Question

In a recent​ year, an author wrote 171 checks. Use the Poisson distribution to find the probability​ that, on a randomly selected​ day, he wrote at least one check. The probability is:_______.

Answer 1

To find the probability of the author writing at least one check on a randomly selected day, we can use the Poisson distribution. By calculating the probability of writing zero checks and subtracting it from 1, we can find the probability of writing at least one check. In this case, the probability is approximately 1, meaning it is highly likely that the author wrote at least one check on a randomly selected day.

Step-by-step explanation:

To find the probability that the author wrote at least one check on a randomly selected day, we can use the Poisson distribution. The Poisson distribution is used to calculate the probability of a certain number of events occurring in a fixed interval of time or space, given a known average rate. In this case, the average rate is the number of checks the author wrote in a year, which is 171.

The formula for the Poisson distribution is:

P(X = k) = (e^-λ * λ^k) / k!

where λ is the average rate and k is the number of events.

To calculate the probability of at least one check being written on a randomly selected day, we can sum up the probabilities of writing 1, 2, 3, and so on checks per day.

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

We can plug in the values into the Poisson distribution formula to calculate the probability:

P(X >= 1) = 1 - (e^-171 * 171^0) / 0!

Using a calculator or statistical software, this evaluates to approximately 1.

Answer 2

Answer: 0.3741

Step-by-step explanation:

Poison probability ;

P(x) = [U(^x) e(^-U)] ÷ x!

Where U = mean

Note: e = exponential symbol

Number of checks that year = 171

Number of days in a year = 365

U = 171/365 = 0.468

Average checks per day = 0.4685

Probability that at least one check was written per day is can be calculated by;

P(not 0) = 1 - P(0)

Therefore,

P(x) = [U(^x) e(^-U)] ÷ x!

P(0) = [ 0.4685^0 * e^-0.4685] ÷ 0!

P(0) = [ 1 * 0.6259] ÷ 1

P(0) = 0.6259

Therefore,

P(not 0) = 1 - 0.6259 = 0.3741