|
Yechimi. Hodisa: A - Nisonga o’q otib tekkiza olmagan mergan.
Gepotezalar: H — Mergan birinchi guruhga tegishli fft — Mergan ikkinchi guruhga tegishli
ff3 — Mergan uchinchi guruhga tegishli
>4 — Mergan to’rtinchi guruhga tegishli
Ehtimollikning klassik ta’rifiga ko’ra, biz gepotezalar ehtimolini topamiz:
5 7
18 18 ’ ( 3) 18
A hodisaning shartli ehtimolini topamiz:
P(A/ > ı) = 1 — 0. 8 = 0. 2 , P(A/H z) = 1 — 0. 7 = 0. 3 ,
P(A/H z) = 1 — 0. 6 = 0. 4 , P(A/ff4) = 1 — 0. 5 = 0. S .
To’liq ehtimollik formulasidan bizga kerakli ehtimollikni topamiz:
5
P(A) —— 0.2 +
18'
7
18'
4
0. 3 +
18
2
0. 4 +
=
0. 5 =
5 0. 2 + 7 0. 3 + 4 0. 4 + 2 0. 5 19
18 60
Bayes formulasidan foydalanib nishonga tekkiza olmagan mergan:
0. 317
guruhga tegishlilik ehtimolini topamiz
P(H 2 fA)
guruhga tegishlilik ehtimolini topamiz
19 '
60
60
57
7
19 0. 368
P(H /A) —— P tH 3) Pt A/H 3 ) 4g ' 0• 4 16 0. 281
3 PtA) 19 57
60
guruhga tegishlilik ehtimolini topamiz
P(H
P H 4) P A/H z) P A)
10
19 =
> 0. 175
z/A)
57
60
Javob: Nishonga o’q otib, tekkiza olmagan mergan 2-guruhga tegishlilik ehtimoli kattaroq.
tashkil qilsa, olingan 200 ta mahsulotning 3 tadan ortig‘i yaroqsiz bo‘1ishi ehtimoli topilsin.
Yechimi: = 200 , p = 0. 01 , = up = 200 0. 01 = 2
Ptk) —— 3 fk-! e°’
3 tadan ortiq yaroqsiz — bu yaroqsizlar 4 ta yoki undan ko’proq bo’lishi mumkin degani.
Qarama-qarshi voqea orqali: 1 minus yaroqsiz. N 3
V
3) = I — P(k N 3) = 1 — P t0) — P t 1) — P(2) — P(3)
P(0) =2'
Ş 2
0!
P(2) =
2!
e* 2 = e*2 0. 135 , P(1) = ' e * 2 = 2 e*2 0. 271 ,
1
3!
e* 2 = 2 e*2 0. 271 , P 3) = 2’ e 2 = 1. 33 e 2 0. 180 .
V
2) = 1 — 0. 135 — 0. 271 — 0. 271 — 0. 180 = 0. 143 .
Javob: Agar yaroqsiz mahsulotlar miqdori, mahsulotlar umumiy miqdorining ha ni tashkil qilsa, olingan 200 ta mahsulotning 3 tadan ortig‘i yaroqsiz bo‘1ishi ehtimoli P(k > 2) = 0. 143 ga teng.
Quyida berilgan tanlanma taqsimotiga ko‘ra: a) Nisbiy chastotalar taqsimoti topilsin. b) Nisbiy chastotalar poligonini yasang. c)Tan1anma o’rta qiymat
*2 va tanlanma dispersiyalarni D2 toping.
z,
|
6
|
8
|
9
|
12
|
ni
|
14
|
10
|
10
|
16
|
Yechimi: a) To’plam hajmi: n = Şt- 1 ni = 14 + 10 + 10 +
16 = 50 Nisbiy chastotalarni hisoblaymiz:
w, = 1 = = 0. 28 , w2 = n¿ _ 10 = 0. 2 , w _ n3 _ 10 = 0.2,
nş 1 5 6
4
' n 50
= 0. 32 .
