Question

In a 24 hours period, the average number of boats arriving at a port is 10. Assuming that boats arrive at a random rate that is the same for all subintervals of equal length (i.e. the probability of a boat arriving during a 1 hour period the same for every 1 hour period no matter what). Calculate the probability that more than 1 boat will arrive during a 1 hour period. (P(X>1) ) Give your answers to 4 decimal places and in a range between 0 and 1

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Fred

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To solve this problem, we will use the Poisson distribution since we are dealing with the arrival of boats over a given time period.

The average number of boats arriving at the port in a 1-hour period is given as 10. Therefore, the average rate parameter, λ, is also 10.

The probability of more than 1 boat arriving during a 1-hour period can be calculated as:

P(X > 1) = 1 - P(X = 0) - P(X = 1)

where P(X = k) represents the probability of k boats arriving during a 1-hour period.

Using the formula for the Poisson distribution:

P(X = k) = (e^(-λ) * λ^k) / k!

we can calculate each term.

P(X = 0) = (e^(-10) * 10^0) / 0! = e^(-10)

P(X = 1) = (e^(-10) * 10^1) / 1! = 10 * e^(-10)

Now we can substitute these values into the formula for P(X > 1):

P(X > 1) = 1 - e^(-10) - 10 * e^(-10)

Calculating this expression, we find:

P(X > 1) ≈ 1 - e^(-10) - 10 * e^(-10) ≈ 1 - 0.00004540 - 0.00045399 ≈ 0.9995

Therefore, the probability that more than 1 boat will arrive during a 1-hour period is approximately 0.9995.

\textbf{Answer: } P(X > 1) \approx 0.9995

The average number of boats arriving at the port in a 1-hour period is given as 10. Therefore, the average rate parameter, λ, is also 10.

The probability of more than 1 boat arriving during a 1-hour period can be calculated as:

P(X > 1) = 1 - P(X = 0) - P(X = 1)

where P(X = k) represents the probability of k boats arriving during a 1-hour period.

Using the formula for the Poisson distribution:

P(X = k) = (e^(-λ) * λ^k) / k!

we can calculate each term.

P(X = 0) = (e^(-10) * 10^0) / 0! = e^(-10)

P(X = 1) = (e^(-10) * 10^1) / 1! = 10 * e^(-10)

Now we can substitute these values into the formula for P(X > 1):

P(X > 1) = 1 - e^(-10) - 10 * e^(-10)

Calculating this expression, we find:

P(X > 1) ≈ 1 - e^(-10) - 10 * e^(-10) ≈ 1 - 0.00004540 - 0.00045399 ≈ 0.9995

Therefore, the probability that more than 1 boat will arrive during a 1-hour period is approximately 0.9995.

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