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QMS 102 (186)
Lecture 7

# Lecture 7 Notes.docx

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School
Department
Quantitative Methods
Course
QMS 102
Professor
Sheila Rosenberg
Semester
Fall

Description
Lecture9Notes Poisson Distribution Characteristics 1. Observing a situation over a period of time. Amount of space is analyzed (length, area, volume, or weight) 2. There will be one success or no successes (must occur randomly) 3. The average rate of success for the time or space is given in symbol lambda; λ Poisson Distribution Examples To get around a problem that states P (X < 3) you would use P (X ≤ 2) A= an event 1. Find the probability that there will be fewer than 3 customers in the next 10 minutes 2. We are given λ= 15/hr i.e. number of customers per hour we need to express λ in terms of 10 minutes 3. An hour has 60 minutes which is 6 * 10 min 4. Λ = 15/HR = 6 * 10 MIN = 15/6 / 10 min = 2.5 / 10 min 5. So λ = 15/ hr. = 2.5 / 10 min 6. P (X < 3) i.e. the probability that there will be fewer than 3 customers in the next 10 minutes 7. Now P (X ≤ 2) = Pcd (2, 2.5) = 0.5438) 8. There is a 54% chance that there will be less than 3 customers in the next 10 minutes Find the probability that there was less than 3 customers in the next 7 minutes Λ = 15/hr. = y/7 Using λ = 15/hr. = 15 / 60 = 15/60 = y / 7 Implies: 15 * 7 = y * 60 Solving for y we get 15 * 7/ 60 = 1.75 Find the probability there will be more than 40 customers in the next 2 hours 15 / hr. = 15 * 2 / 2 hours = 30 / 2 hours So λ = 15 / hr. = 30 / 2 hours P (X > 40) i.e. there will be more than 40 customers in the next 2 hours) P (X >40) = 1 – P (X ≤ 40) = 1 – Pcd (40, 30) = 1 – 0.9677 = 0.0323 Probability involving potholes 450 potholes per 200 km 450 / 200 = y / 3 1350 / 200 = 200y / 200 Y = 6.75 (λ) P (x ≥ 6) = 1 – P (X ≤ 5) = 1 – Pcd (5, 6.75) = 1- 0.3338 = 0.6662 Using CASIO Calculator to computer P (X = x) Using Example 2: X = 7, 8, 9 in LIST 1 Select STAT F5 (DI
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