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(MATH144)[2008](f)final~1487^_10027.pdf
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08 Fall Math 144 Final Exam Suggested Sol
1 (a) (8 points).0.4 y ..1
(i)
fy ( y) . ..0.3 y . 0
.


.0.3 y . 1

(ii)
E(Y) = C1(0.4) + 0.3 = C0.1


(iii) P(X + Y < 1) = 0.2 + 0.1 + 0.1 = 0.4
(iv) E(XY) = C1(0.1) + 0.15 C 2(0.1) + 2(0.05) = C0.05
(b) (4 points)
EX = 40, . X .
24
P(| X . EX |.. X )

. P(40 .
24 . X . 40 .
24)
. P(35.10 . X . 44.90)
. P(36 . X . 44)
. P(35.5 . X . 44.5)
..(0.92) ..(.0.92)
. 0.6424
(c)
(3 points)
X C Y ~ N (C1, 4)
P(X > Y) = P(X C Y > 0)


= P(Z > .) where Z ~ N (0, 1)
= 0.3185


(d)
(3 points)
p-value = P( Z < 0.5) = 0.6915


(e)
(3 points)
p-value = 2 . P(t18 . 2.2)



P(t18 . 2.552) . P(t18 . 2.2) . P(t18 . 2.101)
2 . 0.01 . 2 . P(t18 . 2.2) . 2 . 0.025
.0.02 . p . value . 0.05
Reject H0 at significance level 0.1

(f) (3 points)
we have EX . 0, EX 2 .. 2

1 n 2
EW . E( .X )
n
i.1
n
12
..EX
n
i.1
1
. n. 2
n
. 2
.
08 Fall Math 144 Final Exam Suggested Sol
2. (a) (7 points)
2
1 y 11
(i) fX(x) ..(x .y)dy .xy . |0 .x . 0 .x .1
0 22
f (x, y) x .y

(ii) fY | X ( y) ..
fX (x) x .1/ 2
.fY | X .0.5( y) .y .10 .y .1
2
1
E(Y | X .0.5) ..yfY | X.0.5( y)dy
0
11
. y( y . )dy
.02
7
.
12
(iii) Cov(X, Y) = EXY C EXEY
11 21
EXY (x2 yd .xy )dxdy .
...
00 3
27

EX .EY . 1 x . x dx .
.
02 12
177 1

Cov(X ,Y ) .. . ..
3 12 12 144
(iv) No. X and Y are dependent because .fX (x) fy ( y) . f (x, y)
(b) (5 points)
(i) FY ( y) .P(Y .y)
.P(max(x1, x2) .y)

.P(x1 .x2 .y)
.P(x1 .y, x2 .y)
.y2
.fy( y) .F '( y) .2 y 0 .y .1
Y
(ii) EY 12y2dy .2
..03
08 Fall Math 144 Final Exam Suggested Sol
3. (i) (5 points) Let X = weight of a can of the new soft drink P(11.9 < X < 12.3)
11.9 .12.1 X .12.1 12.3 .12.1
. P( .. )
0.1 0.1 0.1
..(2) ..(.2)
. 0.9544

(ii) (5 points)
Let X = average weight of the cans in a six pack

0.12
X ~ N (12.1, )
6
-Z0.01 = -2.325

x .12.1
..2.325

0.1/ 6
x
. 12.00508

4. (i) (2 points)
.H0: .. 50
.

H : .. 50
. 1
(ii) (5 points)
Under H0, test statistic:
45 . 50

t .
..2
10/ 16

p-value = P(t15 < C2) = P(t15> 2)

0.025 < p-value < 0.05
Reject H0 at a significance level 0.05

(iii) (5 points)
A 95% C.I. for :
45 2.131(10)/4 t(0.025, 15) = 2.131
= (39.6725, 45.3275)

08 Fall Math 144
Final Exam Suggested Sol
5. (i) (6 points)
Let 1 = mean score for the male school students
2 = mean score for the female school students
. . . .. .. 0: 0: 211 210 .. .. H H = 0.05
Test statistic 28 1 34 117.9234 298.5288.5 . .t . 2.863. . sp = 17.9234
p-value = P(t60 < C2.863) = P(t60 > 2.863)
0.01 < p-value < 0.025
Reject H0 at a significance level 0.05
(ii) (6 points)
A 90% C.I. for 1 C 2 :

11
(288.5 C 298.5) 1.671(17.9234)
. t(0.05, 60) = 1.671
34 28
= (C17.6432, C2.3568)
6. (i) (4 points) n . .. Z. /2 .. 2 p(1. p)
. m.e . ..1.645 .21
..