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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.
chart
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.
- 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?
- Find x of 1% or .01, which is the area beyond z.
- So, find the z score in the table that is closest to .01.
- z = 2.33 is closest to.
- z = (2.33)15 + 100 = x or 134.
- 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?
- Find x of 5% or .05, which is the area beyond z.
- 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.
- z = 1.645.
- z = (1.645)300 + 1000 = x or 1,493.
- A person needed to have donated about $1,494 to be in the top 5% of all donors to the football program.
Example 3: What is the score at the 83 percentile? N = 83, M = 4.23, SD = 1.
Ratings ranged on a scale from 1 to 7.
The 83rd percentile means that 83% of the scores fall equal to or below a certain score in the distribution of scores. A z score of 0 means 50% of scores fall below the mean score. So we still need to find another 33% of the area under the standard normal curve (50% + 33% = 83%). In this case, we can work backwards or 100% - 83% (or 17%) to find the area beyond that point. The z score is .95 (area of 17.11% beyond). X = .95(1.02) + 4.23 or a score of 5.199 which is rounded to 5.20. Therefore, 5.20 represents the score with a percentile rank of 83, or 17% of the students scored equal to or higher than a 5.20, or 83% of students scored equal to or lower than a 5.20.
Alternatively, we could have solved this problem by looking at the Percentiles to z Scores Table on the course site and found that the 83rd percentile = a z score of .95 and then: X = .95(1.02) + 4.23 or a score of 5.199, which is rounded to 5..
Example 2:
How many of the 83 students had scores between 4.88 and 5.62?
We need to do is convert both scores into z scores and then find the area within the standard normal curve that lies between those two scores.
So: N = 83, M = 4.23, SD = 1.
- Convert each raw score to z scores
z = 4.88 - 4.
Which is 0.637 or rounded to 0.
z = 5.62 - 4.
Which is 1.362 or rounded to 1.36.
Now we look up both z values in the z table.
z of 1.36 has a proportion of 41.31 between the mean and that value Part D of the chart or the Proportion between the Mean and z.
z of 0.64 has a proportion of 23.89 between the mean and that value.
The proportion between those sections is (41.31 - 23.89) or 17.42 percent. That means that 17.42% of the scores in the data set would fall between 4.88 and 5.62.
- Now we multiply 17.42% x 83 (or N) or 14.45. 6. 14 students have scores between 4.88 and 5.
