9 3 6 2 9 2 6 2 9 1 1 2 6 9 6 9 2 6 2 3 2 2 9 3 6 2 9

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𝐸𝐸 [ 𝑋𝑋 ] = � 𝑥𝑥 � 9 𝑥𝑥 3 � 𝑑𝑑𝑥𝑥 6 2 = 9 𝑥𝑥 2 𝑑𝑑𝑥𝑥 6 2 = 9 𝑥𝑥 −1 1 2 6 = 9 6 + 9 2 = 6 2 = 3 𝐸𝐸 [ 𝑋𝑋 2 ] = � 𝑥𝑥 2 9 𝑥𝑥 3 � 𝑑𝑑𝑥𝑥 6 2 = 9 𝑥𝑥 𝑑𝑑𝑥𝑥 6 2 = 9 ln| 𝑥𝑥 | 2 6 = 9(ln(6) ln(2)) = 9 ln(3) 9.8875 𝑉𝑉𝑎𝑎𝐹𝐹 [ 𝑋𝑋 ] = 𝐸𝐸 [ 𝑋𝑋 2 ] ( 𝐸𝐸 [ 𝑋𝑋 ]) 2 = 9 ln(3) 3 2 9.8875 9 = 𝟎𝟎 . 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖 2. Braking distance for a particular type of vehicles is uniformly distributed with mean 15 ft and variance 12 ft 2 . Find the probability that the braking distance for such vehicle would not exceed 11 ft. 𝐸𝐸 [ 𝑋𝑋 ] = 𝑎𝑎 + 𝑏𝑏 2 = 15 , 𝑉𝑉𝑎𝑎𝐹𝐹 [ 𝑋𝑋 ] = ( 𝑏𝑏 − 𝑎𝑎 ) 2 12 = 12 𝑎𝑎 + 𝑏𝑏 = 30 𝑏𝑏 − 𝑎𝑎 = 12 𝑏𝑏 = 30 − 𝑎𝑎 30 2 𝑎𝑎 = 12 𝑏𝑏 = 30 − 𝑎𝑎 2 𝑎𝑎 = 18 𝑏𝑏 = 30 − 𝑎𝑎 𝑎𝑎 = 9 𝑏𝑏 = 21 𝑎𝑎 = 9 𝑃𝑃 [ 𝑋𝑋 11] = 𝑥𝑥 − 𝑎𝑎 𝑏𝑏 − 𝑎𝑎 = 11 9 21 9 = 2 12 = 1 6 0 .1667 3. Suppose that X has an exponential probability density function. Show that, if a > 0 and b > 0, then 𝑃𝑃 ( 𝑋𝑋 > 𝑎𝑎 + 𝑏𝑏 | 𝑋𝑋 > 𝑎𝑎 ) = 𝑃𝑃 ( 𝑋𝑋 > 𝑏𝑏 ). This is called the memoryless property of exponential distribution. 𝑃𝑃 ( 𝑋𝑋 > 𝑎𝑎 + 𝑏𝑏 | 𝑋𝑋 > 𝑎𝑎 ) = 𝑃𝑃 [ 𝑋𝑋 > 𝑎𝑎 + 𝑏𝑏 ∩ 𝑋𝑋 > 𝑎𝑎 ] 𝑃𝑃 [ 𝑋𝑋 > 𝑎𝑎 ] = 𝑃𝑃 [ 𝑋𝑋 > 𝑎𝑎 + 𝑏𝑏 ] 𝑃𝑃 [ 𝑋𝑋 > 𝑎𝑎 ] = 1 − � 1 − 𝑒𝑒 −𝜆𝜆 ( 𝑎𝑎+𝑏𝑏 ) 1 (1 − 𝑒𝑒 −𝜆𝜆𝑎𝑎 ) = 𝑒𝑒 −𝜆𝜆𝑎𝑎 𝑒𝑒 −𝜆𝜆𝑏𝑏 𝑒𝑒 −𝜆𝜆𝑎𝑎 = 𝑒𝑒 −𝜆𝜆𝑏𝑏 = 1 − � 1 − 𝑒𝑒 −𝜆𝜆𝑏𝑏 = 𝑃𝑃 [ 𝑋𝑋 > 𝑏𝑏 ]
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4. Phone calls that come to the customer service phone number are independent. Average waiting time until the fifth phone call is 10 minutes. a. Find the average waiting time between any two consecutive phone calls. 𝑋𝑋 = 𝑤𝑤𝑎𝑎𝑓𝑓𝑑𝑑𝑓𝑓𝑙𝑙𝑤𝑤 𝑑𝑑𝑓𝑓𝐺𝐺𝑒𝑒 𝑑𝑑𝑙𝑙𝑑𝑑𝑓𝑓𝑙𝑙 𝑑𝑑ℎ𝑒𝑒 5 𝑡𝑡ℎ 𝑝𝑝ℎ𝐹𝐹𝑙𝑙𝑒𝑒 𝑐𝑐𝑎𝑎𝑙𝑙𝑙𝑙 𝑋𝑋 ~ 𝐺𝐺𝑎𝑎𝐺𝐺𝐺𝐺𝑎𝑎 ( 𝛼𝛼 = 5, 𝛽𝛽 ), 𝐸𝐸 [ 𝑋𝑋 ] = 𝛼𝛼𝛽𝛽 = 5 𝛽𝛽 = 10 => 𝛽𝛽 = 10 5 = 2 b. Find the variance of the waiting time until the fifth phone call to customer service. 𝑉𝑉𝑎𝑎𝐹𝐹 [ 𝑋𝑋 ] = 𝛼𝛼𝛽𝛽 2 = 5(2 2 ) = 20 c. Find the variance of the waiting time until the first phone call to customer service. 𝑇𝑇 = 𝑤𝑤𝑎𝑎𝑓𝑓𝑑𝑑𝑓𝑓𝑙𝑙𝑤𝑤 𝑑𝑑𝑓𝑓𝐺𝐺𝑒𝑒 𝑑𝑑𝑙𝑙𝑑𝑑𝑓𝑓𝑙𝑙 𝑑𝑑ℎ𝑒𝑒 1 𝑠𝑠𝑡𝑡 𝑝𝑝ℎ𝐹𝐹𝑙𝑙𝑒𝑒 𝑐𝑐𝑎𝑎𝑙𝑙𝑙𝑙 𝑋𝑋 ~ 𝐸𝐸𝑥𝑥𝑝𝑝𝐹𝐹𝑙𝑙𝑒𝑒𝑙𝑙𝑑𝑑𝑓𝑓𝑎𝑎𝑙𝑙 (
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  • Winter '08
  • moulib
  • probability density function, 10 minutes, 6ft, 11 ft

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