Correct option is A
Solution:
Matching Standard Normal Intervals with ProbabilitiesA. B. C. D. −1≤Z≤+∞−1≤Z≤+1−∞≤Z≤20≤Z≤2=>P(Z≥−1)=1−P(Z<−1)=1−0.1587=0.8413≈0.84=>III=>P(−1≤Z≤1)=0.6826≈0.82=>IV=>P(Z≤2)=0.9772≈0.98=>I=>P(0≤Z≤2)=0.4772≈0.48=>IIFinal Matching:A-III, B-IV, C-I, D-II
