Page 109 - Distribusi peluang binomial
P. 109
5
A D 1
P(1 < X < 5) = න 20 X + 2
1
P(1 < X < 5) = L ABCD
1
= AD + BC AB
2
B C 1 1
1 3 3 = +
AD = f 1 = 20 1 + 2 = 20 2 2
20
7
1 7 1 1
BC = f 5 = 20 5 + 2 = 20 = 2 2 (4)
20
AB = 5 − 1 = 4 = 1
4
