mt1-sample-solutions

mt1-sample-solutions - Y = you understand F = f riend...

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Unformatted text preview: Y = you understand F = f riend understands P (Y ) = 0.5 P (F ) = 0.6 P (Y ∩ F ) = 0.25 P (Y )P (F ) = 0.5 × 0.6 = 0.3 ￿= 0.25 P (Y )P (F ) ￿= P (Y ∩ F ) P (Y C ∩ F ) P (F ) = 0.6 = 0.35 = P (Y ∩ F ) + P (Y C ∩ F ) 0.25 + P (Y C ∩ F ) P (Y C ∩ F ) S T = = taking steroids tests positive f or steroids P (T | S ) = 0.80 = 0.95 P (S ) = 0.10 P (T C C |S ) P (S | T ) P (S | T ) = = = = = = P (S )P (T | S ) P (T ) P (S )P (T | S ) P (T ∩ S ) + P (T ∩ S C ) P (S )P (T | S ) P (S )P (T | S ) + P (S C )P (T | S C ) 0.10(0.80) 0.10(0.80) + 0.90(0.05) 0.08 0.08 + 0.045 0.64 µ = σ2 = 1 = 1.667 6 15 np(1 − p) = 10 × × = 1.389 66 np = 10 × P (X > 0) P (X > 0) = = = = 1 − P (X = 0) ￿￿ 10 1 0 5 10 1− ( )( ) 066 1 − 0.162 0.838 X P (X < 10) = 0.5 P (X < 5) = = 5 − 10 ) 2 P (Z < −2.5) P (Z < = 0.0062 = P (Z > P (X > 12) = 12 − 10 ) 2 P (Z > 1) = 0.1586 P (7 < X < 13) = = = = P (X < 6 ∪ X > 14) = = = = x∗ 7 − 10 13 − 10 <Z< ) 2 2 P (−1.5 < Z < 1.5) P( 1 − 2(0.0668) 0.8664 6 − 10 14 − 10 ∪Z > ) 2 2 P (Z < −2 ∪ Z > 2) P (Z < 2(0.0228) 0.0455 P (X < x∗ ) = 0.95 Z ∗ P (Z < z ) 0.95 z = 1.645 x ∗ = µ + z∗σ = 10 + 1.645(2) = X = ∗ 13.29 4 2 0 Frequency 6 8 0, 1, 3, 3, 4, 4, 5, 6, 8, 12, 12, 18, 20, 20, 24, 27, 28, 34, 37, 39, 56, 71 0 20 40 60 80 x ● 0 10 20 30 ¯ X= 40 1 22 ￿ 50 60 Xi = 19.64 70 X(11) +X(12) 2 = 12+18 2 = 15 ...
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This note was uploaded on 11/20/2011 for the course PSTAT 5E taught by Professor Eduardomontoya during the Fall '08 term at UCSB.

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