Probability of Poisson Distribution: Nr of Customers in Shop

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SUMMARY

The discussion focuses on the application of the Poisson distribution to model customer arrivals at a shop, specifically with an average of 4 customers arriving every 15 minutes. The probability of waiting at least 5 more minutes for another customer after one has just arrived is calculated to be 0.2636. Additionally, the probability of having at least 7 and at most 15 out of 40 non-overlapping 15-minute intervals with at most 2 customers arriving is determined using the binomial distribution, yielding a result of 0.855.

PREREQUISITES
  • Understanding of Poisson distribution and its properties
  • Familiarity with binomial distribution concepts
  • Knowledge of probability theory, specifically memorylessness
  • Ability to perform statistical calculations using cumulative distribution functions
NEXT STEPS
  • Study the properties of the Poisson distribution in detail
  • Learn how to apply the binomial distribution to real-world scenarios
  • Explore the concept of memorylessness in probability distributions
  • Practice calculating cumulative distribution functions for various distributions
USEFUL FOR

Statisticians, data analysts, and anyone involved in probability modeling or customer behavior analysis will benefit from this discussion.

pinto89a
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Nr of customers arriving at a shop follow Poisson.
In 15, an average of 4 customers arrive.

a)
A customer has just arrived. Then a minute passed and no one arrived. What is the probability of it takoing at least 5 more min. until another customer arrives?

b)
Consider 40 non-overlapping periods of 15 min.

What is the probability that
at least 7 and at most 15 of those intervals have at most 2 customers arriving?

In book, answer to
a) is 0.2636
b) is F(2.22) - F(-1.12) = 0.855

In a) although I don't see why, I understande that it's something about memorylessness or something. But how do you get to the answer in question b)?
 
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pinto89a said:
how do you get to the answer in question b)?

via binomial distribution
 
Ok, I see now.Thank you.
 

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