Respuesta :
Answer:
a) 85.90% probability that the instrument does not fail in an 8-hour shift.
b) 36.62% probability of at least one failure in a 24-hour day
Step-by-step explanation:
The only information that we have is a mean during an interval. So we use the Poisson distribution solve this question.
We have that the probability of exactly x events is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*(\mu)^{x}}{x!}[/tex]
In which [tex]\mu[/tex] is the mean.
(a) What is the probability that the instrument does not fail in an 8-hour shift
Mean of 0.019 failures per hour. For 8 hours
[tex]\mu = 8*0.019 = 0.152[/tex].
This probability is P(X = 0). So
[tex]P(X = x) = \frac{e^{-\mu}*(\mu)^{x}}{x!}[/tex]
[tex]P(X = 0) = \frac{e^{-0.152}*(0.152)^{0}}{0!} = 0.8590[/tex]
85.90% probability that the instrument does not fail in an 8-hour shift.
(b) Probability of at least one failure in a 24-hour day?
Mean of 0.019 failures per hour. For 24 hours
[tex]\mu = 24*0.019 = 0.456[/tex].
Either there are no failures, or there is at least one failure. The sum of the probabilities of these events is decimal 1. So
[tex]P(X = 0) + P(X \geq 1) = 1[/tex]
We want to find [tex]P(X \geq 1)[/tex]. So
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
In which
[tex]P(X = x) = \frac{e^{-\mu}*(\mu)^{x}}{x!}[/tex]
[tex]P(X = 0) = \frac{e^{-0.456}*(0.456)^{0}}{0!} = 0.6338[/tex]
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.6338 = 0.3662[/tex]
36.62% probability of at least one failure in a 24-hour day
Answer:
(a) 0.8590
(b) 0.3662
Step-by-step explanation:
We are given that the number of failures of a testing instrument from contamination particles on the product is a Poisson random variable with a mean of 0.019 failures per hour.
Let X = number of failures of a testing instrument
So, X ~ Poisson([tex]\lambda[/tex])
We know that mean of Poisson distribution = [tex]\lambda[/tex] = 0.019 per hour
Hence, X ~ Poisson(0.019)
The probability distribution for Poisson random variable is;
[tex]P(X=x) = \frac{e^{-\lambda} * \lambda^{x} }{x!}; x = 0,1,2,3,....[/tex]
(a) Probability that the instrument does not fail in an 8-hour shift = P(X=0)
In the question, [tex]\lambda[/tex] is given for per hour failures. So, [tex]\lambda[/tex] for 8-hour shift is given by = 0.019 * 8 = 0.152
So, P(X = 0) = [tex]\frac{e^{-0.152} * 0.152^{0} }{0!}[/tex] = [tex]e^{-0.152}[/tex] = 0.8590
Therefore, probability that the instrument does not fail in an 8-hour shift is 0.8590 .
(b) Probability of at least one failure in a 24-hour day = P(X >= 1)
In the question, [tex]\lambda[/tex] is given for per hour failures. So, [tex]\lambda[/tex] for 24-hour shift is given by = 0.019 * 24 = 0.456
So, P(X >= 1) = 1 - P(X = 0)
= 1 - [tex]\frac{e^{-0.456} * 0.456^{0} }{0!}[/tex] = 1 - [tex]e^{-0.456}[/tex] = 0.3662
Therefore, probability of at least one failure in a 24-hour day is 0.3662 .
