Respuesta :
Answer:
The probability that fewer than 60 false alarms are received is 0.0643.
Step-by-step explanation:
Let X denote the number of false alarms received at a certain fire station in a year.
The mean rate of false alarms in a day is 0.20.
Then the mean rate of false alarms in a year will be: [tex]365\times 0.20=73[/tex].
Then the random variable X follows a Poisson distribution with parameter λ = 73.
The Poisson distribution with parameter λ, can be approximated by the Normal distribution, when λ is large.
If X follows Poisson (λ) and λ is large then the distribution of X can be approximated but he Normal distribution.
The mean of the approximated distribution of X is:
μ = λ
The standard deviation of the approximated distribution of X is:
σ = √λ
Thus, if λ is large, then X follows N (μ = λ, σ² = λ).
Compute the mean and standard deviation of X as follows:
[tex]\mu=\lambda=73\\\\\sigma=\sqrt{\lambda}=\sqrt{73}=8.544[/tex]
Compute the probability that fewer than 60 false alarms are received as follows:
[tex]P(X<60)=P(\frac{X-\mu}{\sigma}<\frac{60-73}{8.544})[/tex]
[tex]=P(Z<-1.52)\\\\=0.06426\\\\\approx 0.0643[/tex]
Thus, the probability that fewer than 60 false alarms are received is 0.0643.
