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 10150 particles? Use normal approximation without continuity correction. Round your answer to 3 decimal places. Enter your answer in accordance to the question statement

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Answer:

The probability that 10 squared centimetres of dust contains more than 10150 particles is 0.067.

Step-by-step explanation:

The Poisson distribution with parameter λ, can be approximated by the Normal distribution, when λ is large say λ > 1,000.

If X follows Poisson (λ) and λ > 1,000 then the distribution of X can be approximated but he Normal distribution.

The mean of the approximated distribution of X is:

μ = λ

The standard deviation of the approximated distribution of X is:

σ = √λ

Thus, if λ > 1,000, then [tex]X\sim N(\mu=\lambda,\ \sigma^{2}=\lambda)[/tex].

Let X = number of asbestos particles in a sample of 1 squared centimetre of dust.

The random variable X follows a Poisson distribution with mean, μ = 1000.

Then the average number of asbestos particles in a sample of 10 squared centimetre of dust will be, [tex]\lambda = 10\times \mu=10\times 1000=10,000[/tex].

Compute the probability that 10 squared centimetres of dust contains more than 10150 particles as follows:

[tex]P(X>10150)=P(\frac{X-\mu}{\sigma}>\frac{10150-10000}{\sqrt{10000}})[/tex]

                       [tex]=P(Z>1.50)\\=1-P(Z<1.50)\\=1-0.93319\\=0.06681\\\approx0.067[/tex]

*Use a z-table for the probability.

Thus, the probability that 10 squared centimetres of dust contains more than 10150 particles is 0.067.