1. 40S Applied Math Mr. Knight – Killarney School Slide 1 Unit: Statistics Lesson: ST-4 Standard Scores and Z-Scores Standard Scores and Z-Scores Learning Outcome B-4 ST-L4 Objectives: To use the Standard Normal Distribution to calculate probabilities.

2. 40S Applied Math Mr. Knight – Killarney School Slide 2 Unit: Statistics Lesson: ST-4 Standard Scores and Z-Scores Professor Adams has 140 students who wrote a statistics test. If the marks are approximately normally distributed: how many students should have a 'B' mark (i.e., 70 to 79 percent)? how many students should have failed (i.e., less than 50 percent)? how high must she set the mark for an 'A' if she wants 5 percent of the students to get an A? how high must she set the passing mark if she wants only the top 75 percent of the marks to be passing marks? In this lesson, you will use IT to solve problems similar to the ones mentioned above. Theory – Intro

3. 40S Applied Math Mr. Knight – Killarney School Slide 3 Unit: Statistics Lesson: ST-4 Standard Scores and Z-Scores Theory – The Normal Distribution The graph below represents the marks of a large number of students where the mean mark is 69.3 percent and the standard deviation is 7 percent, and the distribution of marks is approximately normal. We know that if the marks are approximately normally distributed, then approximately: 68 percent of the marks are between 62.3 percent and 76.3 percent (i.e.,  ± 1  ) 34% of the marks are between 69.3 percent and 76.3 percent (i.e., between  and  + 1  ) 50% of the marks are below 69.3 percent (i.e., less than  ) 16% of the marks are above 76.3 percent (i.e., greater than  + 1  )

4. 40S Applied Math Mr. Knight – Killarney School Slide 4 Unit: Statistics Lesson: ST-4 Standard Scores and Z-Scores Theory – Case 1(a): Calculate the Percentage of Scores Between Two Given Scores (Case 1) If we know the mean and standard deviation of a normal distribution, we can find what percentage of the scores lie between any two given scores. The example from the previous page is repeated here: The mean mark for a large number of students is 69.3 percent with a standard deviation of 7 percent. What percent of the students have a 'B' mark (i.e., 70 percent to 79 percent)? Assume that the marks are normally distributed. Solution using Winstats: 1. Open Winstats. 2. Select Window, then Probability, and then Normal. A normal curve should appear in a separate window. 3. In the new window, select Edit > Parameters, and enter values for mean (69.3), st dev (7). 4. In the new window, select Calc > Probabilities and enter values for lo x (70), hi x (79). Click on the Probability button. 5. The answer: prob = 0.37726 indicates that the probability of any mark randomly selected being 70% to 79% is 0.377.

5. 40S Applied Math Mr. Knight – Killarney School Slide 5 Unit: Statistics Lesson: ST-4 Standard Scores and Z-Scores Theory – Case 1(b): Calculate the Percentage of Passing Scores Another example: The mean mark for a large number of students is 69.3 percent with a standard deviation of 7 percent. What percent of the students have a passing mark if they must get 60 percent or better to pass? Assume that the marks are normally distributed. Follow the same procedure as on the previous page. The only new procedure is to write the value for hi x at least four standard deviations above the mean, because there is no upper limit. Therefore, P(passing), or P(x > 60%) = 0.908 Therefore, 90.8 percent of the students are expected to pass.

6. 40S Applied Math Mr. Knight – Killarney School Slide 6 Unit: Statistics Lesson: ST-4 Standard Scores and Z-Scores Theory – Case 1(c): Calculate the Number of Scores

7. 40S Applied Math Mr. Knight – Killarney School Slide 7 Unit: Statistics Lesson: ST-4 Standard Scores and Z-Scores The Standard Normal Distribution The normal distributions used up to this point have all been non- standard normal distributions. This means that the mean and standard deviation of the distribution are the mean and standard deviation of the data being studied. In a standard normal distribution, the scale on the x-axis is the z- score (a standard score) where 'z = 0' is the mean, and the standard deviation is '1'. The distribution is a probability distribution where the area under the curve = 1. This means there is a 100 percent chance of every score being included in this distribution. Other probabilities are illustrated below.

