Question

The Securities and Exchange Commission has determined that the number of companies listed on the NYSE declaring bankruptcy is approximately a Poisson distribution with a mean of 2.6 per month. Find the probability that exactly 4 bankruptcies occur next month.

Answer 1

Answer:  0.1414

Step-by-step explanation:

Poisson distribution formula : $P(X=x)=\dfrac{e^{-\lambda}\lambda^{x}}{x!}$ , where $\lambda$ is the mean for distribution.

It is given that : The Securities and Exchange Commission has determined that the number of companies listed on the NYSE declaring bankruptcy is approximately a Poisson distribution with a mean of 2.6 per month.

i.e . $\lambda$ =  2.6 per month.

Let x be the number of bankruptcies occurs.

Then, the probability that exactly 4 bankruptcies occur next month will be :

$P(X=4)=\dfrac{e^{-2.6}2.6^{4}}{4!}\\\\=\dfrac{0.0742735782143\times45.6976}{24}\approx0.1414$

Hence, the probability that exactly 4 bankruptcies occur next month. is 0.1414.

Answer 2

Using the Poisson distribution, it is found that there is a 0.1414 = 14.14% probability that exactly 4 bankruptcies occur next month.

What is the Poisson distribution?

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.

In this problem, the mean is of $\mu = 2.6$, and the probability that exactly 4 bankruptcies occur next month is P(X = 4), hence:

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

$P(X = 4) = \frac{e^{-2.6}2.6^{4}}{(4)!} = 0.1414$

0.1414 = 14.14% probability that exactly 4 bankruptcies occur next month.

More can be learned about the Poisson distribution at https://brainly.com/question/13971530