Question

The number of accidents in a certain city is modeled by a Poisson random variable with an average rate of 10 accidents per day. Suppose that the number of accidents on different days are independent. Use the central limit theorem to find the probability that there will be more than 3800 accidents in a certain year. Assume that there are 365 days in a year.

Answer 1

Answer:

The probability that there will be more than 3800 accidents in a year is 0,00657

Step-by-step explanation:

1. Consider X as the number of accidents per day. X follow a Poisson distribution with λ=10

2. Consider Y as the humer off accidents per year. Y follow a Poisson distribution, but in this case, λ=(10)(365)=3650

3. The convergency Poisson to Normal distribution equation is:

$P(\lambda )=\frac{Y-\lambda }{\sqrt{\lambda}}$

4. Central Limit Theoreme could be expressed as the equation of the standard normal Z:

$Z=\frac{X- \mu }{\sigma }$

5. In this case we have:

$\mu =\lambda =3650$

$\sigma =\sqrt{\lambda }=\sqrt{3650}$

6. Calculate the probability that Y>3800, transforming Y into Z:

$P(Y>3800)=P(\frac{Y-3650}{\sqrt{3650} } >\frac{3800-3650}{\sqrt{3650}} )$

$P(Z>2,48)=1-\phi(2,48)$

$P=1-0,99343$$P=0,00657$