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A limousine service receives 236 calls over 40 days from customers who need a ride to the airport.Use the Poisson distribution to find the probability that fewer than 2 customers need a ride on a given day.Do not round intermediate computations, and round your answer to three decimal places.(If necessary, consult a list of formulas.)

Question

A limousine service receives 236 calls over 40 days from customers who need a ride to the airport.Use the Poisson distribution to find the probability that fewer than 2 customers need a ride on a given day.Do not round intermediate computations, and round your answer to three decimal places.(If necessary, consult a list of formulas.)

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Solution

To solve this problem, we first need to find the average number of calls per day.

Step 1: Find the average number of calls per day The limousine service receives 236 calls over 40 days. So, the average number of calls per day (λ) is 236/40 = 5.9 calls per day.

Step 2: Use the Poisson distribution formula The Poisson distribution formula is P(x; λ) = (e^-λ * λ^x) / x!, where:

  • P(x; λ) is the probability of x events in an interval,
  • e is the base of the natural logarithm (approximately equal to 2.71828),
  • λ is the average rate of value,
  • x is the actual number of successes that result from the experiment, and
  • x! is the factorial of x.

We want to find the probability that fewer than 2 customers need a ride on a given day. This means we need to find P(0; 5.9) + P(1; 5.9).

Step 3: Calculate P(0; 5.9) P(0; 5.9) = (e^-5.9 * 5.9^0) / 0! = 0.00277

Step 4: Calculate P(1; 5.9) P(1; 5.9) = (e^-5.9 * 5.9^1) / 1! = 0.01635

Step 5: Add the probabilities P(x < 2) = P(0; 5.9) + P(1; 5.9) = 0.00277 + 0.01635 = 0.01912

So, the probability that fewer than 2 customers need a ride on a given day is 0.019, or 1.9% when rounded to three decimal places.

This problem has been solved

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