20071ee131A_1_131a6sol

# 20071ee131A_1_131a6sol - EE 131A Homework#6 Solution Winter...

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EE 131A Homework #6 Winter 2007 Solution K. Yao 1 Let X be the number of small donuts in a package of 12. Then X is a binomial r.v. with b(x,12, 0.02). The probability a package is unacceptable is P = 1 - p X (0) - p X (1) = 1 - C 12 0 ( . 02) 0 ( . 98) 12 - C 12 1 ( . 02)( . 98) 11 = 1 - . 785 - . 192 = 0 . 0231 . 2 The average number of calls in 15 minutes is (1 / 4) × 2 = 1 / 2 = λ . a. P(no calls in 15 minutes) = e - 1 / 2 . b. P(no more than 1 call in 15 minutes) = P(no call) + P(one call) = e - 1 / 2 + (1 / 2) e - 1 / 2 . 3. (37) a. p 0 = λ 0 0! e - λ = e - 15 = 3 . 06 × 10 - 7 , λ = 15 . b. P [ N > 10] = 1 - P [ N 9] = 1 - 9 k =0 (15) k k ! e - 15 = 0 . 8815 . 4. (42) a. P [ X x ] = 1 - e - λx = r 100 , 1 - r 100 = e - λx π ( r ) = x = - 1 λ ln(1 - r 100 ) = 1 λ ln( 100 100 - r ) . π (90) 2 . 3 λ , π (95) 3 λ , π (99) 4 . 6 λ . b. P [ X x ] = 1 - Q ( x σ ) . Using Tables 3.3 and 3.4: 1 - Q ( x σ ) = 0 . 90 x σ = 1 . 28 π (90) = 1 . 28 σ 1 - Q ( x σ ) = 0 . 95 x σ 1 . 5 π (95) 1 . 5 σ 1 - Q ( x σ ) = 0 . 99 x σ 2 . 33 π (90) 2 . 33 σ 5. (44) P [ X < m ] = P [ X m ] = Φ( m - m σ ) = Φ(0) = 1 2 . P [ | X - m | > kσ ] = 1 - P [ - + m X m + ] = 1 - [Φ( m + - m σ ) - Φ( m - - m σ )] = 1 - Φ( k ) | {z } + Φ( - k ) | {z } = Q ( k ) + Q ( k ) = 2 Q ( k ) . From Table 3.3: k = 1 k = 2 k = 3 k = 4 k = 5 2 Q ( k ) 0 . 318 4 . 56 × 10 - 2 4 . 05 × 10 - 3 6 . 34 × 10 - 5 5 . 74 × 10 - 7 P [ X > m + ] = Q ( m + - m σ ) = Q ( k ). From Table 3.4: k = 1 . 28 k = 3 . 09 k = 4 . 26 k = 5 . 20 Q ( k ) 10 - 1 10 - 3 10 - 5 10 - 7 1

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6. (45) P [error | v = - 1] = P [ Y 0 | v = - 1] = P [ - 1 + N 0] = P [ N 1] = Q (1) = 0 . 159 . P [error | v = 1]
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