1. Normal Distribution

2. Normal Distribution Curve A normal distribution curve is symmetrical, bell-shaped curve defined by the mean and standard deviation of a data set. The normal curve is a probability distribution with a total area under the curve of 1.

3. Standard Normal Distribution A standard normal distribution is the set of all z -scores. The mean of the data in a standard normal distribution is 0 and the standard deviation is 1.

4. Variance Variance is the average squared deviation from the mean of a set of data. It is used to find the standard deviation.

5. Variance 1. Find the mean of the data. Hint – mean is the average so add up the values and divide by the number of items. 5. Divide the total by the number of items. 4. Find the sum of the squares. 3. Square each deviation of the mean. 2. Subtract the mean from each value – the result is called the deviation from the mean.

6. Variance Formula The variance formula includes the Sigma Notation, which represents the sum of all the items to the right of Sigma. Mean is represented by and n is the number of items.

7. Standard Deviation Standard Deviation shows the variation in data. If the data is close together, the standard deviation will be small. If the data is spread out, the standard deviation will be large. Standard Deviation is often denoted by the lowercase Greek letter sigma,.

8. The bell curve which represents a normal distribution of data shows what standard deviation represents. One standard deviation away from the mean ( ) in either direction on the horizontal axis accounts for around 68 percent of the data. Two standard deviations away from the mean accounts for roughly 95 percent of the data with three standard deviations representing about 99 percent of the data.

9. Standard Deviation Formula The standard deviation formula can be represented using Sigma Notation: Notice the standard deviation formula is the square root of the variance.

10. Find the variance and standard deviation The math test scores of five students are: 92,88,80,68 and 52. 1) Find the mean: (92+88+80+68+52)/5 = 76. 2) Find the deviation from the mean: 92-76=16 88-76=12 80-76=4 68-76= -8 52-76= -24

11. 3) Square the deviation from the mean: Find the variance and standard deviation The math test scores of five students are: 92,88,80,68 and 52.

12. Find the variance and standard deviation The math test scores of five students are: 92,88,80,68 and 52. 4) Find the sum of the squares of the deviation from the mean: 256+144+16+64+576= 1056 5) Divide by the number of data items to find the variance: 1056/5 = 211.2

13. Find the variance and standard deviation The math test scores of five students are: 92,88,80,68 and 52. 6) Find the square root of the variance: Thus the standard deviation of the test scores is 14.53.

14. Standard Deviation A different math class took the same test with these five test scores: 92,92,92,52,52. Find the standard deviation for this class.

15. Solve: A different math class took the same test with these five test scores: 92,92,92,52,52. Find the standard deviation for this class.

16. The math test scores of five students are: 92,92,92,52 and 52. 1) Find the mean: (92+92+92+52+52)/5 = 76 2) Find the deviation from the mean: 92-76=16 92-76=16 92-76=16 52-76= -24 52-76= -24 4) Find the sum of the squares: 256+256+256+576+576= 1920 3) Square the deviation from the mean:

17. The math test scores of five students are: 92,92,92,52 and 52. 5) Divide the sum of the squares by the number of items : 1920/5 = 384 variance 6) Find the square root of the variance: Thus the standard deviation of the second set of test scores is 19.6.

18. Consider both sets of scores. Both classes have the same mean, 76. However, each class does not have the same scores. Thus we use the standard deviation to show the variation in the scores. With a standard variation of 14.53 for the first class and 19.6 for the second class, what does this tell us? Analyzing the data:

19. Class A: 92,88,80,68,52 Class B: 92,92,92,52,52 With a standard variation of 14.53 for the first class and 19.6 for the second class, the scores from the second class would be more spread out than the scores in the second class.

20. z -scores When a set of data values are normally distributed, we can standardize each score by converting it into a z -score. z -scores make it easier to compare data values measured on different scales.

21. z -scores A z -score reflects how many standard deviations above or below the mean a raw score is. The z -score is positive if the data value lies above the mean and negative if the data value lies below the mean.

22. z -score formula Where x represents an element of the data set, the mean is represented by and standard deviation by.

23. Analyzing the data Suppose SAT scores among college students are normally distributed with a mean of 500 and a standard deviation of 100. If a student scores a 700, what would be her z -score?