8. 40S Applied Math Mr. Knight – Killarney School Slide 8 Unit: Statistics Lesson: ST-4 Standard Scores and Z-Scores Theory - Case 2(a): Calculate the Percentage of Scores Between Two Z-Scores (Case 2).If we know two z-scores of a standard normal distribution, we can find the percentage of scores that lie between them. The procedure using Winstats is the same as in the previous examples, just with a mean of 0 and  of 1. Sample questions: What percent of scores lie between z = 0.87 and z = 2.57?OR What is the probability that a score will fall between z = 0.87 and z = 2.57? OR Find the area between z = 0.87 and z = 2.57 in a standard normal distribution. Answer: P(0.87 < z < 2.57) = 0.187 This means that 18.7 percent of the scores lie between z = 0.87 and z = 2.57.

9. 40S Applied Math Mr. Knight – Killarney School Slide 9 Unit: Statistics Lesson: ST-4 Standard Scores and Z-Scores Case 2(b): Calculate the Percentage of Scores Between Two Z-Scores P(x<0.75) = 77.3%

10. 40S Applied Math Mr. Knight – Killarney School Slide 10 Unit: Statistics Lesson: ST-4 Standard Scores and Z-Scores Case 3(a): Find the Z-Score that Corresponds to a Given Probability (Case 3): If we know the probability of an event, we can find the z-score that corresponds to this probability. This is the reverse of what we did in Case 2. Sample question: What is the z-score if the probability of getting more than this z-score is 0.350? Procedure using Winstats. 1. Open Winstats. 2. Select Window, then Probability, and then Normal. A new window with a normal distribution curve should open. Set mean to 0 and  to 1. 3. Select Calc > Probabilities. In the significance box, type the probability (0.35), and click critical value. A z-score should appear. Answer: The z-score is: z = 0.385

11. 40S Applied Math Mr. Knight – Killarney School Slide 11 Unit: Statistics Lesson: ST-4 Standard Scores and Z-Scores Case 3(b): Find the Z-Value that Corresponds to a Given Probability Note about Winstats: Winstats gives a z-score that represents the area greater than the z- score we are looking for. If the problem asks for a probability less than the given z-score, you must change the sign of the z-score given by Winstats. Sample question: What is the z-score if the probability of getting less than this z-score is 0.750? Answer: The z-value is shown as: z = -0.67448, and since the probability is less than the desired z-value, the answer is rounded to: z = 0.675 Note: To draw the normal distribution shaded properly (as shown in the diagram above), you need to use the z-score as your hi x setting. Winplot calculates this value You convert to get this value

12. 40S Applied Math Mr. Knight – Killarney School Slide 12 Unit: Statistics Lesson: ST-4 Standard Scores and Z-Scores Sample Problem: Income from Oranges An orange producer who calls himself Doctor Juice grows an exclusive variety of oranges which are sorted into three categories and sold at different prices. The diameters of the oranges are distributed normally with a mean of 84 mm and a standard deviation of 12 mm. Answer each question that follows, and then check your answer.

13. 40S Applied Math Mr. Knight – Killarney School Slide 13 Unit: Statistics Lesson: ST-4 Standard Scores and Z-Scores Sample Problem: Income from Oranges

14. 40S Applied Math Mr. Knight – Killarney School Slide 14 Unit: Statistics Lesson: ST-4 Standard Scores and Z-Scores Sample Problem: Income from Oranges x =  + z  = 84 x 1.18(12) = 98 Non-Standard Normal Distribution Standard Normal Distribution

15. 40S Applied Math Mr. Knight – Killarney School Slide 15 Unit: Statistics Lesson: ST-4 Standard Scores and Z-Scores Sample Problem: Income from Oranges