# STAT 225 Spring 2011 Final Exam Your Name: Mike (11:30 ...

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﻿STAT 225 ? Spring 2011 Final Exam

Zhaonan (7:30) Mike (11:30) Lin-Yang(2:30)

Zhaonan(8:30) Chris (12:30) Yong (3:30)

Jeff (9:30) Yen-Ning (1:30)

Jeremy (3:30)

Jeff (10:30) Yen-Ning (2:30)

Jeremy (4:30)

Question 1 2 3 4 5 6 7 8

Cheat Sheet Total

Points Possible

15 9 16 13 16 9 10 10 2 100

1. The Borders Bookstore at West Lafayette used to be open 10 hours per day and 7 days a week. In each of the following situations, there is a random variable involved. Find its name and parameter(s). If an approximation can be made, you must name and label both the exact and approximate distribution with corresponding parameter(s) to receive full credit. The customers are assumed to behave independently of each other and from time to time. (3 points each) a). It takes, on average, 20 minutes for the store to get its first customer. Let A be the amount of time until the first customer visits the store. A ~ Exponential( 20 minutes or =1/20)

b). During a visit, a customer is equally likely to make a purchase between \$15 and \$150. Let B represent the amount a customer spends during a visit. B~U(15,150)

c). On average, 6 Borders customers also visit Panera Bread which operates on the same schedule as Borders during each hour. Let C represent the total number of Borders customers who visit Panera Bread during a week.

C ~ Poisson( 6*10*7 420)

d). A customer used to spend part of his Sunday morning at Borders every week. The amount of time of his visit ranged evenly between 1 and 3 hours. Let D be the average amount of time this customer spent at Borders on each Sunday morning in the previous year (52 weeks). Let S be the amount of time spent on 1 Sunday. S~U(1,3). ES=2 and VarS = 1/3. D is x, so D~N(mean=2, and Var= 1 .00641 or SD=.080064)

3*52

e). A study from Borders headquarters shows that 12% of all customers visiting Borders stores purchase Kobo, Borders' own brand of eReader. On the Black Friday of 2010, a record 1500 customers visited the West Lafayette Borders. Let P be the proportion of those customers who purchased Kobo.

exact: P~ Binomial(n=1500, p=.12) approximation: P*~N(mean= 180, var = 158.4, sd=12.58571).

1

2. The lifetime of a particular brand of light bulb follows an exponential distribution with a mean life of 180 days. a). Suppose you bought one light bulb of this brand. What's the probability it will burn out within 160 days? (3 points)

P( X 160) 1 e160*(1/180) .58.

b). Given that the light bulb you bought is still working after 90, what's the probability that it will keep working after another 50 days? (3 points)

P(X 140 | X 90) P(X 50) e50/180 .7575. This problem takes advantage of the memoryless property of the exponential distribution.

c). If a bulb's life span is shorter than the life of 97.5% of all the bulbs under this brand, it will be considered defective. A store selling this bulb has a policy that they will replace a defective bulb for free. Their ad says: "If your bulb lasts less than ______ days, we will give you a new one for free!" Use your STAT225 knowledge to fill in the blank above. (3 points)

This is looking for the 2.5% percentile for this exponential distribution. So we set F(x) = .025 and solve for x. The answer is 4.5572 days.

2

3. Let X be a continuous random variable with Probability Density Function (PDF):

C(x2 1) :1 x 3

f (x)

, where C is a constant

0 : Else

a). Find the value of C such that f(x) is a valid PDF. (4 points)

It is a valid PDF if 2 things happen. 1) f(x) >=0 for all x. For this we need C to be positive. 2) The integral of f(x) over the range must equal 1. C = 3/20.

b). What is the Cumulative Distribution Function (CDF) of X? (4 points)

F[x] where it is defined as a function is the integral of f[x] from the lower bound of the interval to x. Remember to put in your C from part a to make sure that F[x] is proper.

0

F

[

x]

x3 20

3x 20

1 10

1

x3

The other parts of problem 3 will be continued on the next page 3

Problem 3. Continued c). Find P(X < 2.5 | X > 2) (4 points)

P(X2) = P(2 < X < 2.5) / P(X>2) = (F[2.5]-F)/(1-F) = (.50626-.2)/(1-.2)=.3828.

d). Find E(20X) (4 points)

E[X ] x * f [x]dx

so, E[20X ] 20x * f [x]dx

E[20X ] here is 48.

4

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