A. Determining Proportions with z Scores （答确定比例与z分数）.pdf

A. Determining Proportions with z Scores Example 1: x is normal with mu of 100 and sigma of 15. Find the proportion of people who have IQs of 122 or higher. 122 −100 1.47 ⎯⎯→chart .0708 15 Looking at the z score of 1.47, we go to part C of the chart, or the Proportion in the Tail, and determine that .0708 is beyond the x of 122. Thus, 7% have an IQ of 122 or higher. Example 2: Find the proportion of people with IQs of 80 or less. 80 −100 −1.33 .0918 15 Thus, 9% have IQs of 80 or less Note: For negative values of z, probabilities are found by symmetry. Example 3: Determine the probability that z is less than -.022 or P(z -.022) There is a .4920 or 49% probability that z is -.022. B. Determining Exact Percentiles with z scores Example 1: Using the data above, what IQ must a person have in order to be in the top 1% of IQs? 1. Find x of 1% or .01, which is the area beyond z. 2. So, find the z score in the table that is closest to .01. 3. z = 2.33 is closest to .01 4. z = (2.33)15 + 100 = x or 134.95 5. A person needs an IQ of 135 to be in the top 1% of the group IQs. Example 2: Last year, NIU boosters gave an average of $1,000 (SD = $300) to the athletic department for the school’s football team. How much money would a booster needed to have donated to be considered in the top 5%, or the Huskie Club, of all donors? 1. Find x of 5% or .05, which is the area beyond z. 2. z scores in the table that are closest to .05 are 1.64 and 1.65. They are of equal distance, so we need to find the z score average or 1.64 + 1.65 = 3.29 / 2 = 1.645. 3. z = 1.645. 4. z = (1.645)300 + 1000