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3.2 Nilai Harapan (Mean/Rata–rata)
dan Varians Dist. Diskrit
Nilai harapan suatu variabel acak x ditulis E (x)
didefinisikan
E (x) = x. f (x)
Var (x) = x2 =E [ x – E (x) ] 2 =E (x 2 ) –{E (x) } 2
Jika k suatu bilangan , maka E ( k ) = k
Contoh : E (3) = 3 dan seterusnya.
Latihan Soal
1 .Dua buah dadu dilempar . Jika x = jumlah mata dadu yang timbul ,
berapakah:
a. P (3 < x 6)
b. Rata–rata (Nilai harapan)
Jawab:
a. P (3 < x 6) = P (x = 4) + P (x = 5) + P (x = 6)
= f (4) + f (5) + f (6)
= 3/36 + 4/36 + 5/36 = 12/36 = 1/3
b. E (x) = x . f(x)
= 2.1/36 + 3.2/36 + 4.3/36 + 5.4/36 + 6.5/36 +
7.6/36 + 8 .5/36 + 9 . 4/36 + 10.3/36 +
11.2/36 + 12.1/36 = 252/36 = 7
2 . Jika Nilai E (x) = 1/3 dan E (x 2 ) = 1/3 . Tentukan Nilai
Variansnya.
Jawab : Var (x) = E (x 2 ) – { E (x) } 2
= 1/3 – (1/3) 2 = 1/3 – 1/9 = 2/9
3 . Jika E (x) = 2 , berapa nilai dari : a. E [ 3 (x + 2)]
b. E [x – 3 (x + 2)]
Jawab : a. E [ 3 (x + 2) ] = E [ 3x + 6 ]
= E (3x) + E (6)
= 3. E (x) + 6
= 3 . 2 + 6 = 6 + 6 = 12
b. E [ x – 3 (x + 2) ] = E (x) – E [ 3 (x + 2) ]
= 2 – 12 = -10
4. Jika x mata dadu seimbang , berapa nilai harapan (rata – rata) nya
Jawab :
E (x) = x . f (x)
= 1 .1/6 + 2 .1/6 + 3 .1/6 + 4 .1/6 + 5 .1/6 + 6 . 1/6
= 21/6 = 3,5
II. DISTRIBUSI PROB DISKRET