Soal STATISTIKA
Contoh
1:
Ditentukan data : 6, 6, 3, 2, 2, 2, 2, 5, 4, 8
Tentukan:
a. nilai minimum
b. nilai maksimum
c. jangkauan / rentang / range
d. Kuartil bawah
e. Kuartil Atas
f. Median
g. Jangkauan antar Kuartil
h. Simpangan kuartil
i. Langkah
j. Pagar Dalam
k. Pagar Luar
l. Rata-rata (mean)
m. Simpangan rata-rata
n. Simpangan Baku
o. Varians (Ragam)
Jawab:
6, 6, 3, 2,
2, 2, 2, 5, 4, 8
Diurutkan
menjadi: 2, 2, 2, 2, 3, 4, 5, 6, 6, 8
a. nilai
minimum
Xmin = 2
b. nilai maksimum
Xmaks = 8
c. jangkauan
R = Xmaks - Xmin
R = 8 - 2
R = 6
d. Kuartil
bawah
2, 2, 2, 2, 3 | 4, 5, 6, 7, 8
Q₁ = 2
e. Kuartil
Atas
2, 2, 2, 2, 3 | 4, 5, 6,
7, 8
Q₃ = 6
f. Median
2, 2, 2, 2, 3 | 4, 5,
6, 7, 8
Q₂ = ½(3 + 4)
Q₂ = 3,5
g. Jangkauan
antar Kuartil
H = Q₃ – Q₁
H = 6 – 2
H = 4
h. Simpangan
kuartil
Qd = ½ H
Qd = ½ ∙ 4
Qd = 2
i. Langkah
L = (3/2)H
L = (3/2) ∙ 4
L = 6
j. Pagar
Dalam
Pd = Q₁ – L
Pd = 2 – 6
Pd = –4
k. Pagar
Luar
Pl = Q₃ + L
Pl = 6 + 6
Pl = 12
l. Rata-rata
(mean)
xˉ = (2 + 2 + 2 + 2 + 3 + 4 + 5 + 6 + 6 + 8) / 10
xˉ = 40/10
xˉ = 4
m. Simpangan
rata-rata
SR = (|2 – 4| + |2 – 4| + |2 – 4| + |2 – 4| + |3 – 4| + |4 – 4| + |5 – 4|
+ |6 – 4| + |6 – 4| + |8 – 4|) / 10
+ |6 – 4| + |6 – 4| + |8 – 4|) / 10
SR = (2 + 2 + 2 + 2 + 1 + 0 + 1 + 2 + 2 + 4) / 10
SR = 18/10
SR = 1,8
n. Simpangan
Baku
SB = √[{(2 – 4)² + (2 – 4)² + (2 – 4)² + (2 – 4)² + (3 – 4)² + (4 – 4)²
+ (5 – 4)² + (6 – 4)² + (6 – 4)² + (8 – 4)²} / 10]
+ (5 – 4)² + (6 – 4)² + (6 – 4)² + (8 – 4)²} / 10]
SB = √{(4 + 4 + 4 + 4 + 1 + 0 + 1 + 4 + 4 +
16)/10}
SB = √(42/10)
SB = √4,2
SB = 2,05
o. Varians
(Ragam)
R = SB²
R = 4,2
Contoh 2:
Contoh 2:
|
tinggi (cm)
|
frek.
|
|
141 – 145
|
4
|
|
146 – 150
|
7
|
|
151 –155
|
12
|
|
156 – 160
|
13
|
|
161 – 165
|
10
|
|
166 – 170
|
6
|
|
171 – 175
|
3
|
Dari data di atas,
tentukan:
a. Mean
b. Modus
c. Kuartil bawah
d. Kuartil atas
e. Media
Jawab:
a. Mean
|
tinggi (cm)
|
frek.
