1) In a normal distribution, what is the proportion of are between the z scores - 2.09 and +.53?

2) In a normal distributio, find the probability expressed as a percentage that is greater than a z score of 2.33

1.) P (-2.09 < z < 0.53) = P (0.53) - P (-2.09) = P (0.53) - [1 - P (2.09) ] = P (0.53) + P (2.09) - 1 = 0.70194 + 0.98169 - 1 = 0.68363 = 68.36%

2.) P (z > 2.33) = 1 - P (z < 2.33) = 1 - 0.9901 = 0.0099 = 0.99%