Question

Suppose that the number of asbestos particles in a sample of 1 squared centimeter of dust is a Poisson random variable with a mean of 1000. What is the probability that 10 squared centimeters of dust contains more than 10110 particles?

Answer 1

Answer:

The approximate probability that 10 squared centimeters of dust contains more than 10110 particles is 0.1357

Step-by-step explanation:

For a sample of 10 sqaured centimeters of dust, the total amount of asbestos particles has a Poisson distribution with a mean of 1000/1 * 10 = 10000.

We will approximate this probability to a normal distribution. The variance is also 10000 (because it is poisson), therefore the standard deviation is √10000 = 100. Lets call X the distribution, and W its standarization, given by

$W = \frac{X-\mu}{\sigma} = \frac{X-10000}{100}$

We have

$P(X>10110) = P(\frac{X-10000}{100} > \frac{10110-10000}{100}) = P(W > 1,1) = 1-\phi(1,1) = 1-0.8643 = 0.1357$

Where $\phi$ is the cummulative distribution function of a standard normal distribution. The values of $\phi$ are well known and they can be found in the attached file.

We conclude that the approximate probability that 10 squared centimeters of dust contains more than 10110 particles is 0.1357.