Tugas 2.xlsx

Problem Find the probability distribution of the random variable X if X~Bin(10,0.4). Find also P(X=5) and P(X<2). Then find the mean and variance for X. Solution a. The random variable X X~Bin(10,0.4) x= 4 n= 10 x= 3 p= 0.4 x= 2 So, the probability from the random variables X are as follows : (■8(𝑛@𝑥)) 𝑝^𝑥 (1−𝑝)^(𝑛 −𝑥) P (x) = P (x = P (x = P (x =

b. P (x=5) x=

4) 3) 2)

= = =

0.25082 0.2150 0.12093

X~Bin(10,0.4) n= 10 p= 0.4 So, the probability from X=5 is as follows : 5

(■8(𝑛@𝑥)) 𝑝^𝑥 (1−𝑝)^(𝑛 −𝑥) P (x) = P (x =

P (x<2) x= x=

5)

=

0.20066

0 1

X~Bin(10,0.4) n= 10 p= 0.4

So, the probability from X<2 are as follows : (■8(𝑛@𝑥)) 𝑝^𝑥 (1−𝑝)^(𝑛 −𝑥) P (x) = P (x = P (x =

0) 1)

= =

0.0060 0.0403

total

0.0464

Problem A fair die is rolled 8 times. Find the probability that no more than 2 sixes comes up. Then find the mean and variance for X. Solution a. The probability that no more than 2 sixes comes up is as follows : X<2 So, the probability that comes up 0 and 1 of six are : n(S) = 48 x=0 n(A) = 40 P = 0.83333 x=1 n(A) = 8 P = 0.16667 Total = 1 OK b. Mean and variance for X Xx P (X x) X.P(X x) X².P(X x) x-µ (x - µ)² (x - µ)².P(Xx)

0 0.83333 0 0 -0.1667 0.02778 0.02315

1 0.16667 0.16667 0.16667 0.83333 0.69444 0.11574

So, The mean for X is as follows : Ex = µx∑▒ = 〖𝑥 . 〗 � _((𝑋𝑥)) = 0.16667

So, The variance for X is as follows : Vx = σx∑▒ = 〖 (𝑥 − 𝜇)^2.�_((𝑋𝑥)) 〗 = 0.13889

Problem A large industrial firm allows a discount on any invoice that is paid within 30 days. Of all invoices, 10% receive the discount. In a company audit, 12 invoices are sampled at random. What is probability that fewer than 4 of 12 sampled invoices receive the discount? Then, what is probability that more than 1 of the 12 sampled invoices received a discount. Solution a. X < 4 X~Bin(12,0.1) x= 3 n= 12 x= 2 p= 0.1 x= 1 x= 0 So, the probability from the random variables X are as follows : (■8(𝑛@𝑥)) 𝑝^𝑥 (1−𝑝)^(𝑛 −𝑥) P (x) = P (x = P (x = P (x = P (x =

3) 2) 1) 0) Total

= = = = =

0.08523 0.23013 0.37657 0.28243 0.97436

b. More than 1 (1< X ≤12) So, the probability from the random variables X are as follows : (■8(𝑛@𝑥)) 𝑝^𝑥 (1−𝑝)^(𝑛 −𝑥) P (x) = No 1 2 3 4 5 6 7 8 9 10 11

X 2 3 4 5 6 7 8 9 10 11 12 Total

P (x) 0.230127770466 0.085232507580 0.021308126895 0.003788111448 0.000491051484 0.000046766808 0.000003247695 0.000000160380 0.000000005346 0.000000000108 0.000000000001 0.34099774821

10% receive the han 4 of 12 sampled