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Exercise: Emergency hospital admissions in USA with herat disease are found to followPoisson distribution. An investigation showed that there are four admissions per day(mean). Find the probability that exactly three emergency admissions will occur in a givenday?

Question

Exercise: Emergency hospital admissions in USA with herat disease are found to followPoisson distribution. An investigation showed that there are four admissions per day(mean). Find the probability that exactly three emergency admissions will occur in a givenday?

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Solution

To solve this problem, we will use the formula for the Poisson distribution, which is:

P(X=k) = λ^k * e^-λ / k!

where:

  • P(X=k) is the probability of k events in an interval,
  • λ is the average rate of value (mean),
  • e is the base of the natural logarithm (approximately equal to 2.71828),
  • k! is the factorial of k.

Given in the problem, we know that λ = 4 (the average number of admissions per day) and we want to find the probability of k = 3 admissions in a day.

Substituting these values into the formula, we get:

P(X=3) = 4^3 * e^-4 / 3!

Calculating the values, we get:

P(X=3) = 64 * e^-4 / 6

Now, we just need to calculate this expression to get the probability. Using a calculator, we find that e^-4 is approximately 0.0183.

So, P(X=3) = 64 * 0.0183 / 6 = 0.195

Therefore, the probability that exactly three emergency admissions will occur in a given day is approximately 0.195 or 19.5%.

This problem has been solved

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