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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:_______.

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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

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