Answer:
a
[tex] P(X = 0) = 0.6065 [/tex]
b
[tex]P(x < 25 ) = 1.18 *10^{-33} [/tex]
c
[tex] P(x \le 5 ) = 0.9994 [/tex]
Step-by-step explanation:
From the question we are told that
The rate is [tex]\lambda = \frac{1}{2}\ hr^{-1}[/tex] = 0.5 / hr
Generally Poisson distribution formula is mathematically represented as
[tex]P(X = x) = \frac{(\lambda t) ^x e^{-\lambda t }}{x!}[/tex]
Generally the probability that no error occurred during a day is mathematically represented as
Here t = 1 hour according to question a
So
[tex]P(X = x) = \frac{\lambda^x e^{-\lambda}}{x!}[/tex]
Hence
[tex][tex]P(X = 0) = \frac{\frac{1}{2} ^0 e^{-\frac{1}{2}}}{0!}[/tex]
=> [tex] P(X = 0) = 0.6065 [/tex]
Generally the probability that a critical error occurs since the start of a day is mathematically represented as
Here t = 1 hour according to question a
So
[tex]P(X = x) = \frac{\lambda^x e^{-\lambda}}{x!}[/tex]
Hence
[tex]P(x \ge 25 ) = 1 - P(x < 25 )[/tex]
Here
[tex]P(x < 25 ) = \sum_{x=0}^{24} \frac{e^{-\lambda} * \lambda^{x}}{x!}[/tex]
=> [tex]P(x < 25 ) = \frac{e^{-0.5} *0.5^{0}}{0!} + \cdots + \frac{e^{-0.5} *0.5^{24}}{24!}[/tex]
[tex]P(x < 25 ) = 0.6065 + \cdots + \frac{e^{-0.5} *0.5^{24}}{6.204484 * 10^{23}}[/tex]
[tex]P(x < 25 ) = 0.6065 + \cdots + 6.0*10^{-32}[/tex]
[tex]P(x < 25 ) = 1.18 *10^{-33} [/tex]
Considering question c
Here t = 2
Gnerally given that the system just started up and an error occurred the probability the next reset will occur within 2 hours
[tex]P(x \le 5 ) = \sum_{n=0}^{5} \frac{(\lambda t) ^x e^{-\lambda t }}{x!}[/tex]
=> [tex]P(x \le 5 ) = \frac{(0.5 * 2) ^ 0 e^{- 0.5 * 2 }}{0!} + \cdots + \frac{(0.5 * 2) ^ 5 e^{- 0.5 * 2 }}{5!}[/tex]
=> [tex]P(x \le 5 ) = \frac{1* 2.7183 }{1 } + \cdots + \frac{1 *2.7183 }{120}[/tex]
=> [tex]P(x \le 5 ) = 2.7183 + \cdots + 0.0226525[/tex]
[tex] P(x \le 5 ) = 0.9994 [/tex]
