a) To find the probability of a variable, X, between two limits, a and b when the variable is normally distributed:P(a < X < b) = Î¦[(bÂµ)/Ïƒ]  Î¦[(aÂµ)/Ïƒ] where the values within the square brackets are standard normal zvalues. Phi (Î¦) is and operator that means "find the probability associated with the Zvalues in the square brackets". So evaluating:P(550 < X < 650) = Î¦[(650550)/80]  Î¦[(550550)/80] P(550 < X < 650) = Î¦[(650550)/80] = Î¦[1.25]  Î¦[0] = 0.8944  0.50 = 0.3944b) You want to eliminate the bottom 20%. So the Zvalue associated with the BOTTOM 20% is 0.84, and the limit for the bottom 20% is found by:0.84 = (x  550)/80; solving for x = 482.8. If an integer is required, then the lowest acceptable score would be 483.c) Find the probability that the sample mean is less than 650:Z = (650550)/(80/âˆš50) = 100/35.78) = 2.79Then Î¦[2.79]  Î¦[0] = 0.9974 0.50 = 0.4974