24. Analyzing the data Suppose SAT scores among college students are normally distributed with a mean of 500 and a standard deviation of 100. If a student scores a 700, what would be her z -score? Her z -score would be 2 which means her score is two standard deviations above the mean.

25. Analyzing the data: Class A: 92,88,80,68,52 Class B: 92,92,92,52,52 Class C: 77,76,76,76,75 Estimate the standard deviation for Class C. a) Standard deviation will be less than 14.53. b) Standard deviation will be greater than 19.6. c) Standard deviation will be between 14.53 and 19.6. d) Can not make an estimate of the standard deviation.

26. Class A: 92,88,80,68,52 Class B: 92,92,92,52,52 Class C: 77,76,76,76,75 Estimate the standard deviation for Class C. a) Standard deviation will be less than 14.53. Analyzing the data: Answer: A The scores in class C have the same mean of 76 as the other two classes. However, the scores in Class C are all much closer to the mean than the other classes so the standard deviation will be smaller than for the other classes.

27. Summary: As we have seen, standard deviation measures the dispersion of data. The greater the value of the standard deviation, the further the data tend to be dispersed from the mean.

28. Analyzing the data A set of math test scores has a mean of 70 and a standard deviation of 8. A set of English test scores has a mean of 74 and a standard deviation of 16. For which test would a score of 78 have a higher standing?

29. Analyzing the data To solve: Find the z -score for each test. A set of math test scores has a mean of 70 and a standard deviation of 8. A set of English test scores has a mean of 74 and a standard deviation of 16. For which test would a score of 78 have a higher standing? The math score would have the highest standing since it is 1 standard deviation above the mean while the English score is only.25 standard deviation above the mean.

30. Analyzing the data What will be the miles per gallon for a Toyota Camry when the average mpg is 23, it has a z value of 1.5 and a standard deviation of 2?

31. Analyzing the data What will be the miles per gallon for a Toyota Camry when the average mpg is 23, it has a z value of 1.5 and a standard deviation of 2? The Toyota Camry would be expected to use 26 mpg of gasoline. Using the formula for z -scores:

32. Normal Distribution Probability With a graphing calculator, we can calculate the probability of normal distribution data falling between two specific values using the mean and standard deviation of the data

33. Normal Distribution Probability A Calculus exam is given to 500 students. The scores have a normal distribution with a mean of 78 and a standard deviation of 5. What percent of the students have scores between 82 and 90? Example:

34. Normal Distribution Probability A Calculus exam is given to 500 students. The scores have a normal distribution with a mean of 78 and a standard deviation of 5. What percent of the students have scores between 82 and 90? Example: TI 83/84 directions: a. Press [2nd][VARS](DISTR) [2] (normalcdf) b. Press [82] [,] [90] [,] [78] [,] [5] [)][Enter] normalcdf(82,90, 78,5).2036578048 There is a 20.37% probability that a student scored between 82 and 90 on the Calculus exam. TI 83/84

35. Normal Distribution Probability A Calculus exam is given to 500 students. The scores have a normal distribution with a mean of 78 and a standard deviation of 5. How many students have scores between 82 and 90? Extension: Using the probability previously found: 500 *.2037 = 101.85 There are about 102 students who scored between 82 and 90 on the Calculus exam.

36. Normal Distribution Probability A Calculus exam is given to 500 students. The scores have a normal distribution with a mean of 78 and a standard deviation of 5. What percent of the students have scores above 70? Practice: Hint: Use 1E99 for upper limit; [2 nd ][,] on T I

37. Normal Distribution Probability A Calculus exam is given to 500 students. The scores have a normal distribution with a mean of 78 and a standard deviation of 5. How many students have scores above 70? Practice: Normal C.D prob=0.9452 Normalcdf(70,1E9 9,78,5).9452007106 TI 84 500*.9452= 472.6 About 473 students have a score above 70 on the Calculus exam.

38. Normal Distribution Probability Find the probability of scoring below a 1400 on the SAT if the scores are normal distributed with a mean of 1500 and a standard deviation of 200. Practice: Hint: Use -1E99 for lower limit; [2 nd ][,] on T I

39. Normal Distribution Probability Find the probability of scoring below a 1400 on the SAT if the scores are normal distributed with a mean of 1500 and a standard deviation of 200. Practice: Normalcdf(-1E99, 1400,1500,200).3085375322 TI 84 There is a 30.85% probability that a student will score below a 1400 on the SAT.