After interpolating, the probability value for z = -1.645 is 0.45. This means that 0.45, or 45 percent, of the total area under the normal curve lies between xN and E(RN), and it implies a profit probability for the newspaper promotion of 0.45 + 0.5 = 0.95, or 95 percent. In terms of profit probability, the newspaper advertisement is the less risky alternative.

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