1. 9.2.2016 2.2 Normal Distributions

2. Homework Answers 1. a) Her percentile is.25, or the 25 th percentile. 25% of girls have fewer pairs of shoes b) His percentile is.85, or the 85 th percentile. 85% of boys have fewer pairs of shoes c) The boy is more unusual, because he is further from the 50 th percentile (or the median). He is 35 percentiles away from the 50 th, with only 15% being more extreme than he is, while the girl is 25 percentiles away from the 50 th, with 25% being more extreme than she is.

3. 5. The girl is taller than 78% of girls her age, but weighs more than only 48%. She is therefore probably fairly thin. 9. a) Since the first quartile is the 25 th percentile, you would find 25 on the y- axis, and find that your first quartile is about $19. Then find the 75 th percentile (third quartile) using the same method—it is about $50. So we know that our IQR is about $31 (50-19) b) Approximately the 25 th percentile c)

4. 10. a) To find the 60 th percentile, find 60 on the y-axis, and find the x-value where y=60. It is approximately 1,000 hours b) Approximately the 35 th percentile 11. For Eleanor, z=1.8. Gerald’s z-score is 1.5. Therefore, Eleanor has the higher score. 12. For Cobb, z=4.15. For Williams, z=4.26, and for Brett, z=4.07. All three hitters were more than 4 standard deviations above the mean, but Williams had the best season. 13. a) Judy’s bone density is about one and one half standard deviations below the mean for women her age. The negative value indicates that it is below average, and the magnitude of the value indicates how many standard deviations above/below the mean it is. b) 5.52 grams/cm^2

5. 14. a) Mary’s score (0.5) indicates that her bone density is about half of a standard deviation above average for women her age. Even though the two bone density scores are exactly the same, Mary is 10 years older, so her bone density is better in comparison to the women her age. b) 8 grams/cm^2 15. a) Since 22 salaries were less than Lidge’s salary, his salary was at the 75.86 percentile. b) Lidge’s salary was.79 standard deviations above the mean salary of $3,388,617

6. 19. a) The mean and the median both increase by 18, so the mean is 87.188, and the median is 87.5. The distribution of heights just shifts by 18 inches. b) The standard deviation and IQR do not change 20. a) The mean and median salaries will each increase by $1000 (the distribution of salaries just shifts by $1000) b) The extremes and quartiles will also each increase by $1000. The standard deviation will not change. 21. a) To give the heights in feet, not inches, we need to divide each observation by 12 (12 inches=1 foot). Thus, the mean and the median are divided by 12. The new mean is 5.77 feet, and the new median is 5.79 feet. b) We simply divide the old standard deviation by 12 to get.27 feet. Similarly, we divide the old first and third quartiles by 12 to get 5.65 feet for Q1 and 5.92 feet for Q3, leaving us with an IQR of.27 feet

7. 22. a) The mean and the median will each increase by 5% b) The 5% raise will increase the distance of the quartiles from the median. The quartiles and the standard deviation will each increase by 5% (the original values multiplied by 1.05) 23. Mean in degrees Fahrenheit is 77. Standard deviation in degrees Fahrenheit is 3.6 31. a) Mean is C, median is B b) Mean is B, median is B

8. 33. C 34. B 35. C 36. B 37. D 38. E

9. New Material Density Curves and Normal Distributions

10. + Describing Location in a Distribution Density Curves In Chapter 1, we developed a kit of graphical and numericaltools for describing distributions. Now, we ’ ll add one more step to the strategy. So far our strategy for exploring data is : 1. Graph the data to get an idea of the overall pattern 2. Calculate an appropriate numerical summary todescribe the center and spread of the distribution. Sometimes the overall pattern of a large number of observations is so regular, that we can describe it by asmooth curve, called a density curve.

11. + Describing Location in a Distribution Density Curve Definition: A density curve is a curve that is always on or above the horizontal axis, and has area exactly 1 underneath it. A density curve describes the overall pattern of a distribution. The area under the curve and above any interval of values on the horizontal axis is the proportion of all observations that fall in that interval. The overall pattern of this histogram of the scores of all 947 seventh-grade students in Gary, Indiana, on the vocabulary part of the Iowa Test of Basic Skills (ITBS) can be described by a smooth curve drawn through the tops of the bars.

12. + Here’s the idea behind density curves:

13. + Density Curves NOTE: No real set of data is described by a density curve. It is simply an approximation that is easy to use and accurateenough for practical use. Describing Density Curves The median of the density curve is the “equal-areas” point. The mean of the density curve is the “balance” point.

