Answer:
[tex]P(X=8) = 0.0194[/tex]
[tex]P(X \ge 8)= 0.9820[/tex]
[tex]P(X \ge 1)= 1[/tex]
[tex]P(X\le 1) = 0.0000049[/tex]
Step-by-step explanation:
Given
[tex]\lambda = 15[/tex]
Poisson distribution is given by:
[tex]P(X=x) = \frac{\lambda^x e^{-\lambda}}{x!}[/tex]
Solving (a): 8 bugs
This implies that:
[tex]x = 8[/tex]
So, we have:
[tex]P(X=8) = \frac{15^8 * e^{-15}}{8!}[/tex]
[tex]P(X=8) = \frac{783.99418938}{40320}[/tex]
[tex]P(X=8) = 0.0194[/tex]
Solving (b): At least 8 bugs
This is represented as: [tex]P(X \ge 8)[/tex]
Using complement rule:
[tex]P(X \ge 8)= 1 - P(X<8)[/tex]
Where
[tex]P(X<8) = P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7)[/tex]
[tex]P(X<8) = \frac{15^1 * e^{-15}}{1!} + \frac{15^2 * e^{-15}}{2!} + \frac{15^3 * e^{-15}}{3!} + \frac{15^4 * e^{-15}}{4!} + \frac{15^5 * e^{-15}}{5!} + \frac{15^6 * e^{-15}}{6!} + \frac{15^7 * e^{-15}}{7!}[/tex]
[tex]P(X<8) = (\frac{15^1}{1!} + \frac{15^2}{2!} + \frac{15^3 }{3!} + \frac{15^4}{4!} + \frac{15^5}{5!} + \frac{15^6}{6!} + \frac{15^7}{7!}) e^{-15}[/tex]
[tex]P(X<8) = (15 + 112.5 + 562.5 + 2109.375 + 6328.125 + 15820.3125 + 33900.6696429) *e^{-15}[/tex]
[tex]P(X<8) = 58848.4821429 *e^{-15}[/tex]
[tex]P(X<8) = 0.0180[/tex]
So:
[tex]P(X \ge 8)= 1 - P(X<8)[/tex]
[tex]P(X \ge 8)= 1 - 0.0180[/tex]
[tex]P(X \ge 8)= 0.9820[/tex]
Solving (c): At least 1
This is represented as: [tex]P(X \ge 1)[/tex]
Using complement rule:
[tex]P(X \ge 1)= 1 - P(X<1)[/tex]
[tex]P(X<1) = P(X = 0)[/tex]
[tex]P(X<1) = \frac{15^0 e^{-15}}{0!}[/tex]
[tex]P(X<1) = \frac{e^{-15}}{1}[/tex]
[tex]P(X<1) = e^{-15}[/tex]
So:
[tex]P(X \ge 1)= 1 - P(X<1)[/tex]
[tex]P(X \ge 1)= 1 - e^{-15[/tex]
[tex]P(X \ge 1)= 0.99999969409[/tex]
[tex]P(X \ge 1)= 1[/tex]
Solving (d): Not more than 1
This implies at most 1.
It is represented as:
[tex]P(X\le 1)[/tex]
It is calculated using:
[tex]P(X\le 1) = P(X = 0) + P(X =1)[/tex]
[tex]P(X = 0) = e^{-15}[/tex]
[tex]P(X=1) = \frac{15^1 * e^{-15}}{1!}[/tex]
[tex]P(X=1) = 15 * e^{-15}[/tex]
So:
[tex]P(X\le 1) = e^{-15} + 15 * e^{-15}[/tex]
[tex]P(X\le 1) = 0.00000489443[/tex]
[tex]P(X\le 1) = 0.0000049[/tex]
