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(MATH244)midterm01F_L1sol.pdf
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Mid-term Examination Solution
1.
(a)
Positively skewed.

(b)

(c)
Min = 0.2 , Max = 8.0 For QL , p = 0.25 , np = 10000 0.25 = 2500 ,


CPU time (s) Frequency Cumulative Frequency
0.2 C 1.0 960 960
1.0 C 1.8 2880 3840
1.8 C 2.8 3000 6840
2.8 C 3.8 1600 8440
3.8 C 5.6 1080 9520
5.6 C 8.0 480 10000
Total 10000

(1.8 .1.0)( 2500 . 960)
QL = 1.0 += 1.43
2880
For median, p = 0.5 , np = 10000 0.5 = 5000 ,

(2.8 .1.8)( 5000 . 3840)
meidan =1.8 += 2.19
3000
For QU, p = 0.75 , np = 10000 0.75 = 7500 ,

(3.8 . 2.8)( 7500 . 6840)
Q = 2.8 += 3.21
U1600 Hence the five number summary is : 0.2, 1.43, 2.19, 3.21, 8.0 .
(d) Boxplot for the data :
0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0
2. Joint and marginal distributions of X and Y:
Y
X 0 1 2 Marginal
1 0 0.05 0.1 0.15
2 0.15 0.1 0.15 0.4
3 0.3 0.15 0 0.45
Marginal 0.45 0.3 0.25 1

(a) Pr(X1 Pr Y 0) 0.15 0.45 0.0675 0 Pr X 1,Y 0)
=)( == = =(= =
Hence X and Y are not independent.
P. 1
Mid-term Examination Solution

2. (b) E()X =10.15 +2 0.4 +30.45 =2.3
22 22
E(X )=1 0.15 +2 0.4 +3 0.45 =5.8
2
Var()X =5.8 .2.3 =0.51
() 0 0.45 1 0.3 2 0.25 0.8

EY = ++ =
22 22
E(Y )=0 0.45 +1 0.3 +2 0.25 =1.3
2
Var()Y =1.3 .0.8 =0.66
( ) ()()+1 ()1 0.05)+3 + ()=1.5

E XY 320
Cov(X ,Y )=1.5 .2.30.8 =.0.34

=10 () ()0 ( ()()
=.0.5860
(c) Pr(at least 2 machines available)=Pr(X 2)=0.4 +0.45 =0.85
Pr(exactly one operation at least 2 machines available)=Pr(Y =1, X 2)
=0.1+0.15 =0.25

Pr(exactly one operation | at least 2 machines available)
Pr(Y =1, X 2) 0.25

=Pr(Y =1| X 2)= ==0.2941
Pr(X 2) 0.85
3. Let X be the number of jobs submitted to the system from 3:00pm to 3:30pm. Then
X ~ .()6.
.

Pr(X 3)=1.e.6 .1+6 +()62 ..=0.9380
..
2!
..
4. =200 , =55 Let Xi , i =1,2,...,45 be the weight of the 45 boxes. Then by normal approximation,
. 552 .
X~. N..200, ...
45
.. .45 .. 10000 .
Pr(all boxes can be loaded)=Pr.Xi 10000.=Pr.X .
.i=1 .. 45 .

.10000
45 .200 ..=(2.71)=0.9966
.
.
.
55
45
.
.
5. (a) Pr(lucky ticket)=0.001+0.002 +0.004 +0.01+0.03 =0.047
(b) Let W be the prize of one ticket. Then X =W .10 .
E()W =2000 0.001+1000 0.002 +250 0.004 +100 0.01+50 0.03 =7.5

22 2222
E(W )=2000 0.001+1000 0.002 +250 0.004 +100 0.01+50 0.03 =6425
2
Var()W =6425 .7.5 =6368.75
E()X=() .10 =. ()=() =

EW 2.5 , Var X VarW 6368.75
(c) Let Y be the number of lucky tickets in 100 ticket purchased. Then Y ~ b(100,0.047).
100 99 964
Pr(Y <5) ( =0.953)+100 C1 (0.953)( 0.047)+3+100 C4 (0.953)( 0.047)=0.4915
P. 2
Mid-term Examination Solution
(d) Let Xi , i = 1,2,...,100 be the net gains by each of the 100 tickets. Then by normal approximation,
.
6368.75 .
X~ N... 2.5, . .
. 100 .
Hence

. 100 . ..1. (. 2.5).
..Pr. X >.100.= Pr(X >.1) 1.