EXAM P SAMPLE SOLUTIONS
EXAM P SAMPLE SOLUTIONS
EXAM P SAMPLE SOLUTIONS
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( X < ) ∪ ( Y < )<br />
Pr ⎡⎣ 1 1 ⎤⎦<br />
2<br />
3<br />
3 3 3 3<br />
x+ y x + 2xy 1<br />
= 1− ∫∫ dx dy = 1− dy 1 ( 9 6y1 2y)<br />
dy<br />
27 ∫ = − + − −<br />
54 54 ∫<br />
1 1 1 1<br />
1<br />
3<br />
1 3<br />
1 2 1 32 11<br />
( ) ( ) ( )<br />
54 ∫ y dy y y<br />
1<br />
54 1 54 54 27<br />
= 1− 8+ 4 = 1− 8 + 2 = 1− 24+ 18−8− 2 = 1− = = 0.41<br />
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79. Solution: E<br />
The domain of s and t is pictured below.<br />
Note that the shaded region is the portion of the domain of s and t over which the device<br />
fails sometime during the first half hour. Therefore,<br />
⎡⎛ 1⎞ ⎛ 1⎞⎤<br />
1/2 1 1 1/2<br />
Pr ⎢⎜S ≤ ⎟∪⎜T ≤ ⎟ = f ( s, t) dsdt+ f ( s, t) d<br />
2 2<br />
⎥ ∫0 ∫1/2 ∫0∫ sdt<br />
0<br />
⎣⎝ ⎠ ⎝ ⎠⎦<br />
(where the first integral covers A and the second integral covers B).<br />
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80. Solution: C<br />
By the central limit theorem, the total contributions are approximately normally<br />
nμ = 2025 3125 = 6,328,125 and standard deviation<br />
distributed with mean ( )( )<br />
σ n = 250 2025 = 11,250 . From the tables, the 90 th percentile for a standard normal<br />
random variable is 1.282 . Letting p be the 90 th percentile for total contributions,<br />
p−nμ = 1.282, and so p= nμ + 1.282σ n = 6,328,125 + ( 1.282)( 11, 250) = 6,342,548 .<br />
σ n<br />
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