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Unformatted text preview: Solutions to Exam One Review Problems 1. (a) 12 (b) 12 3 . 2 = 8 . 8 (These are approximate answers!) (c) Approximately 0 . 50 since the median is at approximately 6. (d) The mean is greater than the median– the data set is positively skewed. 2. (a) Approximately 0 . 10 + 0 . 05 + 0 . 10 = 0 . 25 (b) Approximately 29 since 40% of the data is less than or equal to 29. (c) The median is located between 29 and 30. (d) The proportion of cars getting at least 30 mph is approximately 0 . 30 + . 025 + 0 . 05 + 0 . 025 = 0 . 40. This makes the number of such cars approxi mately (0 . 40)(300) = 120. 3. Units: Tens, Leaves: Ones 9 8 , 9 10 2 , 4 , 6 , 6 11 4 , 5 , 7 , 8 , 8 , 9 12 2 , 4 , 5 , 7 13 1 , 2 14 4 15 1 4. summationdisplay x i = 53 , summationdisplay x 2 i = 1167 , n = 15 So, the sample mean is x = 53 5 10 . 6 and the sample variance is s 2 = ∑ x 2 i ( ∑ x i ) 2 n n 1 = 1167 53 2 15 14 = 69 . 98095 . So, the standard deviation is s = √ s 2 = √ 69 . 98095 ≈ 8 . 365462 5. Let A be the event that the soldier is in the standing position. Let B be the event that the soldier is in the sitting position. Let C be the event that the soldier is in the prone position. If we randomly select one of the rounds, P ( A ) = 1 3 , P ( B ) = 1 3 , P ( C ) = 1 3 . Let H be the probability that the soldier hits the target. Then P ( H  A ) = 18 30 , P ( H  B ) = 27 30 , P ( H  C ) = 24 30 . (a) P ( H ) = P ( H  A ) P ( A ) + P ( H  B ) P ( B ) + P ( H  C ) P ( C ) = 18 30 1 3 + 27 30 1 3 + 24 30 1 3 = 69 90 = 23 30 (b) P ( C ) = 1 3 (c) P ( C  H ) = P ( H  C ) P ( C ) P ( H ) = 24 30 1 3 23 30 = 8 23 6. Let D be the event that the randomly selected individual has the disease . Then P ( D ) = 0 . 01. Let T 1 be the event that the first test is positive and let T 2 be the event that the second test is positive. Note: This is a tricky problem! The test results for a particular individual are not independent! The tests may have been administered independently, but if the first test is positive, it is likely the person has the disease, so it is likely that their second test will also be positive. So, we can’t just assume that P ( T 1 ∩ T 2 ) = P ( T 1 ) P ( T 2 ), for example. However, given that the person has the disease (or not), the tests are indepen dent: P ( T 1 ∩ T 2  D ) = P ( T 1  D ) P ( T 2  D ) and P ( T 1 ∩ T 2  D ′ ) = P ( T 1  D ′ ) P ( T 2  D ′ ) . (a) We want both positive or both negative: P (( T 1 ∩ T 2 ) ∪ ( T ′ 1 ∩ T ′ 2 )) disjoint = P ( T 1 ∩ T 2 ) + P ( T ′ 1 ∩ T ′ 2 ) But, P ( T 1 ∩ T 2 ) = P ( T 1 ∩ T 2  D ) P ( D ) + P ( T 1 ∩ T 2  D ′ ) P ( D ′ ) = (0 . 9)(0 . 9)(0 . 1) + (0 . 05)(0...
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 '08
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 Poisson Distribution, Probability theory, Binomial distribution, CDF

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