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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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