2. standard scores
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The document outlines the concepts of standard scores and z-scores, explaining how to convert a normal random variable to a standard normal variable and compute probabilities using the standard normal table. It provides various examples demonstrating how to calculate z-scores based on different scenarios, including examination scores and height distributions. Additionally, it includes a quiz to assess understanding of calculating probabilities based on given mean and standard deviation values.
2. standard scores
- 2. Learning Competencies The learner will be able to: 1. Convert a normal random variable to a standard normal variable and vice versa; and 2. Compute probabilities and percentiles using the standard normal table.
- 3. The standard score or z-score measure how many standard deviation a given value (x) is above or below the mean. The z-scores are useful in comparing observed values. A positive z-score indicates that the score or observed value is above the mean. Whereas a negative z-score indicates that the score or an observed value is below the mean.
- 4. The z-score is found by using the following equations. A. For Sample where z=standard score, x=raw score or observed value, =sample mean and s=sample standard deviation. B. For population where z=standard score, x= raw score or observed value, =population mean and =population standard
- 5. Example 1. On a final examination on Biology, the mean was 75 and the standard deviation was 12. Determine the standard score of a student who received a score of 60 assuming that the scores are normally distributed and sketch the graph.
- 6. Solution: This indicates that 60 is 1.25 standard deviation below the mean
- 7. Example 2. On the first periodic exam in Statistics, the population mean was 70 and the population standard deviation was 9. Determine the standard score of a student who got a score of 88 assuming that the scores are normally distributed.
- 8. Solution: This indicates that 88 is 2 standard deviations above the mean.
- 9. Example 3. Luz scored 90 in an English test and 70 in a Physics test. Scores in the English test have a mean of 80 and a standard deviation of 10. Scores in the Physics test have a mean of 60 and standard deviation of 8. In which subject was her standing better assuming that the scores in her English and Physics class are normally distributes?
- 10. Solution: For English For Physics Her standing in physics was better than her standing in English. Her score in English was one standard deviation above the mean of the scores in English whereas in Physics, her score was 1.25 standard deviation above the mean.
- 11. Example 4. In a Science test, the mean score is 42 and the standard deviation is 5. Assuming the scores are normally distributed, what percent of the score is a.) greater than 48? b.) less than 50? c.) between 30 and 48?
- 12. Solution: From the table, A1=0.3849 A=A2-A1 =0.5-0.3849=0.1151 Hence, 11.51% of the scores is greater than 48.
- 13. From the table, A2=0.4452 A=A1+A2 =0.5-0.4452=0.9452 or 94.52% Hence, 94.52% is less than 50
- 14. From the table, A1=0.4918 and A2=0.3849 A=A1+A2=0.4918+0.3849=0.8767 or 87.67% Hence, 87.67% of the scores are between 30 and 48.
- 15. Example 5. The mean height of grade 9 students at a certain high school is 164 centimeters and the standard deviation is 10 centimeters. Assuming the heights are normally are normally distributed, what percent of the heights is greater than 168 centimeters? Solution
- 16. From the table, A1=0.1554 A=A2-A1=0.5-0.1554=0.3446 Hence, 34.46% of the heights are greater than 168 centimetres.
- 17. Quiz: In a math test, the mean score is 45 and the standard deviation is 4. Assuming normality, what is the probability that a score picked random will lie a. Above score 50? b. Below score 38?