2. SHOWTIME!

3. STATISTICAL TOOLS FOR EVALUATION THE NORMAL CURVE AND PROBABILITY

4. THE NORMAL CURVE SMOOTH, BILATERALLY SYMMETRICAL CURVE CENTERED AROUND A POINT THAT IS SIMULTANEOUSLY THE MODE, MEDIAN, AND MEAN HENCE, THE CENTER POINT IS BOTH THE MOST FREQUENT SCORE AND THE SCORE BELOW AND ABOVE WHICH HALF THE SCORES FALL

5. THE NORMAL CURVE THE NORMAL CURVE HAS A MEAN = 0 AND A STANDARD DEVIATION = 1 THE NORMAL CURVE IS THE GRAPH OF AN INFINITE NUMBER OF Z- SCORES TO USE THE NORMAL CURVE TO MAKE PROBABILITY STATEMENTS, THINK OF THE AREA UNDER THE CURVE AS 100 EQUAL PORTIONS 50 LIE ON EACH SIDE OF THE MEAN

6. THE NORMAL CURVE AND PROBABILITY

7. WHAT IS THE PROBABILITY OF A Z SCORE EQUAL TO OR GREATER THAN 0? P (Z > 0 ) = 50 / 100 = 1/2 OR 50%

8. THE NORMAL CURVE AND PROBABILITY WHAT PERCENTAGE OF THE AREA UNDER THE NORMAL CURVE LIES BETWEEN O (Z = 0) AND 1.36 (Z = 1.36)?

9. THE NORMAL CURVE AND PROBABILITY

10. WHAT PERCENTAGE OF THE AREA UNDER THE NORMAL CURVE LIES BETWEEN O (Z = 0) AND 1.36 (Z = 1.36)? USING TABLE 2.9, Z OF 0 = 0 AND Z OF 1.36 = 41.31 P (Z BETWEEN 0 AND 1.36) = 41.31 / 100 P =.4131 OR 41%

11. THE NORMAL CURVE AND PROBABILITY WHAT IS THE PROBABILITY THAT Z IS EQUAL TO OR GREATER THAN 1.03?

12. THE NORMAL CURVE AND PROBABILITY

13. WHAT IS THE PROBABILITY THAT Z IS EQUAL TO OR GREATER THAN 1.03? P (Z ≥ 0) = 50 / 100 0R.50 P (Z BETWEEN 0 AND 1.03) = 34.85 / 100 =.3485 P (Z > 1.03) =.50 -.3485 =.1515 =.15 OR 15%

14. THE NORMAL CURVE AND PROBABILITY A TEACHER ALWAYS ADMINISTERS 100-POINT TESTS AND ALWAYS GIVES A’S TO SCORES OF 93 AND ABOVE. ON THE LAST EXAM THE MEAN WAS 72 AND THE STANDARD DEVIATION WAS 9. ASSUMING THAT THE TEST SCORES WERE NORMALLY DISTRIBUTED, WHAT WAS THE PROBABILITY OF RECEIVING AN A ON THAT TEST? FIRST NEED TO CALCULATE THE Z-SCORE Z = (SCORE - MEAN) / STANDARD DEVIATION Z = (93 - 72) / 9 = 21 / 9 = 2.33

15. THE NORMAL CURVE AND PROBABILITY

16. A TEACHER ALWAYS ADMINISTERS 100-POINT TESTS AND ALWAYS GIVES A’S TO SCORES OF 93 AND ABOVE. ON THE LAST EXAM THE MEAN WAS 72 AND THE STANDARD DEVIATION WAS 9. ASSUMING THAT THE TEST SCORES WERE NORMALLY DISTRIBUTED, WHAT WAS THE PROBABILITY OF RECEIVING AN A ON THAT TEST? P (Z ≥ 0) = 50 / 100 0R.50 P (Z BETWEEN 0 AND 2.33) = 49.01 / 100 =.4901 P (Z > 2.33) =.50 -.4901 =.0099 =.01 OR 1%

17. THE NORMAL CURVE AND PROBABILITY TO DEVELOP SOME PERFORMANCE STANDARDS, A TEACHER DECIDES TO USE THE NORMAL CURVE TO DETERMINE THE TEST SCORE ABOVE WHICH 7% OF THE SCORES SHOULD FALL

18. THE NORMAL CURVE AND PROBABILITY TO DEVELOP SOME PERFORMANCE STANDARDS, A TEACHER DECIDES TO USE THE NORMAL CURVE TO DETERMINE THE TEST SCORE ABOVE WHICH 7% OF THE SCORES SHOULD FALL FIRST DETERMINE THE Z-SCORE ABOVE WHICH 7% OF THE AREA UNDER THE NORMAL CURVE FALLS P (Z > ?) = 7 / 100 50% OF TEST SCORES LIE ABOVE THE MEAN WHERE Z = 0 IF WE ARE LOOKING FOR THE TOP 7%, 43% OF THE TEST SCORES LIE BETWEEN THE MEAN AND THE SCORE FOR WHICH THE TEACHER IS LOOKING USING TABLE 2.9, WHAT Z-SCORE CORRESPONDS CLOSEST TO 43%? Z-SCORE OF 1.48 CORRESPONDING TO.4306 LIES THE CLOSET TO.43

19. THE NORMAL CURVE AND PROBABILITY

20. THEN DETERMINE THE TEST SCORE (X) ABOVE WHICH 7% OF THE SCORES SHOULD FALL USING THE FOLLOWING FORMULA TO CALCULATE X: X = MEAN + Z(STANDARD DEVIATION) ASSUMING THE MEAN IS 31.25, STANDARD DEVIATION IS 5, AND THE CALCULATED Z-SCORE = 1.48: X = 31.25 + 1.48(5) = 31.25 + 7.4 = 38.65

21. QUESTIONS OR COMMENTS??

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