Use the fact that the variance of a Poisson distribution is σ2=μ. The mean number of bankruptcies filed per hour by businesses in a country was about five. (a) Find the variance and the standard deviation. Interpret the results. (b) Find the probability that at most three businesses will file bankruptcy in any given hour.

Using the Poisson distribution, it is found that:

a) The variance is of 5 while the standard deviation is of 2.24. It means that over many hours, the number of bankruptcies should be within 2.24 of 5.

b) 0.265 = 26.5% probability that at most three businesses will file bankruptcy in any given hour.

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

The parameters are:

In this problem, the mean is of .

Item a:

The variance is of 5 while the standard deviation is of 2.24. It means that over many hours, the number of bankruptcies should be within 2.24 of 5.

Item b:

This probability is:

In which

Then:

0.265 = 26.5% probability that at most three businesses will file bankruptcy in any given hour.

A similar problem is given at link

Variance = 5

Standard deviation = 2.236

0.2650257

Step-by-step explanation:

For a Poisson distribution is σ² = μ.

Given that:

Mean , μ of bankruptcies files per hour = 5

μ = 5

For a Poisson distribution :

P(x = x) = (μ^x * e^-μ) / x!

The Variance and standard deviation :

Variance : σ² = μ = 5

Standard deviation = sqrt(variance) = sqrt(5) = 2.236

B.) Find the probability that at most three businesses will file bankruptcy in any given hour.

P( x ≤ 3) = P(0) + P(1) + P(2) + P(3)

P(x = 0) = (5^0 * e^-5) / 0! = 0.0067379

P(x = 1) = (5^1 * e^-5) / 1! = 0.0336897

P(x = 2) = (5^2 * e^-5) / 2! = 0.0842243

P(x = 3) = (5^3 * e^-5) / 3! = 0.1403738

0.0067379 + 0.0336897 + 0.0842243 + 0.1403738

= 0.2650257