1. INTRODUCTORY STATISTICS FOR CRIMINAL JUSTICE Chapter 5: z-Scores
COURSE: JUST 3900
INTRODUCTORY STATISTICS
FOR CRIMINAL JUSTICE
Chapter 5: z-Scores
Peer Tutor Slides
Instructor:
Mr. Ethan W. Cooper, Lead Tutor
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2. Key Terms and Formulas: Don’t Forget Notecards
z-Score (p. 141)
Standardized Distribution (p. 147)
Formulas for populations:
𝑧= 𝑋−𝜇 𝜎
𝑋=𝜇+𝑧𝜎
Formulas for Samples:
𝑧= 𝑋−𝑀 𝑠
𝑋=𝑀+𝑧𝑠

3. Describing z-Scores
Question 1: Identify the z-score value corresponding to each of the following locations in a distribution.
Below the mean by 3 standard deviations.
Above the mean by 1/4 standard deviations.
Below the mean by 2.50 standard deviations.

4. Describing z-Scores
Question 1 Answer:
z = -3.00
z = 0.25
z = -2.50

5. Describing z-Scores
Question 2: Describe the location in the distribution for each of the following z-scores.
z =
z = 0.25
z =
z = 1.75

6. Describing z-Scores Question 2 Answer:
Below the mean by 1.50 standard deviations.
Above the mean by ¼ standard deviations.
Below the mean by 3.50 standard deviations.
Above the mean by 1.75 standard deviations.

7. Describing z-Scores
The numerator in our z-score formula (𝑧= 𝑋−𝜇 𝜎 ) describes the difference between X and µ. Therefore, if a question asks you to calculate the z-score for a score that is above the mean by 4 points and has a standard deviation of σ = 2, you cannot calculate (X - µ) - - i.e., you cannot find the original values for X or µ and calculate the difference between the two. In this case, it has already been provided for you because the question tells you the distance between X and µ (X - µ) = 4. Thus, your formula should read z = 4/2, which comes out to be z = 2.00.

8. Transforming X-Values into z-Scores
Question 3: For a distribution of µ = 40 and σ = 12, find the z-score for each of the following scores.
X = 36
X = 46
X = 56

9. Transforming X-Values into z-Scores
Question 3 Answer:
𝑧= 𝑋−𝜇 𝜎 = 36−40 12 = −4 12 =−0.25
𝑧= 𝑋−𝜇 𝜎 = 46−40 12 = 6 12 =0.50
𝑧= 𝑋−𝜇 𝜎 = 56−40 12 = =1.33

10. Transforming X-Values into z-Scores
Question 4: For a population with µ = 30 and σ = 8, find the z-score for each of the following scores.
X = 32
X = 26
X = 42

11. Transforming X-Values into z-Scores
Question 4 Answer:
𝑧= 𝑋−𝜇 𝜎 = 32−30 8 = 2 8 =0.25
𝑧= 𝑋−𝜇 𝜎 = 26−30 8 = −4 8 =−0.50
𝑧= 𝑋−𝜇 𝜎 = 42−30 8 = 12 8 =1.50

12. Using z-Scores to Compare Different Populations
Question 5: A distribution of English exam scores has µ = 70 and σ = 4. A distribution of history exam scores has µ = 60 and σ = 20. For which exam would a score of X = 78 have a higher standing? Explain your answer.

13. Using z-Scores to Compare Different Populations
Question 5 Answer:
𝑧= 𝑋−𝜇 𝜎 = 78−70 4 = =2.00
𝑧= 𝑋−𝜇 𝜎 = 78− = =0.90
For the English exam, X = 78 corresponds to z = 2.00, which is a higher standing than z = 0.90 for the history exam.
Remember that 95% of all scores fall between ± Thus, a score
of means that over 95% of the class scored below 78 on the English exam.

