CJ 301 Assignment 5: Z Score – Please Answer Each Of The ✓ Solved
CJ 301 Assignment #5 Z SCORE –Please answer each of the
CJ 301 Assignment #5 Z SCORE –Please answer each of the following questions. Also, INTERPRET your answers. You must use the lecture for this week while working on this assignment – Z score module – under the Learning Modules. ( Also, use the Z score table).
A state department of corrections has a policy whereby it accepts as correctional officers only those who score in the top 5 % of a qualifying exam. The mean of this test is 80. Standard deviation is 10. Would a person with a raw score of 95 be accepted? (Calculate a Z score: score – mean/st.dev.=)
Given a normal distribution of raw scores with a mean of 60 and a standard deviation of 10, what proportion of cases fall: between a raw score of 40 and 80? (3 points) between a raw score of 45 and 50? (3 points) Find the z-score corresponding to a raw score of 90 from a normal distribution with mean 60 and standard deviation 8. (2 points) For a normal distribution where the mean is 50 and the standard deviation is 8, what is the area: Between the scores of 30 and 65? (2 points)
Assume that the distribution of a college entrance exam is normal with a mean of 500 and a standard deviation of 100. For each score below, find the equivalent Z score, the percentage of the area above the score, and the percentage of the area below the score. (3 + 3 points) Score Z score % Area Above % Area Below a) 375 b) 437
Paper For Above Instructions
The Z score, also known as the standard score, indicates how many standard deviations a data point is from the mean. This paper will cover various statistical evaluations based on the normal distribution using Z scores, including interpretations for competence assessments and their applicability in educational systems.
1. Correctional Officer Acceptance Criteria
The state department of corrections requires that prospective correctional officers achieve scores in the top 5% of a qualifying exam, which has a mean (μ) of 80 and a standard deviation (σ) of 10. To determine if a candidate with a raw score of 95 would be accepted, we calculate the Z score using the formula:
Z = (X - μ) / σ
Where X represents the raw score. Plugging in the values:
Z = (95 - 80) / 10 = 1.5
A Z score of 1.5 corresponds to a percentile in the normal distribution. Referring to the Z score table, a Z score of 1.5 is approximately equal to 0.9332, which means that a score of 95 is higher than 93.32% of the test-takers. Therefore, since 93.32% exceeds 95% of the applicants, the candidate with a raw score of 95 would indeed be accepted. Given this, we can conclude that the raw score of 95 categorically places an individual well within the top 5% threshold needed for acceptance.
2. Proportions of Cases Falling Between Raw Scores
Next, we analyze a normal distribution with a mean of 60 and standard deviation of 10. To find the proportion of cases that fall between a raw score of 40 and 80, we first find the Z scores for these values:
- For a raw score of 40: Z = (40 - 60) / 10 = -2.0
- For a raw score of 80: Z = (80 - 60) / 10 = 2.0
Using the Z score table, a Z score of -2.0 corresponds to a cumulative probability of 0.0228, and a Z score of 2.0 corresponds to 0.9772. The proportion of cases falling between these scores is:
P(40
Thus, approximately 95.44% of cases fall between raw scores of 40 and 80.
2a. Proportion between Raw Scores 45 and 50
Next, we find the proportion between the raw scores of 45 and 50:
- For a raw score of 45: Z = (45 - 60) / 10 = -1.5
- For a raw score of 50: Z = (50 - 60) / 10 = -1.0
The cumulative probabilities are approximately:
- P(Z
- P(Z
Therefore, the proportion between these two scores is:
P(45
This means approximately 9.19% of cases fall between raw scores of 45 and 50.
3. Z-Score for Raw Score of 90
For a raw score of 90 from a normal distribution with a mean of 60 and a standard deviation of 8, we calculate:
Z = (90 - 60) / 8 = 3.75
A Z score of 3.75 indicates that this score is significantly higher than the average, corresponding to a very small percentile in terms of the distribution of scores.
4. Area Between Scores 30 and 65
For the distribution where the mean is 50 and the standard deviation is 8, we find the area between the scores of 30 and 65:
- For a raw score of 30: Z = (30 - 50) / 8 = -2.5
- For a raw score of 65: Z = (65 - 50) / 8 = 1.875
Using the Z table:
- P(Z
- P(Z
The area between these scores is:
P(30
This means that approximately 96.37% of observations fall between the scores of 30 and 65.
5. College Entrance Exam Scores
Finally, we address a college entrance exam distributed normally with a mean of 500 and a standard deviation of 100. We will calculate the Z scores and area percentages for the scores of 375 and 437:
- For a raw score of 375:
- Z = (375 - 500) / 100 = -1.25
- % Area Above = 1 - P(Z
- % Area Below = P(Z
- For a raw score of 437:
- Z = (437 - 500) / 100 = -0.63
- % Area Above = 1 - P(Z
- % Area Below = P(Z
In summary, the calculations and interpretations of Z scores provide valuable insights on performance in relation to the mean of different distributions, allowing for effective assessments in various contexts including corrections and education.
References
- Bluman, A.G. (2018). Elementary Statistics: A Step By Step Approach. McGraw-Hill Education.
- Hinton, P.R., Brownlow, C., McMurray, I., & Cozens, B. (2004). SPSS Explained. Routledge.
- Moore, D.S., & McCabe, G.P. (2006). Introduction to the Practice of Statistics. W.H. Freeman.
- De Veaux, R.D., Velleman, P.F., & Bock, D.E. (2016). Stats: Data and Models. Pearson.
- Field, A. (2013). Discovering Statistics Using IBM SPSS Statistics. SAGE Publications.
- Triola, M.F. (2018). Essentials of Statistics. Pearson.
- Tavakol, M., & Dennick, R. (2011). Making Sense of Cronbach's Alpha. International Journal of Medical Education, 2, 53-55.
- Schneider, G. (2011). Quantitative Analysis for Management. Pearson.
- Newman, A.J., & Cramer, D. (2010). Introduction to Statistics: A Problem Solving Approach. Wiley.
- Rice, J.A. (2006). Mathematical Statistics and Data Analysis. Cengage Learning.