Suppose a random variable, x, follows a Poisson distribution. Let μ = 2.5 every minute, find the P(X ≥ 125) over an hour. Round answer to 4 decimal places.

Respuesta :

Answer:

P(X ≥ 125) = 0.9812

Step-by-step explanation:

To solve this question, we need to understand the Poisson distribution and the normal distribution.

Poisson distribution:

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

[tex]P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\lambda[/tex] is the mean in the given interval, which is the same as the variance.

Normal distribution:

When the distribution is normal, we use the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

The Poisson can be approximated to the normal, with [tex]\mu = \lambda, \sigma = \sqrt{\lambda}[/tex]

Let μ = 2.5 every minute

This is the mean of the Poisson, so [tex]\lambda = 2.5n[/tex], in which n is the number of minutes.

P(X ≥ 125) over an hour

An hour has 60 minutes, so [tex]n = 60, \lambda = 2.5*60 = 150, \sigma = \sqrt{150} = 12.25[/tex]

Using continuity correction, this is [tex]P(X \geq 125 - 0.5) = P(X \geq 124.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 124.5. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{124.5 - 150}{12.25}[/tex]

[tex]Z = -2.08[/tex]

[tex]Z = -2.08[/tex] has a pvalue of 0.0188

1 - 0.0188 = 0.9812

So

P(X ≥ 125) = 0.9812

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