14. Section 2.2 Normal Distributions Learning Objectives After this section, you should be able to… DESCRIBE and APPLY the 68-95-99.7 Rule DESCRIBE the standard Normal Distribution PERFORM Normal distribution calculations ASSESS Normality

15. Normal Distributions One particularly important class of density curves are the Normal curves, which describe Normal distributions. All Normal curves are symmetric, single-peaked, and bell-shaped A Specific Normal curve is described by giving its mean µ and standard deviation σ. Two Normal curves, showing the mean µ and standard deviation σ.

16. Normal Distributions Definition: A Normal distribution is described by a Normal density curve. Any particular Normal distribution is completely specified by two numbers: its mean µ and standard deviation σ. The mean of a Normal distribution is the center of the symmetric Normal curve. The standard deviation is the distance from the center to the change-of-curvature points on either side. We abbreviate the Normal distribution with mean µ and standard deviation σ as N(µ, σ ). Normal distributions are good descriptions for some distributions of real data. Normal distributions are good approximations of the results of many kinds of chance outcomes. Many statistical inference procedures are based on Normal distributions.

17. Normal Distributions Although there are many Normal curves, they all have properties in common. The 68-95-99.7 Rule Definition: The 68-95-99.7 Rule (“The Empirical Rule”) In the Normal distribution with mean µ and standard deviation σ : Approximately 68% of the observations fall within σ of µ. Approximately 95% of the observations fall within 2 σ of µ. Approximately 99.7% of the observations fall within 3 σ of µ.

18. Normal Distributions The distribution of Iowa Test of Basic Skills (ITBS) vocabulary scores for 7 th grade students in Gary, Indiana, is close to Normal. Suppose the distribution is N (6.84, 1.55). a) Sketch the Normal density curve for this distribution. b) What percent of ITBS vocabulary scores are less than 3.74? c) What percent of the scores are between 5.29 and 9.94? Example, p. 113

19. Normal Distributions The Standard Normal Distribution All Normal distributions are the same if we measure in units of size σ from the mean µ as center. Definition: The standard Normal distribution is the Normal distribution with mean 0 and standard deviation 1. If a variable x has any Normal distribution N(µ, σ ) with mean µ and standard deviation σ, then the standardized variable has the standard Normal distribution, N(0,1).

20. Normal Distributions The Standard Normal Table Because all Normal distributions are the same when we standardize, we can find areas under any Normal curve from a single table. Definition: The Standard Normal Table Table A is a table of areas under the standard Normal curve. The table entry for each value z is the area under the curve to the left of z. Z.00.01.02 0.7.7580.7611.7642 0.8.7881.7910.7939 0.9.8159.8186.8212 P(z < 0.81) =.7910 Suppose we want to find the proportion of observations from the standard Normal distribution that are less than 0.81. We can use Table A:

21. Normal Distributions Finding Areas Under the Standard Normal Curve Find the proportion of observations from the standard Normal distribution that are between -1.25 and 0.81. Example, p. 117 Can you find the same proportion using a different approach? 1 - (0.1056+0.2090)= 1 – 0.3146 = 0.6854

22. Normal Distributions Normal Distribution Calculations State: Express the problem in terms of the observed variable x. Plan: Draw a picture of the distribution and shade the area of interest under the curve. Do: Perform calculations. Standardize x to restate the problem in terms of a standard Normal variable z. Use Table A and the fact that the total area under the curve is 1 to find the required area under the standard Normal curve. Conclude: Write your conclusion in the context of the problem.

23. Normal Distributions Normal Distribution Calculations When Tiger Woods hits his driver, the distance the ball travels can be described by N(304, 8). What percent of Tiger’s drives travel between 305 and 325 yards? Using Table A, we can find the area to the left of z=2.63 and the area to the left of z=0.13. 0.9957 – 0.5517 = 0.4440. About 44% of Tiger’s drives travel between 305 and 325 yards.

24. Using Technology There are two useful “technology corner” sections of your textbook that will help you use your calculator to do these types of calculations Page 118 and Page 123 Let’s focus on the one on Page 123 You can use your calculator as a replacement for Table A You will use the normalcdf command on your TI 83/84 Let’s try it, using our height data:

25. Let’s assume that our mean height was 67 inches, and the standard deviation of our heights was 5 inches We know that roughly 68 percent of people should fall between 62 inches and 72 inches. How do we know this? Let’s confirm it You will hit ‘2 nd ’ and then ‘VARS’ and then go to ‘DISTR’ Choose #2: ‘normalcdf’ The syntax of this command is normalcdf(lower bound, upper bound, mean, standard deviation) What should we choose as lower and upper bounds? Depends—if we want to know the percentage BETWEEN, we choose our two endpoints So our command would look like this: normalcdf(62,72,67,5) What is the result? What if we wanted to know what percent of people fall below 5 feet Normalcdf(0,60,67,5) What if we wanted to know the percent above 6 feet?