For a hypothetical normal distribution of test scores, approximately 95% fall between 38 and 62, 2.5% are below 38, and 2.5% are above 62. Given this information, (a) the mode=_____ and (b) the standard deviation=_____. Solution X: test scores (X - ?)/? = Z P(Z < - z ) = 0.025 P(Z < - 1.96) = 0.025 - 1.96 = (38 - ?)/? 1.96 = (62 - ?)/? - 1.96? = 38 - ? 1.96? = 62 - ? ----------------------- 0 = 100 - 2? ? = 50 1.96? = 62 - ? ? = 12/1.96 = 6.12 mode = 50.