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Quiz 6 Solutions - Probability for Business | MATH 218, Quizzes of Mathematics

Material Type: Quiz; Professor: Lin; Class: Probability for Business; Subject: Mathematics; University: University of Southern California; Term: Fall 2008;

Typology: Quizzes

Pre 2010

Uploaded on 11/08/2009

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Math 218 (Fall 2008) Quiz 6 Solutions TA: Wei Lin
Name: Section (circle one): 2pm 3pm 4pm
1. (6 pts) Customer arrivals at a bank teller’s window follow a Poisson distribution with a
mean rate of 2 customers per 10 minutes.
(a) Find the probability that there will be 2 or more customers arriving between 9:30am
and 9:45am.
Solution. Let Xbe the number of customers arriving between 9:30am and 9:45am.
Then Xis Poisson with µ= (2/10)15 = 3, and P(X2) = 1 P(X= 0) P(X=
1) = 1 e3e33.8009.
(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 Xthe number of customers arriving between 9:00am and noon. Then
EX= (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 Xbe the number of customers arriving in 20 minutes. Then Xis
Poisson with µ= (2/10)20 = 4, and P(X1) = 1 P(X= 0) = 1 e4.9817.
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Math 218 (Fall 2008) Quiz 6 Solutions TA: Wei Lin

Name: Section (circle one): 2pm 3pm 4pm

  1. (6 pts) Customer arrivals at a bank teller’s window follow a Poisson distribution with a mean rate of 2 customers per 10 minutes.

(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 =

  1. = 1 − e−^3 − e−^33 ≈ .8009.

(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.

  1. (4 pts) The water level of a certain reservoir decreases at a constant rate of 20 cm per day. The reservoir is refilled by rainfalls which occur according to a Poisson distribution with a mean rate of one rainfall per 5 days. Each rainfall adds 15 cm to the water level. The present water level is 1.2 m.

(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.