n 50 3 50
Demak,
i
|
6
|
8
|
9
|
12
|
w,
|
0.28
|
0.2
|
0.2
|
0.32
|
4
i=1
b)
w; —— 0. 28 -1- 0. 2 -1- 0. 2 -1- 0. 32 = 1
n SO ((6 — 8. 92)' 14 + (8 — 8. 92)' 10 + (9 — 8. 92)' 10 + (12 — 8. 92)' 16) =
Bosh to‘plamdan hajmi n = 50 ga teng tanlanma olingan va taqsimoti quyidagicha:
z;
|
2
|
4
|
7
|
12
|
ni
|
18
|
10
|
11
|
11
|
Bosh to‘plamning tuzatilgan dispersiya s’ ni toping.
( 218+410+Tl1+12.11) 285 5.7
50 SO
\ t(zi—* )'ni ((2— 5. 7)'- 18 + (4 — 5.7)- 2 10 + (7 — 5. 7)- 2 11 + (12 — 5. 7)- 2 11) 730. 5
" n
o » 1-4. 61
5500
50 730. 5
s5o0 ' 14 61
5'
Javob: s2 < 14. 91
(n — 1) (50 — 1) 49 14. 91
Quyida berilgan tıi - empirik chastotalar va unga mos tt, - nazariy chastotalar hisoblangan.
n;
|
5
|
13
|
30
|
72
|
99
|
81
|
31
|
12
|
n,
|
6
|
14
|
33
|
71
|
96
|
83
|
40
|
14
|
Pirsonning muvofiqlik kiriteriyasiga ko’ra , n = 0, 05 qiymatdorlik darajasida Bosh to’plam normal taqsimlanganligi haqida gipoteza tekshiring.
Yechimi: Quyidagi hisoblash jadvalini to’ldiramiz:
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
i
|
ni
|
n,
|
ni — n,
|
(n; — n,)'
|
(ni— n,) 2
ni
|
n§
|
n{
n,
|
1
|
5
|
6
|
-1
|
1
|
0,2
|
25
|
4,17
|
2
|
13
|
14
|
-1
|
1
|
0,08
|
169
|
12,07
|
3
|
30
|
33
|
•3
|
9
|
0,3
|
900
|
27,27
|
4
|
72
|
71
|
1
|
1
|
0,01
|
5184
|
73,01
|
5
|
99
|
96
|
3
|
9
|
0,09
|
9801
|
102,09
|
6
|
81
|
83
|
-2
|
4
|
0,05
|
6561
|
79,05
|
7
|
31
|
40
|
-9
|
81
|
2,61
|
961
|
24,03
|
8
|
12
|
14
|
-2
|
4
|
0,33
|
144
|
10,29
|
|
343
|
357
|
|
|
X$q = 3, 68
|
|
331,98
|
Tekshirish: Xçq = 3, 68 — n = 331, 98 — 343 = —11, 02
Jadvaldagi ma’lumotlar notug’ri berilgan.
Tanlanma guruhlari soni s —— 8. Demak k = 8 — 3 = 5.
X' taqsimotining kritik nuqtalari jadvalidan n = 0, 05 va k = 5 ga mos keluvchi XŞ, qiymatini topamiz:
X g(0, 05; 5) = II, 1
X k gg < Xç, bo’lgani uchun ff 0 gipotezani rad etishga asos yo’q.
Boshqacha aytganda, empirik va nazariy chastotalar farqi muhim emas (tasodifiy). Demak, kuzatishlar natijasi bilan bosh to’plam normal taqsimlangan degan gepoteza muvofiq keladi.
Quyida berilgan korreliatsion jadval malumotlari asosida Y ning X ga to‘g‘ri chiziqli regressiya tenglamasini tuzing va tanlanma korreliatsiya koeffitsentini (r2)ni toping.
