Deck 5: Standardization and Z Scores
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Deck 5: Standardization and Z Scores
1
Suppose that in the population of chickens, the average number of feathers per chicken is 1000 with a standard deviation of 200. What is the probability of randomly selecting an individual from this population that has at least 1100 feathers?
(1100-1000)/200 = .50. Probability = .3085.
2
Suppose that on a talent show like American Idol, the only contestants who actually make it onto the show must be at least 3 standard deviations above the mean in terms of their talent. What percentage of those who audition would you expect to actually make it onto the show?
Three standard deviations above the mean is a z score of 3. In Appendix A, the box on the second page shows that when the z score is three, the area probability of getting a value between that z score and infinity is .001350, or .135 percent.
3
Suppose that in a class with 50 students in it with an average score on an exam of 30 and a standard deviation of 10, John gets a score of 45. Calculate his z score and explain what it tells you.
John's z score is (45-30)/10 15/20 = 1.5 So John's z score is 1.5, and this tells us that John's test score is 1.5 standard deviations above the mean for his class.
4
If a test has a mean of 500 and a standard deviation of 50 (and is normally distributed), what score marks the 15th percentile?
a. On this same test, what proportion of students would be expected to receive scores between 475 and 550?
b. What scores fall at the extreme 10% of this distribution (two-tailed)?
a. On this same test, what proportion of students would be expected to receive scores between 475 and 550?
b. What scores fall at the extreme 10% of this distribution (two-tailed)?
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k this deck