Topic4_M9118_Data_Analytics.pdf
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Dr. Ahmed Bakri, Dr. Krisztina Soreg & Mr. Antonio Moya o o o o σ σ = 2 = 1.290 = 1.135 P(0)= n C x ( ) x (1 − ) n − x = 5 C0 (0.20) 0 (1 − 0.20) 5−0 = (1)(1)(.3277) = 0.3277 P(1)= n C x ( ) x (1 − ) n − x = 5 C1 (0.20)1 (1 − 0.20) 5−1 = (5)(0.20)(.4096) = 0.4096 P(X>2)= 1-P(X≤2) opposit...
Dr. Ahmed Bakri, Dr. Krisztina Soreg & Mr. Antonio Moya o o o o σ σ = 2 = 1.290 = 1.135 P(0)= n C x ( ) x (1 − ) n − x = 5 C0 (0.20) 0 (1 − 0.20) 5−0 = (1)(1)(.3277) = 0.3277 P(1)= n C x ( ) x (1 − ) n − x = 5 C1 (0.20)1 (1 − 0.20) 5−1 = (5)(0.20)(.4096) = 0.4096 P(X>2)= 1-P(X≤2) opposite