Answer:
a. [tex]P(x\le 3) = 0.9957[/tex] and [tex]P(x < 3) = 0.965[/tex]
b. [tex]P(X \ge 4) = 0.0043[/tex]
c. [tex]P(1 \le x \le 3)= 0.5313[/tex]
d. [tex]E(x) = 0.75[/tex] and [tex]\sigma(x)= 0.8441[/tex]
e. [tex]P(0) = 0.0769[/tex]
Step-by-step explanation:
Given
X ~ Bin(15, 0.05)
This implies that:
[tex]n = 15[/tex]
[tex]p =0.05[/tex]
A binomial distribution is represented as:
[tex]P(x) = ^nC_x\ p^x\ (1- p)^{n-x}[/tex]
Solving (a):
(i)
[tex]P(x\le 3)[/tex]
This is solved as:
[tex]P(x\le 3) = P(0) + P(1) + P(2) + P(3)[/tex]
[tex]P(0) = ^{15}C_0\ * 0.05^0\ (1- 0.05)^{15-0} = 1 * 1 * 0.95^{15} = 0.4633[/tex]
[tex]P(1) = ^{15}C_1\ * 0.05^1\ (1- 0.05)^{15-1} = 15 * 0.05 * 0.95^{14} = 0.3658[/tex]
[tex]P(2) = ^{15}C_2\ * 0.05^2\ (1- 0.05)^{15-2} = 105 * 0.05^2 * 0.95^{13} = 0.1348[/tex]
[tex]P(3) = ^{15}C_3\ * 0.05^3\ (1- 0.05)^{15-3} = 455 * 0.05^3 * 0.95^{12} = 0.0307[/tex]
So:
[tex]P(x\le 3) = 0.4644 + 0.3658 + 0.1348 + 0.0307[/tex]
[tex]P(x\le 3) = 0.9957[/tex]
(ii)
[tex]P(x < 3) = P(0) + P(1) + P(2)[/tex]
[tex]P(x < 3) = 0.4644 + 0.3658 + 0.1348[/tex]
[tex]P(x < 3) = 0.965[/tex]
Solving (b):
[tex]P(X \ge 4)[/tex]
This is represented as:
[tex]P(x \le 3) + P(X \ge 4) = 1[/tex]
[tex]P(X \ge 4) = 1 - P(x \le 3)[/tex]
[tex]P(X \ge 4) = 1 - 0.9957[/tex]
[tex]P(X \ge 4) = 0.0043[/tex]
Solving (c):
[tex]P(1 \le x \le 3)[/tex]
This is represented as:
[tex]P(1 \le x \le 3) =P(1) + P(2) + P(3)[/tex]
[tex]P(1 \le x \le 3) = P(1) + P(2) + P(3)[/tex]
[tex]P(1 \le x \le 3)= 0.3658 + 0.1348 + 0.0307[/tex]
[tex]P(1 \le x \le 3)= 0.5313[/tex]
Solving (d): E(x) and Sigma(x)
[tex]E(x) = n * p[/tex]
[tex]E(x) = 15 * 0.05[/tex]
[tex]E(x) = 0.75[/tex]
[tex]\sigma(x)= \sqrt{E(x)* (1-p)}[/tex]
[tex]\sigma(x)= \sqrt{0.75* (1 - 0.05)}[/tex]
[tex]\sigma(x)= \sqrt{0.7125}[/tex]
[tex]\sigma(x)= 0.8441[/tex]
Solving (e):
Calculate P(0) when n= 50
Using: [tex]P(x) = ^nC_x\ p^x\ (1- p)^{n-x}[/tex]
[tex]P(0) = ^{50}C_0 * 0.05^0 * (1- 0.05)^{50-0}[/tex]
[tex]P(0) = 1 * 1 * (0.95)^{50}[/tex]
[tex]P(0) = 1 * 1 * 0.0769[/tex]
[tex]P(0) = 0.0769[/tex]
