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CS404_Exam_1_09292005

# CS404_Exam_1_09292005 - when v = 6… v’ =(6 – 6(82 –...

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1. i)Normalization from [-25,20] to [0,25]. Basically taking the absolute value of each value ii) I do not see another transformation 2. min = 6, max = 82, new_min = 0.0, new_max = 5.0 v’ = ((v – min) / (max – min))*(new_max – new_min) + new_min when v = 52… v’ = ((52 – 6) / (82 – 6))*(5 – 0) + 0 = 22/57= 3.02 when v = 41… v’ = ((41 – 6) / (82 – 6))*(5 – 0) + 0 = 22/57= 2.30 when v = 13… v’ = ((13 – 6) / (82 – 6))*(5 – 0) + 0 = 22/57= 0.46
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Unformatted text preview: when v = 6… v’ = ((6 – 6) / (82 – 6))*(5 – 0) + 0 = 22/57= 0 when v = 82… v’ = ((82 – 6) / (82 – 6))*(5 – 0) + 0 = 22/57= 5 3. <> 4. 1 2 3 4 5 6 Histogram 1 2 3 4 6 7 8 9 5. i) a. Age b. Number of previous accidents c. Speeding tickets/points on license ii) 6. Expected Values Matrix No Yes Neutral 107 42 Not Concerned 64 25 Very Concerned 223 88...
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