Pembahasan serta soal jawaban lks matematika bab statiska hal 43

Unsur - unsur atau bagian - bagian suatu tabel distribusi frekuensi :

a. Kelas interval = 31- 40
b. Batas Bawah (BB) = 31
c. Batas Atas (BA) = 40
d. Tepi bawah (TB) = BB-0.5
                      = 31-0.5
                      = 30.5
e. Tepi atas (TA)   = BA-0.5
                      = 40 +0.5
                      = 40.5
f. Titik tengah (Xi) =Bb+Ba/2
                      = 31+40/2
                      = 71 /2
                      = 35.5
G. Byk Kelas (K) = byk kelompok
                     data / 1+ 3,3
                     log n
             
H. Distribusi Frekuensi relatif 
   Frelatif = fi/total fi x 100%
 
Pemantapan

1. Jangkauan
= Data terbesar - data terkecil = 98 - 71 = 27

2.Banyak kelas
= 1 + 3,3 log n
= 1 + 3,3 log 30
= 1 + 3,3 .1,48
= 1 + 4,8
= 5,8 dibulatkan menjadi 6

3.Panjang kelas
= Jangkauan/Byk kelas
= 27 / 6
= 4,3 dibulatkan menjadi 5

4.Interval pertama
Ba= Bb +(p -1)
   = 71 + (5-1)
   = 71 + 4
   = 75

5. interval kedua
Ba = Bb + (p-1)
    = 75 + (5-1)
    = 75 + 4
    = 79

Uk 1

1. Panjang kelas
= Ba - Bb + 1
= 47- 42 + 1
= 5 + 1
= 6

2.byk siswa yg nilainya dibwh 60
42- 47 = 8
48- 53 = 6
54- 59 = 3
           --- +
           17 org

3. Slsh byk siswa yg nilainya <67 & >72

< 67 = 8+6+3+7 = 24
> 72 = 9+5.      = 14
                      ---- -
                     10 org
4.TB & TA kls 4
56 - 62
Tb = Bb -0.5
    = 56-0.5 = 55.5
Ta = Ba + 0.5
    = 62 +0.5 = 62.5

5. Presents byk siswa >62 kg
fi = 9+2 = 11
Zigma fi = 3+8+6+12+9+2 = 40

Frelatif = 11/40 × 100= 27.5

6.histogram menujukkan nilai yg byk diperoleh = 60.5 -70.5
                = 61 - 70

7. Byk buah = 100
  Perbandingan 8-10&14-16 = 1:5
  Berusia 17- 19 sbyk

10+ 20+x+15+5x+5x = 100
45 + x + 10x = 100
45 + 11x = 100
11x = 100-45
11x = 55
x = 55/11 = 5

5x = 5.5 = 25

8.byk ssw >66 = 19 org
40-48 = ?

(P+ 4) (p -3) + 2 = 19
2p +1+2 = 19
2p = 19-3
2p = 16
p = 16/2 = 8

2p-1 = 2.8 -1 = 16 -1 = 15

9. 148.5 - 142.5 = 6/2 = 3
  154,5 - 3 = 151.5 + 0.5 = 152
  154.5 + 3 = 157.5 - 0.5 = 157
  Ada 6 anak

10. 160.5 -3 = 157.5 +0,5= 158
    160.5 + 3 = 163.5 -0.5 = 163

Pemusatan data tunggal
mean =zigma fi xi / fi
Me ganjil = xn+1/2
Me genap = xn/2+ xn/2+1
              -------------
                    n
Mo = dgn fi yang plg byk
Qi = i/4 (n+1)
Di = i /10 (n+1)
Pi = i /100 (n+1)

1.mean = 1+3+4+3.5+6+7+2.8/10
        = 8+15+13+16/10
        = 52/10 = 5.2
Mo = 5

2.median = n = 20
  Me genap = x20/2+x20/2+1
                ---------------
                      2
            = x10 +x11/2
            = 7+8 /2 = 15/2 = 7.5