Question

According to the Ministry of Public Education, schools in Samarkand have closed an average of 24 days each year due to some type of emergencies. Assume that the number of school closure days can be approximated by a Poisson distribution. What is the probability that a randomly chosen school in Samarkand will close for 1 or 2 days next month?

Answer 1

Using the Poisson distribution, it is found that there is a 0.5414 = 54.14% probability that a randomly chosen school in Samarkand will close for 1 or 2 days next month.

We have the mean in an interval, hence, the Poisson distribution is used.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

The parameters are:

• x is the number of successes

• e = 2.71828 is the Euler number

• $\mu$ is the mean in the given interval.

Average of 24 days each year, considering that a year has 12 months, the monthly mean is of $\mu = \frac{24}{12} = 2$

Then:

$P(1 \leq X \leq 2) = P(X = 1) + P(X = 2)$

In which

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

$P(X = 1) = \frac{e^{-2}2^{1}}{(1)!} = 0.2707$

$P(X = 2) = \frac{e^{-2}2^{2}}{(2)!} = 0.2707$

$P(1 \leq X \leq 2) = P(X = 1) + P(X = 2) = 0.2707 + 0.2707 = 0.5414$

0.5414 = 54.14% probability that a randomly chosen school in Samarkand will close for 1 or 2 days next month.

To learn more about the Poisson distribution, you can take a look at https://brainly.com/question/13971530