
2
, P( K) = -5Ï  +
7
4
B1 Afi P(M) =
1
4
.
N(M)
N(ø)
=
1
4
N(ø) = 4N(M) Î·È ¤ÙÛÈ ·fi .
64  < N( ø)  < 72  .  64  < 4N( M)  < 72  .  16  < N( M)  < 18  .  N( M)  = 17
ηıfiÙÈ N(M) ·Î¤Ú·ÈÔ˜. ŒÙÛÈ N(ø) = 68
Â›Û˘ Î·È Ì N(ø) = ÔÏ4 ·Ó¿ÌÂÛ· ÛÙÔ˘˜ 64 Î·È 72 ...
B2 Afi  P( A)  + P( K)  + P( M)  = 1  .  4Ï
2
-5Ï +
7
4
+
1
4
= 1  .  4Ï
2
-5Ï + 1 = 0
·fi fiÔ˘ Ï = 1 ‹ Ï =
1
4
.
ŸÌˆ˜ ÌÂ Ï = 1 Â›Ó·È P(A) = 4 ·‰‡Ó·ÙË. ŒÙÛÈ Ï =
1
4
B3 MÂ Ï =
1
4
ÚÔ·ÙÔ˘Ó P(A) =
1
4
= P(M) Î·È P(K) =
1
2
, ÔfiÙÂ ÌÂ N(ø) = 68:
N(A)
N(ø)
=
N(M)
N(ø)
=
1
4
N(A) = N(M) = 17 Î·È .
N(K)
N(ø)
=
1
2
N( K) = 34 .
B4 £¤Ïˆ ÙËÓ Èı·ÓfiÙËÙ· ÁÈ· ÕÛÚË ‹ M·‡ÚË ‰ËÏ. ÙËÓ
P A,M =
N A,M
N(ø)
=
N(A) + N(M)
N(ø)
=
17 + 17
68
=
1
2
.
Z ‹ Ù Ë Ì · !
Z ‹ Ù Ë Ì · @Z ‹ Ù Ë Ì · #
°1 AÊÔ‡ ·fi ˘fiıÂÛË ÙÔ ÙÌ‹Ì· ¢E // xãx ı· Â›Ó·È y¢
= yE . f3% = f4%.
ŒÙÛÈ ·fi f1% + f2% + f3% + f4% + f5% = 100 . 10 + 20 + f3% + f3% + 10 = 100
¢f3% + f3% = 60  .  f3% = f4% = 30 ‰ËÏ y .
= yE
= 30.
T· ÛËÌ›· A Î·È H ·ÓÙÈÛÙÔÈ¯Ô‡Ó ÛÙȘ ÎÂÓÙÚÈΤ˜ ÙÈ̤˜ ÙˆÓ ˘ÔıÂÙÈÎÒÓ ÎÏ¿ÛˆÓ.
Œ¯Ô˘Ì ÏÔÈfiÓ 5 ÎÏ¿ÛÂȘ Ì ÎÂÓÙÚÈΤ˜ ÙÈ̤˜ ÙȘ ÙÂÙÌË̤Ó˜ ÙˆÓ B, °, ¢, E, Z.
O x
f
i
i
%
10
20
30
10 8 12 14 18 20 16
B Z
°
¢ E
A H
K§A™EI™ xi
f
i%
9 - 11 10 10
11 - 13 12 20
13 - 15 14 30
15 - 17 16 30
17 - 19 18 10
°4 ¶ˆÏ‹ÛÂȘ ¿Óˆ ·fi 15 ¯ÈÏ. ·ÓÙÈÛÙÔÈ¯Ô‡Ó ÛÙȘ 4Ë Î·È 5Ë ÎÏ¿ÛË.
ŒÙÛÈ ÙÔ ÔÛÔÛÙfi ÙˆÓ ˆÏËÙÒÓ Â›Ó·È f4 % + f5 % = 40%
°5 TÔ ÂÌ‚·‰fiÓ ÙÔ˘ ¯ˆÚ›Ô˘ ÙÔ˘ ÔÏ˘ÁÒÓÔ˘ Û˘¯ÓÔÙ‹ÙˆÓ ÈÛÔ‡Ù·È Ì ÙÔ
Û˘ÓÔÏÈÎfi ̤ÁÂıÔ˜ Ó. ŒÙÛÈ ¤¯Ô˘ÌÂ Ó = 80.
ÕÚ· Ô ·ÚÈıÌfi˜ ÙˆÓ ·Ú·¿Óˆ ˆÏËÙÒÓ Â›Ó·È Ó4,5
=
40
100
Ó . Ó4,5
= 32
°2 °3¢1 f(x) = e
1
3
x# x
2
-
11
10
x +
2
5 ÌÂ ¶.O ÙÔ ®.
E›Ó·È f(x) = e
1
3
x# x
2
-
11
10
x +
2
5
1
3
x
3
-
11
30
x
2
+
2
15
x ã = e
1
3
x# x
2
-
11
10
x +
2
5 x
2
-
11
15
x +
2
5
.
Afi f(x) = 0 . x
2
-
11
15
x +
2
5
= 0
x1
=
5
15
x2
=
6
15
-ß
5
15
6
15
+ß
Z ‹ Ù Ë Ì · $
x
f
fã + -     +
¢2 AÊÔ‡ ·fi ˘fiıÂÛË A ⊆ B, ı· Â›Ó·È P(A) ≤ P(B) Î·È ¤ÙÛÈ
P(A) =
5
15
P(B) =
6
15
E›Û˘ P A ü B = P( A)  =
5
15
Î·È P A © B = P( B)  =
6
15
.
ŒÙÛÈ Ë P A - B = P( A)  - P  A  ü  B = 0  Î·È P  B - A = P( B)  - P  A  ü  B =
1
15
¢3 h(x) = e
1
5
x#
3x
2
2
- x  -
1
3 ÌÂ ¶.O ÙÔ ®.
·) Afi f(x) = h(x) . e
1
3
x# x
2
-
11
10
x +
2
5 = e
1
5
x#
3x
2
2
- x  -
1
3 .
1
3
x# x
2
-
11
10
x +
2
5
=
1
5
x#
3x
2
2
- x  -
1
3
·fi fiÔ˘ x1
= 0 ‹
1
3
x
2
-
11
30
x +
2
15
=
3x
2
10
-
1
5
x -
1
15
10x .
2
- 11x  + 4  = 9x
2
- 6x  - 2  .
x
2
- 5x  + 6  = 0
x2
= 2
x3
= 3
‚) E›Ó·È vi
= 2xi
+ 1 ÔfiÙÂ
ÌÂ x1
= 0  Ë  Ó1
= 1
ÌÂ x2
= 2  Ë  Ó2
= 5
ÌÂ x3
= 3  Ë  Ó3
= 7
E›Ó·È x =
xi Σ #vi
i = 1
3
v
=
0#1 + 2#5 + 3#7
1 + 5 + 7
=
31
13http://exams.ert.gr/images/stories/themata/Geniki.Paidia-2011.pdf