¥/X
|
1
|
7
|
13
|
19
|
25
|
tig
|
11
|
4
|
6
|
-
|
-
|
-
|
10
|
16
|
-
|
4
|
6
|
-
|
-
|
10
|
21
|
3
|
-
|
20
|
-
|
9
|
32
|
26
|
-
|
-
|
4
|
12
|
6
|
22
|
31
|
-
|
-
|
-
|
1
|
5
|
6
|
ng
|
7
|
10
|
30
|
13
|
20
|
n = 80
|
Yechimi: Soxta nollar sifatida Cm - 13 va C2' 21 ni, qadamlari sifatida fi =
6 va fi2 = S ni tanlab shartli variantalardagi 2-korreliatsion jadvalni tuzamiz, bu yerda
Ai — f 1 Y¡—C 2
(1)
Endi (u; r) tanlanma ma’1umot1ari bo’yicha korreliatsion jadval tuzamiz.
r/u
|
-2
|
-1
|
0
|
1
|
2
|
up
|
-2
|
4
|
6
|
-
|
-
|
-
|
10
|
-1
|
-
|
4
|
6
|
-
|
-
|
10
|
0
|
3
|
-
|
20
|
-
|
9
|
32
|
1
|
-
|
-
|
4
|
12
|
6
|
22
|
2
|
-
|
-
|
-
|
1
|
5
|
6
|
nq
|
7
|
10
|
30
|
13
|
20
|
n=80
|
Ushbu jadval malumotlariga ko’ra o’rta qiymat va o’rtacha kvadratik chetlanishlarni hisoblaymiz.
\ nq u _ 7 (—2) + 10 (—1) + 30 0 + 13 1 + 20 2
n 80 - 0, 36
\ nq- r I0- (—2) + I0- (—1) + 3 2-0 + 2 2- 1 + 6- 2
n 80 = 0, 05
Yordamchi u 2 va v2 kattaliklarni topamiz
2 -\ “ u2 7 .4-+ 10 1 -+ 30 0-+ 13 1 + 20- 4
n 80 < 1, 64
2 - ‘” v' - 10 4-+ 10 1-+ 32
80
trp va wp kattaliklarni topamiz:
0-+ 22 -1+ 6 4
= 1, 2
og —— = 1 64 0 3 6 2 = 51 I, 23
= = 1 2 0 05 2 = 1 1975 1, 09
' **
|
+ * '
|
0,36 6 -F 13 = 15, 16
|
Y — **
|
z + |
0, 05 ' 5 + 21 = 21, 2S
|
°x
|
= >ı °
|
= 6 1, 23 = 7, 38
|
r'
|
z ' r
|
- 5 ' 1, 09 = 5, 45
|
U holda (1) almashtirishlarni qo’1lab quyidagi tengliklarni hosil qilamiz:
(2)
Shundan so’ng l-jadval ma’lumotlariga ko’ra xy — ni hisoblaymiz. Buning uchun avval quyidagi shartli o’rtachalarni hisoblaymiz:
n, zY; 11- 4 + 16- 0 + 21- 3 + 26- 0 + 31- 0
107
< 15, 29
/S2
f - n , 2Y; 11- 6 + 16-4 + 21- 0 + 26- 0 + 31-
0 130
n 2 10
_ -1 ni3ri- 11 0-+ 16 6-+ 21 20-+ 26 4-+ 31 0 620
\ - 1 i4ri 11 0 + 16 0 + 21 0 + 26 12 + 31 1 343
' ng'4
> 26, 39
13 ' ' 13
=1 i5Y; 11- 0 + 16- 0 + 21- 9 + 26- 6 + 31- 5 500
= =25
20 20
U holda:
(2)
va (3) tengliklardan foydalangan holda tanlanmaning regressiya tenglamasini va tanlanma korreliatsiya koeffitsentlarinitopamiz:
XY — Ğ- Y 3 51, 18 — 15, 16 21, 25
tr ' 7,
> 0, 53
25
< 0, 72
Y z — 21, 25 = 0, 53 (X — 15, 16) Y z —— 0, 53X + 13, 22
Bundan 0, S 1 bo’lganligi uchun X va Y tanlanma malumotlari kuchli bog’langan ekan.
F ning X ga regressiya to‘g‘ri chizig‘ining tanlanma tenglamasini l-korreliatsion jadvalda keltirilgan malumotlar bo'yicha toping va tanlanma korreliatsiya koeffitsenti ry ni aniqlang.