(fi)
|
Titik Tengah
(xi)
|
Simpangan
(di)
|
fi ∙ di
|
|
141 – 145
|
4
|
143
|
–15
|
–60
|
|
146 – 150
|
7
|
148
|
–10
|
–70
|
|
151 –155
|
12
|
153
|
–5
|
–60
|
|
156 – 160
|
13
|
158
|
0
|
0
|
|
161 – 165
|
10
|
163
|
5
|
50
|
|
166 – 170
|
6
|
168
|
10
|
60
|
|
171 – 175
|
3
|
173
|
15
|
45
|
|
Jumlah
|
55
|
–35
|
xˉ = 158 + (–35)/55
xˉ = 158 –7/11
xˉ = 157 + 4/11
xˉ = 157,36
b. Modus
|
tinggi (cm)
|
frek.
|
|
141 – 145
|
4
|
|
146 – 150
|
7
|
|
151 –155
|
12
|
|
156 – 160
|
13
|
|
161 – 165
|
10
|
|
166 – 170
|
6
|
|
171 – 175
|
3
|
Tb = 156 – 0,5 = 155.5
I = 146 – 141 = 5
d₁ = 13 – 12 = 1
d₂ = 13 – 10 = 3
Mo = Tb + I{d₁ / (d₁ + d₂)
Mo = 155,5 + 5(1/4)
Mo = 155,5 + 1,25
Mo = 156,75
c. Kuartil bawah
Mo = 155,5 + 5(1/4)
Mo = 155,5 + 1,25
Mo = 156,75
c. Kuartil bawah
|
tinggi (cm)
|
frek.
|
data ke –
|
|
141 – 145
|
4
|
1 sd 4
|
|
146 – 150
|
7
|
5 sd 11
|
|
151 –155
|
12
|
12 sd 23
|
|
156 – 160
|
13
|
24 sd 36
|
|
161 – 165
|
10
|
37 sd. 46
|
|
166 – 170
|
6
|
47 sd. 52
|
|
171 – 175
|
3
|
53 sd 55
|
¼n = ¼(55) = 13,75 ⇒ data ke-14 ⇒ kelas ke-3
F = 7 + 4 = 11
f = 12
Tb = 150,5
Q₁ = Tb + I {(¼n – F)/f}
Q₁ = 150,5 + 5 {(14 – 11)/12}
Q₁ = 150,5 + 1,25
Q₁ = 151,75
d. Kuartil atas
|
tinggi (cm)
|
frek.
|
data ke –
|
|
141 – 145
|
4
|
1 sd 4
|
|
146 – 150
|
7
|
5 sd 11
|
|
151 –155
|
12
|
12 sd 23
|
|
156 – 160
|
13
|
24 sd 36
|
|
161 – 165
|
10
|
37 sd. 46
|
|
166 – 170
|
6
|
47 sd. 52
|
|
171 – 175
|
3
|
53 sd 55
|
¾n = ¾(55) = 41,25 ⇒ data ke-41 ⇒ kelas ke-5
F = 4 + 7 + 12 + 13 = 36
f =10
Tb = 160,5
Q₃ = Tb + I {(¾n – F)/f}
Q₃ = 160,5 + 5 {(41 – 36)/10}
Q₃ = 160,5 + 2,5
Q₃ = 163,0
e. Median
|
tinggi (cm)
|
frek.
|
data ke –
|
|
141 – 145
|
4
|
1 sd 4
|
|
146 – 150
|
7
|
5 sd 11
|
|
151 –155
|
12
|
12 sd 23
|
|
156 – 160
|
13
|
24 sd 36
|
|
161 – 165
|
10
|
37 sd. 46
|
|
166 – 170
|
6
|
47 sd. 52
|
|
171 – 175
|
3
|
53 sd 55
|
½n = ½(55) = 27,5 ⇒ data ke-28 ⇒ kelas ke-4
F = 4 + 7 + 12 = 23
f =13
Tb = 155,5
Q₂ = Tb + I {(½n – F)/f}
Q₂ = 155,5 + 5 {(28 – 23)/13}
Q₂ = 155,5 + 1,92
Q₂ = 157,42
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