The correct answer is D
Calculating z-values, z1 = (110 ? 120) / 20 = ?0.5. z2 = (170 ? 120) / 20 = 2.5. Using the z-table, P(?0.5) = (1 ? 0.6915) = 0.3085. P(2.5) = 0.9938. P(?0.5 < X < 2.5) = 0.9938 ? 0.3085 = 0.6853. Note that on the exam, you will not have access to z-tables, so you would have to reason this one out using the normal distribution approximations. You know that the probability within +/? 1 standard deviation of the mean is approximately 68%, meaning that the area within ?1 standard deviation of the mean is 34%. Since ?0.5 is half of ?1, the area under ?0.5 to 0 standard deviations under the mean is approximately 34% / 2 = 17%. The probability under +/? 2 standard deviations of the mean is approximately 99%. The value $170 is mid way between +2 and +3 standard deviations, so the probability between these values must be (99% / 2) = 2%. The value from 0 to 2.5 standard deviations must therefore be (99% / 2) ? (2% / 2) = 48.5%. Adding these values gives us an approximate probability of (48.5% + 17%) = 65.5%. |