1. 11/7/2005 1 Statistics: Normal Curve CSCE 115

2. 11/7/2005 2 Normal distribution The bell shaped curve The bell shaped curve Many physical quantities are distributed in such a way that their histograms can be approximated by a normal curve Many physical quantities are distributed in such a way that their histograms can be approximated by a normal curve

3. 11/7/2005 3 Examples Examples of normal distributions: Examples of normal distributions: Height of 10 year old girls and most other body measurements Height of 10 year old girls and most other body measurements Lengths of rattle snakes Lengths of rattle snakes Sizes of oranges Sizes of oranges Distributions that are not normal: Distributions that are not normal: Flipping a coin and count of number of heads flips before getting the first tail Flipping a coin and count of number of heads flips before getting the first tail Rolling 1 die Rolling 1 die

4. 11/7/2005 4 Flipping Coins Experiment Experiment: Flip a coin 10 times. Count the number of heads Experiment: Flip a coin 10 times. Count the number of heads Expected results: If we flip a coin 10 times, on the average we would expect 5 heads. Expected results: If we flip a coin 10 times, on the average we would expect 5 heads. Because tossing a coin is a random experiment, the number of heads may be more or less than expected. Because tossing a coin is a random experiment, the number of heads may be more or less than expected.

5. 11/7/2005 5 Flipping Coins Experiment Carry out the experiment Carry out the experiment But the average was supposed to be 5 But the average was supposed to be 5 What can we do to improve the results? What can we do to improve the results? How many times do we have to carry out the experiment to get good results? How many times do we have to carry out the experiment to get good results? Faster: Use a computer simulation Faster: Use a computer simulation

6. 11/7/2005 6 Experiment: Flipping coins

7. 11/7/2005 7 Experiment: Flipping coins

8. 11/7/2005 8 Experiment: Flipping coins

9. 11/7/2005 9 Experiment: Flipping coins

10. 11/7/2005 10 Theory If the number of trials of this type increase, the histogram begins to approximate a normal curve If the number of trials of this type increase, the histogram begins to approximate a normal curve

11. 11/7/2005 11 Normal Curve One can specify mean and standard deviation One can specify mean and standard deviation The shape of the curve does not depend on mean. The curve moves so it is always centered on the mean The shape of the curve does not depend on mean. The curve moves so it is always centered on the mean If the st. dev. is large, the curve is lower and fatter If the st. dev. is large, the curve is lower and fatter If the st. dev, is small, the curve is taller and skinnier If the st. dev, is small, the curve is taller and skinnier

12. 11/7/2005 12 Normal Curve The normal curve with mean = 0 and std. dev. = 1 is often referred to as the z distribution

13. 11/7/2005 13 Normal Curve

14. 11/7/2005 14 Normal Curve

15. 11/7/2005 15 Normal Curve

16. 11/7/2005 16 Normal Curve 50% of the area is right of the mean 50% of the area is right of the mean 50% of the area is left of the mean 50% of the area is left of the mean

17. 11/7/2005 17 Normal Curve The total area under the curve is always 1. The total area under the curve is always 1. 68% (about 2/3) of the area is between 1 st. dev. left of the mean to 1 st. dev. right of the mean. 68% (about 2/3) of the area is between 1 st. dev. left of the mean to 1 st. dev. right of the mean. 95% of the area is between 2 st. dev. left of the mean to 2 st. dev. right of the mean. 95% of the area is between 2 st. dev. left of the mean to 2 st. dev. right of the mean.

18. 11/7/2005 18 Normal Curve 99.7% 95.4% 68.2% 50%

19. 11/7/2005 19 Example: Test Scores Several hundred students take an exam. The average is 70 with a standard deviation of 10. Several hundred students take an exam. The average is 70 with a standard deviation of 10. About 2/3 of the scores are between 60 and 80 About 2/3 of the scores are between 60 and 80 95% of the scores are between 50 and 90 95% of the scores are between 50 and 90 About 50% of the students scored above 70 About 50% of the students scored above 70

20. 11/7/2005 20 Example: Test Scores

21. 11/7/2005 21 Example: Test Scores Suppose that passing is set at 60. What percent of the students would be expected to pass? Suppose that passing is set at 60. What percent of the students would be expected to pass? Solution: (60 - 70)/10 = -1. Hence passing is 1 standard deviation below the mean. Solution: (60 - 70)/10 = -1. Hence passing is 1 standard deviation below the mean. 34.1% of the scores are between 60 and 70 34.1% of the scores are between 60 and 70 50% of the scores are above 70 50% of the scores are above 70 84.1% of students would expected to pass 84.1% of students would expected to pass

22. 11/7/2005 22 Example: Test Scores Alternate solution Suppose that passing is set at 60. What percent of the students would be expected to pass? Suppose that passing is set at 60. What percent of the students would be expected to pass? Solution: According to our charts, 15.9% of the scores will be less than 60 (less than –1 standard deviations below the mean). Solution: According to our charts, 15.9% of the scores will be less than 60 (less than –1 standard deviations below the mean). 100% – 15.9% = 84.1% of the students pass because they are right of the –1 st. dev. line. 100% – 15.9% = 84.1% of the students pass because they are right of the –1 st. dev. line.

