

Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
Material Type: Quiz; Professor: Lin; Class: Probability for Business; Subject: Mathematics; University: University of Southern California; Term: Fall 2008;
Typology: Quizzes
1 / 2
This page cannot be seen from the preview
Don't miss anything!
(a) Find the probability that there will be 2 or more customers arriving between 9:30am and 9:45am. Solution. Let X be the number of customers arriving between 9:30am and 9:45am. Then X is Poisson with μ = (2/10)15 = 3, and P (X ≥ 2) = 1 − P (X = 0) − P (X =
(b) A clerk’s working hours are from 9:00am to noon. Find the expected number and standard deviation of the number of customers arriving during her working hours on a particular day. Solution. Let X the number of customers arriving between 9:00am and noon. Then E X = (2/10)180 = 36 (customers) and σX =
36 = 6 (customers).
(c) Since there is no customer waiting at the window, the clerk decides to leave 20 minutes for a break. Find the probability that the next customer will arrive before she is back. Solution. We just need to find the probability that at least one customer will arrive in 20 minutes. Let X be the number of customers arriving in 20 minutes. Then X is Poisson with μ = (2/10)20 = 4, and P (X ≥ 1) = 1 − P (X = 0) = 1 − e−^4 ≈ .9817.
(a) Find the probability that the reservoir will be empty after 6 days. Solution. Since 120/20 = 6, the reservoir will be empty if and only if no rainfall occurs in 6 days. Let X be the number of rainfalls in 6 days. Then X is Poisson with μ = (1/5)6 = 1.2, and P (X = 0) = e−^1.^2 ≈ .3012.
(b) Find the expected amount of water (in cm) added by rainfalls in the next 10 days. Solution. Let X be the number of rainfalls in 10 days, an d the water added is 15X. Since E X = (1/5)10 = 2, E(15X) = 15 E X = 15 · 2 = 30 (cm).
(c) (Extra credit!) Find the probability that the reservoir will be empty sometime within one week. Solution. If no rainfall occurs in 6 days, the reservoir will be empty after 6 days. This probability has been obtained in part (a), which is e−^1.^2. If exactly one rainfall occurs in 6 days, the 15 cm water added by the rainfall can prevent the reservoir from being empty for another 15/20 = 3/4 day. If no rainfall occurs in this 3/4 day, the reservoir will be empty. This probability is P (1 rainfall in 6 days)P (no rainfall in 3/4 day) = e−^1.^2 (1.2) · e−.2(3/4). Since these are the only two possible cases for the reservoir to be empty, the probability is e−^1.^2 + e−^1.^2 (1.2) · e−.2(3/4)^ ≈ .6123.