14. Using z-Scores to Compare Different Populations
Question 6: A distribution of English exam scores has µ = 50 and σ = 12. A distribution of history exam scores has µ = 58 and σ = 4. For which exam would a score of X = 62 have a higher standing? Explain your answer.

15. Using z-Scores to Compare Different Populations
Question 6 Answer:
𝑧= 𝑋−𝜇 𝜎 = 62− = =1.00
𝑧= 𝑋−𝜇 𝜎 = 62−58 4 = =1.00
The score X = 62 corresponds to z = 1.00 in both distributions. The score has exactly the same standing for both exams.

16. z-Scores and Standardized Scores
Question 7: A population of scores has µ = 73 and σ = 8. If the distribution is standardized to create a new distribution with µ = 100 and σ = 20, what are the new values for each of the following scores from the original distribution?
X = 65
X = 71
X = 81
X = 83

17. z-Scores and Standardized Scores

18. z-Scores and Standardized Scores
Question 8: A population with a mean of µ = 44 and a standard deviation of σ = 6 is standardized to create a new distribution with µ = 50 and σ = 10.
What is the new standardized value for a score of X = 47 from the original distribution?
One individual has a new standardized score of X = 65. What was this person’s score in the original distribution?

19. z-Scores and Standardized Scores
X = 47
z = 0.50
X = 55
Old Distribution 32 38 44 50 56
z-Score Distribution
New Standardized Distribution

20. z-Scores and Standardized Scores
X = 53
z = 1.50
X = 65
Old Distribution 32 38 44 50 56
z-Score Distribution
New Standardized Distribution

21. Measure of Relative Location and Detecting Outliers
Question 9: A sample has a mean of M = 30 and a standard deviation of s = 8.
Would a score of X = 36 be considered a central score or an extreme score in the sample?
If the standard deviation were s = 2, would X = 36 be central or extreme?

22. Measure of Relative Location and Detecting Outliers
Question 9 Answer:
𝑧= 𝑋−𝑀 𝑠 = 36−30 8 = =0.75
X = 36 is not an extreme score because it is within two standard deviations of the mean.
𝑧= 𝑋−𝑀 𝑠 = 36−30 2 = =3.00
In this case, X = 36 is an extreme score because it is more than two standard deviations above the mean.

23. WARNING!!!
The book defines an extreme score as being more than TWO standard deviations away from the mean. However, Aplia defines extreme scores as being more than THREE standard deviations from the mean.
When using Aplia, use the THREE definition of standard deviation for extreme scores.
On in class exercises and on the test, use the TWO definition of standard deviation for extreme scores.

24. Frequently Asked Questions FAQs
How do I find the z-scores from a raw set of scores?
X = 11, 0, 2, 9, 9, 5
Find the mean:
𝜇= 𝑋 𝑁 = = =6
Find SS:
𝑆𝑆= 𝑋−𝜇 2 X
X - µ
(X - µ)2 11
11 – 6 = 5
(5)2 = 25
0 – 6 = -6
(-6)2 = 36 2
2 – 6 = -4
(-4)2 = 16
9 – 6 = 3
(3)2 = 9 9 5
5 – 6 = -1
(-1)2 = 1
𝑆𝑆= 𝑋−𝜇 2 = 96

25. Frequently Asked Questions FAQs
Find σ2:
𝜎 2 = 𝑆𝑆 𝑁 = =16
Find σ:
𝜎= 𝑆𝑆 𝑁 = = =4
Find z-score for each X:
𝑧= 𝑋−𝜇 𝜎 = 11−6 4 = =1.25
𝑧= 𝑋−𝜇 𝜎 = 0−6 4 = −6 4 =−1.50
𝑧= 𝑋−𝜇 𝜎 = 2−6 4 = −4 4 =−1.00
𝑧= 𝑋−𝜇 𝜎 = 9−6 4 = =0.75
𝑧= 𝑋−𝜇 𝜎 = 5−6 4 = −1 4 =−0.25