estatistica

Respostas:
A) xi (3, 5, 7, 9); fi (11, 13, 16, 9) x1 = 0+2/2 = 1; x2 = 4+2/2 = 3; x3 = 6+4/2 = 5; x4 = 8+6/2 = 7; x5 = 10+8/2 = 9;
B) 1.AA = x(Max) – x(min) 2. 0-9 = 9 3. AT: L(max) – l(min) 4. K= 5 ( K representa o numero de classes) 5. l = 6 6. L = 4

C) REPOSTA: h3 = 2; n = 50; l1 = 0; L3 = 6; x2 = 3; f5 = 9

Fi – Fi = fi já que o Fi = f1 + f2 f1 = 2
F2 – F1= f2
9 -2 = 2
F3 - F2= f3
21 – 9 = 12
F4 - F3 = f4
29 - 21 = 8
F5 - F4 = f5
34 – 29 = 5
RESPOSTA: fi (2, 7, 12, 8, 5)

- Primeiro calcularemos o número de classe:

K 1 + 3,3log50
K 7 classes
- Amplitude da Amostra:
AA = X(máx) – X(mín)
98 - 33 = 65.
- Amplitude de cada intervalo de classe:

65 /7
Montando a Tabela:
30 |--------40 = 33, 35, 35, 39
40 |--------50 = 41, 41, 42, 45, 47, 48
50 |--------60 = 50, 52, 53, 54, 55, 55, 56, 57, 59
60 |--------70 = 60, 61, 64, 65, 65, 65, 66, 67, 68, 68, 69
70 |--------80 = 71, 73, 73, 73, 74, 74, 76, 77, 78
80 |--------90 = 80, 81, 84, 85, 85, 88, 89
90 |-------100 = 91, 94, 94, 98 fi = 4,6,9,11,9,7,4

Xi = 1 2 3 4 5 6 fi = 6 8 9 7 10 10 fr1 = 6/50 = 0.12 fr2 =8 /50 = 0.16 fr3 = 9/50 = 0.18 fr4 = 7/50 = 0.14 fr5 = 10/50 = 0.2 fr6 = 10/50 = 0.2 ∑ = 1,00

- N = 100
K 1 + 3,3 * lon(n)
K 1+3,3 + log 100 k8 - Amplitude de amostra
AA = 108 – 62 = 46
- Amplitude de classe h 46/8 6 i 1 62| ----68
2 68|----74
3 74|---80
4 80|----86
5 86|----92
6 92|----98
7 98|----104
8 104|----110 fi = 5,14,16,24,16,13,10,2.

xi = 10 11 12 13 14 15 16 17 fi = 1 3 4 5 7 2 1 1

fr1 = 4/40 = 0,1 fr2 =10/40 = 0,25 fr3 = 14/40 = 0,35 fr4 = 9/40 = 0,225 fr5 = 3/40 = 0,075 ∑=1,000
Fi: 4, 14, 28, 37, 40 – Somatórios
Fr1 = 4/40 = 0,1
Fr2 = 14/40 = 0,35
Fr3 = 28/40 = 0,7
Fr4 = 37/40 = 0,925
Fr5 = 40/40 = 1

A) 2 + 5 + 12 + 10 + 8 + 3 = 40
B) Fr= fi/∑fi
2/40 = 0,05; 5/40 = 0,125; 12/40 = 0,3; 10/40 = 0,25; 8/40 = 0,2; 3/40 = 0,075
C) 2, 2+5 =7 ; 7+12 = 19; 19+10 = 29; 29+8 = 37;37+3=40
D)