The rate at which customers are served at an airport check-in counter is a Poisson process with a rate of 11.9 per hour. The probability that more than 50 customers are served at the counter in the next 5 hours is P(Xp>50). If this is solved as a Poisson variable, the calculations will be tedious. So we use the normal approximation. Now, P(Xp > 50)=P(Z > a), where Z is the standard normal variable. What is the value of a? Please report your answer in 3 decimal places.

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Asked Jul 11, 2019
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The rate at which customers are served at an airport check-in counter is a Poisson process with a rate of 11.9 per hour. The probability that more than 50 customers are served at the counter in the next 5 hours is P(Xp>50). If this is solved as a Poisson variable, the calculations will be tedious. So we use the normal approximation. Now, P(Xp > 50)=P(Z > a), where Z is the standard normal variable. What is the value of a? Please report your answer in 3 decimal places.

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Expert Answer

Step 1

The average number of customers served per hour is 11.9.

So, for 5 hours, the average is (11.9 x 5) = 59.5 customers.

Let Xp be the number of customers served in next 5 hours, which fo...

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a =59.5 =7.714

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