23. 11/7/2005 23 Grading on the Curve Assumes scores are normal Assumes scores are normal Grades are based on how many standard deviations the score is above or below the mean Grades are based on how many standard deviations the score is above or below the mean The grading curve is determined in advanced The grading curve is determined in advanced

24. 11/7/2005 24 Example: Grading on the Curve A 1 st. dev. or greater above the mean A 1 st. dev. or greater above the mean B From the mean to one st. dev. above the mean B From the mean to one st. dev. above the mean C From one st. dev. below mean to mean C From one st. dev. below mean to mean D From two st. dev below mean to one st. dev. below mean D From two st. dev below mean to one st. dev. below mean F More than two st. dev. below mean F More than two st. dev. below mean Selecting these breaks is completely arbitrary. One could assign other grade break downs.

25. 11/7/2005 25

26. 11/7/2005 26 Example: Grading on the Curve Range (in st. dev.) Expected percent Grade from to of scores A +1 50% - 34.1% = 15.9% B 0 +1 34.1% C -1 0 34.1% D -2 -1 13.6% F -2 50% - 34.1 - 13.6% = 2.3% Range (in st. dev.) Expected percent Grade from to of scores A +1 50% - 34.1% = 15.9% B 0 +1 34.1% C -1 0 34.1% D -2 -1 13.6% F -2 50% - 34.1 - 13.6% = 2.3%

27. 11/7/2005 27 Example: Grading on a Curve Assume that the mean is 70, st. dev. is 10 Assume that the mean is 70, st. dev. is 10 Joan scores 82. What is her grade? (82-70)/10 = 12/10 = 1.2 She scored 1.2 st. dev. above the mean. She gets an A Joan scores 82. What is her grade? (82-70)/10 = 12/10 = 1.2 She scored 1.2 st. dev. above the mean. She gets an A

28. 11/7/2005 28 Example: Grading on a Curve Assume that the mean is 70, st. dev. 10 Assume that the mean is 70, st. dev. 10 Tom scores 55. What is his grade? (55 - 70)/10 = -15/10 = -1.5 He scored 1.5 st. dev. below the mean. He gets a D Tom scores 55. What is his grade? (55 - 70)/10 = -15/10 = -1.5 He scored 1.5 st. dev. below the mean. He gets a D

29. 11/7/2005 29 Example: Light bulbs Assume that the life span of a light bulb is normally distributed with a mean of 1000 hours and a standard deviation of 100. The manufacturer guaranties that the bulbs will last 800 hours. What percent of the light bulbs will fail before the guarantee is up? Assume that the life span of a light bulb is normally distributed with a mean of 1000 hours and a standard deviation of 100. The manufacturer guaranties that the bulbs will last 800 hours. What percent of the light bulbs will fail before the guarantee is up?

30. 11/7/2005 30 Example: Light bulbs Solution: (800-1000)/100 = -2 Solution: (800-1000)/100 = -2 The percentage of successes is 13.6% + 34.1% + 50% = 97.7% The percentage of successes is 13.6% + 34.1% + 50% = 97.7% The percentage of failures is 100% - 97.7% = 2.3% The percentage of failures is 100% - 97.7% = 2.3%

31. 11/7/2005 31 Example: Mens socks Assume that the length of men’s feet are normally distributed with a mean of 12 inches and a standard deviation of.5 inch. A certain brand of “one size fits all” strechy socks will fit feet from 11 inches to 13.5 inches. What % of all men cannot wear the socks? Assume that the length of men’s feet are normally distributed with a mean of 12 inches and a standard deviation of.5 inch. A certain brand of “one size fits all” strechy socks will fit feet from 11 inches to 13.5 inches. What % of all men cannot wear the socks?

32. 11/7/2005 32 Example: Mens socks (11-12)/.5 = -2 (11-12)/.5 = -2 (13.5 – 12)/.5 = 3 (13.5 – 12)/.5 = 3 Hence men from 2 standard deviations below the mean and 3 standard deviations above the mean can wear the socks Hence men from 2 standard deviations below the mean and 3 standard deviations above the mean can wear the socks -2 to +2 st. dev. 95.4% +2 to +3 st. dev. 2.2% -2 to + 3 st. dev. 97.6% of men can wear the socks. 100% - 97.6% = 2.4% cannot wear them.

