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Unformatted text preview: Stat 340 Winter 2008 STAT340 — Winter 2008 — Test 1 Family (last) Name: Given (ﬁrst) Name: ID #: User Name: Marks Available: 25 Please circle the correct section number below: Name Section Lecture Time R. Metzger 1 TR 10:0011:20
R. Metzger 2 MWF 1:302:20 Part of your mark will be assigned for the clarity of your solution. An answer without justiﬁcation is worth 0
marks. The only permitted aid is a math faculty approved calculator. Page 1 of 5 Stat 340 Winter 2008 1. At a particular Tim Hortons there are 3 tills inside and 1 drive through outside. Questions 1 a to 1 e refer to the 3 tills inside. (a) [5 Marks] Suppose that we have random variables X1,X2, ._., X" which are mutually independent and
have distribution functions F1, F2, ..., Fn respectively. Show that the distribution of the minimum, X0) = T = min(X1, X2, ,Xn) is given by: F(t) : 1 — HI; (1 — Fi(t)) by starting with the deﬁnition of the CDF of T. F(t) : Pr (T < t) Pr (Xu) < t) 1— Pr (Xu) > t) 1 — Pr (the minimum is bigger than t) 1 i Pr (every number bigger than the minimum is bigger than t)
1— Pr (X1,X2, ...,X,, > t) 1—Pr(X1 > t,X2 >t,...,X,1 >15) 1 — Pr (X1 > t) Pr (X2 > t) Pr (Xn > t) Independence
1 i (1 A Pr (X1 < t)) (1 — Pr (X2 < t)) (1  Pr(Xn < t)) cdfs 1 — 11L] (1 — Fi(t)) (b) [4 Marks] It is known that the time between departures is independent and EX P(i2 + 1) where 2' is the
till/line—up number and i E {1Y 2= 3} and i2 + 1 is the rate with units people /minute. 0 Determine the distribution of the random variable representing the ﬁrst person to leave from the 3 tills. Show your work. The ﬁrst person to leave will do so as the minimum of the 3 times. The distribution of a till is given by: The distribution of the ﬁrst person is given by: PM F(t) = l—exp(—t(i2+1)) F05) = 1—H?:1(1*Fi(t))
: virgin—Fm» = 1— r1221(1—(1—exp(—t(i2+1>))) = 1* H?=1(1*(lieXP(t(i2+1)))) : l—exp(—t(1+1))exp(—t(4+1))exp(—t(9+1))
= 1—exp(—17t) Therefore the distribution is EX P(17) Page 2 of 5 Stat 340 Winter 2008 (c) [1 Marks] If Greg is at the front of till 1, his friend Clarisc is at the front of till 2 and I am at the front
of till 3, ﬁnd the probability that the ﬁrst of us to exit, leaves the line in the next 0.1 minutes. This is the minimum of the 3 till times. min(G, C, I) ~ EXP (2 +10 + 5) = EXP(17) Pr (X < 0.1) 2 1 —exp(—17(0.1))
= 81.7% (d) [2 Marks] If Greg is at the front of till 1, and I am right behind him when he leaves at 0.2 minutes, ﬁnd
the probability that I leave more than 0.3 minutes after him. The exponential distribution is memoryless thus Pr (X > 0.5)X : 0.2) = Pr (X > 0.3) Pr (X > 0.3) : 1 —— Pr (X > 0.3)
: exp (—0.3(2))
: 54.9% (e) [Bonus] If Greg is at the front of till 1, his friend Clarise is at the front of till 2 and I am at the front
of till 3, ﬁnd the probability that Greg leaves ﬁrst. You may assume that for 2 independent exponential Ax
random variables X ~ EXP (AZ) , and Y N EXP (Ag) the Pr (X < Y) = A + A .
I y 1 Bonus mark if you get the answer: /\G 2 m (f) At the drive through, it is thought that the arrivals arrive according to a geometric distribution. For
50 one hour (60 minute) intervals throughout the month, the number of cars going through the drive
through was recorded. The sorted numbers from R were as follows: x<c(17, 22, 22, 23, 23, 24, 25, 25, 25, 26, 27, 27, 27, 27, 27, 28, 28, 29, 29, 29, 30, 30,30, 30, 31, 31, 31,
31, 31, 31,31, 32, 32, 32, 32,32, 32, 32,33, 33, 33,34, 34, 35,35, 36, 36,38, 39, 44) mean(x) length(x)
[1] 30.02 [1] 50 Page 3 of 5 Stat 340 Winter 2008 o [7 Marks] Calculate the observed test statistic (OTS) for a. suitable test using the least number
of bins needed based on degrees of freedom. One of the bins has, as its interval [25, 33]. The We estimate p by: Bin 1: Bin 2: Bin 3: Expected Values: OTS 1
CDF ofa 0190(2)) is F(x) : 1 — (1 —p)“” where x 6 {1,2,3, ...}. E(X) = E.
E(X)=l :> 5:3: Bins=3
P P
Ail ﬂ 1
’“s ‘ 30.02
_ _ _. _ _ 1 25 N
[124] p1 — F(24) F(0) d N24) _ 1 (1 —30'02)
[25,33] p2 : F(33)—F(24) : F(33) — 57.1%e11.6%
[34,00] p3 : l—pl—pz % 32.6%
61 = pin = 50(55.6%) : 27.8 Observed: 01:6
62 = p271 : 50(11.6%) = 5.8 02:35
53 = P3” = 50(32.6%) = 16.3 03:9
$031100?
i=1 61:
2 2 2
(27.8 6) (5.8 35) +(16.3 9) 2167
27.8 5.8 16.3 55.6% o [4 Marks] There are 1500 interarrival times for the above arrivals, denoted by a1...a1500. The analyst A 1
assumes that they follow from an exponential distribution with rate A = Z and has calculated that the largest distance between the CDF and the empiracle CDF amongst the other 1499 observations
is 0.15. Calculate the K—S distance if the only unknown observation is (11000 = 8. 1 1
F(a1000) = 1— exp(—Za1000) = l — exp(—Z d1 = P011000)  F(a1ooo)‘ = 1 — 6XP(—2) —
d2 : lﬁﬁlmool — F (01000 — 1)‘ = 1 ‘ eXP(—2) —
d = max(d1, d2, 0.15) 0.199 Page 4 of 5 1000
1500 999
1500 1 — exp(—2) 22 22 0.198 0.199 ~
~ 63% Stat 340 Winter 2008 o [2 Marks / 1 Mark Each] Based on your answers above (to part f) please clearly answer the following
True or False questions: — With a pvalue of 0.001 we rejected the exponential distribution. This means that the data is not
exponential at all. TRUE ~ We rejected the hypothesis that the data could have arrived according to a the geometric distri—
bution. FALSE GRADE 2 Page 5 of 5 ...
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This note was uploaded on 06/10/2010 for the course STAT 340 taught by Professor Xu(sunny)wang during the Spring '09 term at Waterloo.
 Spring '09
 Xu(Sunny)Wang

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