Question

At a certain fire station, false alarms are received at a mean rate of 0.2 per day. In a year, what is the probability that fewer than 60 false alarms are received? (Consider 365 days a year. Round your standard deviation value to 3 decimal places and z-value to 2 decimal places. Round your final answer to 4 decimal places.)

Answer 1

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: $365\times 0.20=73$.

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:

$\mu=\lambda=73\\\\\sigma=\sqrt{\lambda}=\sqrt{73}=8.544$

Compute the probability that fewer than 60 false alarms are received as follows:

$P(X<60)=P(\frac{X-\mu}{\sigma}<\frac{60-73}{8.544})$

                  $=P(Z<-1.52)\\\\=0.06426\\\\\approx 0.0643$

Thus, the probability that fewer than 60 false alarms are received is 0.0643.