33. 11/7/2005 33 Normal Distributions We might be interested in: We might be interested in: The normal probability (or frequency) The normal probability (or frequency) The cumulative area below the curve The cumulative area below the curve

34. 11/7/2005 34 Normal Curves

35. 11/7/2005 35 Normal Curves

36. 11/7/2005 36 Excel and the Normal Distribution Cumulative probabilities are measured from “- infinity”. Cumulative probabilities are measured from “- infinity”. NORMDIST(x, mean, st dev, cumulative) cumulative - true cumulative false probability NORMDIST(x, mean, st dev, cumulative) cumulative - true cumulative false probability NORMINV(probability, mean, st dev) returns the x value that gives the specified cumulative probability NORMINV(probability, mean, st dev) returns the x value that gives the specified cumulative probability

37. 11/7/2005 37 Excel and the Normal Distribution

38. 11/7/2005 38 Excel and the Normal Distribution The following functions assume mean = 0 and st. dev. = 1 The following functions assume mean = 0 and st. dev. = 1 NORMSDIST(x) returns cumulative distribution NORMSDIST(x) returns cumulative distribution NORMSINV(probability) returns the x value that obtains the specify cumulative probability NORMSINV(probability) returns the x value that obtains the specify cumulative probability

39. 11/7/2005 39 Example: Excel Assume mean = 12, st. dev. =.5 and the men’s foot distribution used earlier. Assume mean = 12, st. dev. =.5 and the men’s foot distribution used earlier. What percent of the men’s feet would be expected to be 13 inches long or shorter? =NORMDIST(13, 12,.5, TRUE) =.977 or 97.7% What percent of the men’s feet would be expected to be 13 inches long or shorter? =NORMDIST(13, 12,.5, TRUE) =.977 or 97.7% What is the probability that a randomly chosen foot would be exactly 13 inches long? =NORMDIST(13, 12,.5, FALSE) =.108 What is the probability that a randomly chosen foot would be exactly 13 inches long? =NORMDIST(13, 12,.5, FALSE) =.108

40. 11/7/2005 40 Example: Excel Assume mean = 70, st. dev. = 10 and the grade distribution used earlier. Assume mean = 70, st. dev. = 10 and the grade distribution used earlier. What percent of the students would be expected to score 90 or below? =NORMDIST(90, 70, 10, TRUE) =.977 or 97.7% What percent of the students would be expected to score 90 or below? =NORMDIST(90, 70, 10, TRUE) =.977 or 97.7% What is the probability that a randomly chosen person scores (exactly) 90? Because this is a discrete distribution =NORMDIST(90.5, 70, 10, TRUE) -NORMDIST(89.5, 70, 10, TRUE) =.0054 What is the probability that a randomly chosen person scores (exactly) 90? Because this is a discrete distribution =NORMDIST(90.5, 70, 10, TRUE) -NORMDIST(89.5, 70, 10, TRUE) =.0054

41. 11/7/2005 41 Example: Excel Assume the grade distribution used earlier with a mean of 70 and st. dev. of 10. What percentage of the students are expected to pass with a D or C? (That is, with scores between 50 and 70)? Passing with a D 70 - 2 * 10 = 50 Passing with a B- 70 We are looking for 50 <= scores < 70 = NORMDIST(70, 70, 10, TRUE) - NORMDIST(50, 70, 10, TRUE) =.5 - NORMDIST(50, 70, 10, TRUE) =.477 or 47.7% Assume the grade distribution used earlier with a mean of 70 and st. dev. of 10. What percentage of the students are expected to pass with a D or C? (That is, with scores between 50 and 70)? Passing with a D 70 - 2 * 10 = 50 Passing with a B- 70 We are looking for 50 <= scores < 70 = NORMDIST(70, 70, 10, TRUE) - NORMDIST(50, 70, 10, TRUE) =.5 - NORMDIST(50, 70, 10, TRUE) =.477 or 47.7%

42. 11/7/2005 42 Example: Excel

43. 11/7/2005 43 Example: Excel What score does one need to be in the top 10%? To be in the top 10%, the student must do better than 100% - 10% = 90% of the students. =NORMINV(90%, 70, 10) = 82.8 (or 83) What score does one need to be in the top 10%? To be in the top 10%, the student must do better than 100% - 10% = 90% of the students. =NORMINV(90%, 70, 10) = 82.8 (or 83)

44. 11/7/2005 44 Example: Excel

45. 11/7/2005 45