V\X
|
20
|
26
|
30
|
36
|
40
|
n„
|
16
|
4
|
6
|
-
|
-
|
-
|
10
|
26
|
-
|
8
|
10
|
-
|
-
|
18
|
36
|
-
|
-
|
32
|
3
|
9
|
44
|
46
|
-
|
-
|
4
|
12
|
6
|
22
|
66
|
-
|
-
|
-
|
1
|
5
|
6
|
n,
|
4
|
14
|
46
|
16
|
20
|
n = 100
|
jadva1.
Yechish. Soxta nollar sifatida Ui = 30 va C2 36 ni, qadamlari sifatida ht 5 va 2 10 ni tanlab
shartli variantalardagi 2-korreliatsion jadvalni tuzamiz. bu yerda
va quyidagicha bog'1anish1ar mavj ud:
Î ÛÛ1 /¿, ÛÛ2 // ,Ay U ’O Oy Ü2’ O .
Endi (u; r) tanlanma ma'1umot1ari b'yicha korreliatsion jadval tuzamiz.
(3.1)
r\u
|
-2
|
-1
|
0
|
1
|
2
|
nt
|
-2
|
4
|
6
|
-
|
-
|
-
|
10
|
-1
|
-
|
8
|
10
|
-
|
-
|
18
|
0
|
-
|
-
|
32
|
3
|
9
|
44
|
1
|
-
|
-
|
4
|
12
|
6
|
22
|
2
|
-
|
-
|
-
|
16
|
|
6
|
nq
|
4
|
14
|
46
|
16
|
20
|
n = 100
|
jadva1.
Ushbu jadval malumotlariga ko'ra o’r ta qiymat va o'rtacha kvadratik chetlanishlar ni hisoblaymiz.
nq u 4 (—2) + 14 (— 1) + 46 0 + 16 1 + 20
n 100
2 = 0. 34;
nv r 10 (—2) -I- 18 (—1) + 44 0 + 22 1 -I- 6 2 0, 04;
n 100
Yordamchi u 2 va r 2 kattaliklarni topamiz
oq va e kattaliklarni topamiz:
v = u ( ) 2 = 2 0 34 1. 0T; = ) 1 04 0 04) 1. 02.
U holda (3.1) almashtirishlarni qo’llab quyidagi tengliklarni hosil qilamiz:
— z —— 0.34 5 + 30 = 31. T; = —0, 04 10 + 36 = 35. 6; = 1. 0T 5 = 5.35; = 1, 02 10 = 10.2. (3.2)
Shundan so’ng l-j adval ma’lumotlariga ko’ra zp— ni hisoblaymiz. Buning uchun avval quyidagi shartli o’rtachalarni hisoblaymiz:
16 4 + 26 0 + 36 0 + 46 0 + 56 0 16,
4
5
- i- 1 16 6 + 26 8 + 36 0 + 46 0 + 56 0 152
*2 U mz
|
|
|
14
|
|
7
|
|
|
16
|
0 + 26 10 + 36 32 + 46 4 + 56
|
0
|
798
|
|
|
|
46
|
|
23
|
- i-1
|
|
16
|
0 + 26 0 + 36 3 + 46 12 + 56
|
1
|
179
|
|
|
|
l6
|
|
4
|
|
|
|
46 6 + 56
|
= 44.
|
U holda:
|
|
|
|
|
j=1 1170, 2. (3.3)
(3.2) va (3.3) tengliklardan foydalangan holda tanlanmaning regressiya tenglamasini((2.14) va (2.15)) va tanlanma korreliatsiya koefhtsentlarini topamiz ((2.16))
zp — z p 1170, 2 — 31, 7 35, o 2 (5, 35)2
1, 46;
0, 76;
Bundan 0, 5 c< rz
¿, — 35, 6 = 1, 46 (z — 31, 7) j, 1, 46z — 10, 628.
1 bo’lganligi uchun A va V tanlanma malumotlari kuchli bog’langan ekan.
31
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