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Modeling Distributions of Data

Modeling Distributions of Data. 2.1 Describing Location in a Distribution 2.2 Normal Distributions. 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

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Modeling Distributions of Data

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  1. Modeling Distributions of Data 2.1Describing Location in a Distribution 2.2Normal Distributions

  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

  3. Normal Distributions • 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.

  4. Density curves A density curve is a mathematical model of a distribution. The total area under the curve, by definition, is equal to 1, or 100%. The area under the curve for a range of values is the proportion of all observations for that range. Histogram of a sample with the smoothed, density curve describing theoretically the population.

  5. Density curves come in any imaginable shape. Some are well known mathematically and others aren’t.

  6. A family of density curves Here means are the same (m = 15) while standard deviations are different (s = 2, 4, and 6). Here means are different (m = 10, 15, and 20) while standard deviations are the same (s = 3)

  7. Describing Location in a Distribution Density Curves In Chapter 1, we developed a kit of graphical and numerical tools for describing distributions. Now, we’ll add one more step to the strategy. Exploring Quantitative Data • Always plot your data: make a graph. • Look for the overall pattern (shape, center, and spread) and for striking departures such as outliers. • Calculate a numerical summary to briefly describe center and spread. 4. Sometimes the overall pattern of a large number of observations is so regular that we can describe it by a smooth curve.

  8. 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.

  9. Normal Distributions • 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.

  10. The 68-95-99.7 Rule Normal Distributions Although there are many Normal curves, they all have properties in common. • 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 2σ of µ. • Approximately 99.7% of the observations fall within 3σ of µ. 99.7%

  11. Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule) • About 68% of the data lie within one standard deviation of the mean. • About 95% of the data lie within two standard deviations of the mean. • About 99.7%of the data lie within three standard deviations of the mean. 11 For data with a (symmetric) bell-shaped distribution, the standard deviation has the following characteristics:

  12. Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule) 99.7% within 3 standard deviations 95% within 2 standard deviations 68% within 1 standard deviation 34% 34% 2.35% 2.35% 13.5% 13.5% 12

  13. All Normal curves N(m,s) share the same properties • About 68% of all observations are within 1 standard deviation (s) of the mean (m). • About 95% of all observations are within 2 s of the mean m. • Almost all (99.7%) observations are within 3 s of the mean. Inflection point mean µ = 64.5 standard deviation s = 2.5 N(µ, s) = N(64.5, 2.5) Reminder: µ (mu) is the mean of the idealized curve, while is the mean of a sample. s (sigma) is the standard deviation of the idealized curve, while s is the s.d. of a sample.

  14. N(64.5, 2.5) N(0,1) => Standardized height (no units) The standard Normal distribution Because all Normal distributions share the same properties, we can standardize our data to transform any Normal curve N(m,s) into the standard Normal curve N(0,1). For each x we calculate a new value, z (called a z-score).

  15. The cool thing about working with normally distributed data is that we can manipulate it and then find answers to questions that involve comparing seemingly non-comparable distributions. We do this by “standardizing” the data. All this involves is changing the scale so that the mean now = 0 and the standard deviation =1. If you do this to different distributions it makes them comparable. N(0,1)

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

  17. 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).

  18. Describing Location in a Distribution Measuring Position: z-Scores A z-score tells us how many standard deviations from the mean an observation falls, and in what direction. Definition: If x is an observation from a distribution that has known mean and standard deviation, the standardized value of x is: A standardized value is often called a z-score. Jenny earned a score of 86 on her test. The class mean is 80 and the standard deviation is 6.07. What is her standardized score?

  19. Describing Location in a Distribution Using z-scores for Comparison We can use z-scores to compare the position of individuals in different distributions. Jenny earned a score of 86 on her statistics test. The class mean was 80 and the standard deviation was 6.07. She earned a score of 82 on her chemistry test. The chemistry scores had a fairly symmetric distribution with a mean 76 and standard deviation of 4. On which test did Jenny perform better relative to the rest of her class?

  20. .0082 is the area under N(0,1) left of z =-2.40 0.0069 is the area under N(0,1) left of z = -2.46 .0080 is the area under N(0,1) left of z = -2.41 Using Table A Table A gives the area under the standard Normal curve to the left of any z value. (…)

  21. 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. 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: P(z < 0.81) = .7910

  22. 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. Can you find the same proportion using a different approach? 1 - (0.1056+0.2090) = 1 – 0.3146 = 0.6854

  23. Normal Distributions • Normal Distribution Calculations How to Solve Problems Involving Normal Distributions • 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. • Standardizex 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.

  24. Ex. Women heights N(µ, s) = N(64.5, 2.5) Women heights follow the N(64.5”,2.5”) distribution. What percent of women are shorter than 67 inches tall (that’s 5’6”)? Area= ??? Area = ??? mean µ = 64.5" standard deviation s = 2.5" x (height) = 67" m = 64.5” x = 67” z = 0 z = 1 We calculate z, the standardized value of x: Because of the 68-95-99.7 rule, we can conclude that the percent of women shorter than 67” should be, approximately, .68 + half of (1 - .68) = .84 or 84%.

  25. Percent of women shorter than 67” For z = 1.00, the area under the standard Normal curve to the left of z is 0.8413. N(µ, s) = N(64.5”, 2.5”) Area ≈ 0.84 Conclusion: 84.13% of women are shorter than 67”. By subtraction, 1 - 0.8413, or 15.87% of women are taller than 67". Area ≈ 0.16 m = 64.5” x = 67” z = 1

  26. Tips on using Table A Because the Normal distribution is symmetrical, there are 2 ways that you can calculate the area under the standard Normal curve to the right of a z value. Area = 0.9901 Area = 0.0099 z = -2.33 area right of z = area left of -z area right of z = 1 - area left of z

  27. Tips on using Table A To calculate the area between 2 z- values, first get the area under N(0,1) to the left for each z-value from Table A. Then subtract the smaller area from the larger area. A common mistake made by students is to subtract both z values. But the Normal curve is not uniform. area between z1 and z2 = area left of z1 – area left of z2  The area under N(0,1) for a single value of z is zero (Try calculating the area to the left of z minus that same area!)

  28. The National Collegiate Athletic Association (NCAA) requires Division I athletes to score at least 820 on the combined math and verbal SAT exam to compete in their first college year. The SAT scores of 2003 were approximately normal with mean 1026 and standard deviation 209. What proportion of all students would be NCAA qualifiers (SAT ≥ 820)? area right of 820 = total area - area left of 820 = 1 - 0.1611 ≈ 84% Note: The actual data may contain students who scored exactly 820 on the SAT. However, the proportion of scores exactly equal to 820 is 0 for a normal distribution is a consequence of the idealized smoothing of density curves.

  29. The NCAA defines a “partial qualifier” eligible to practice and receive an athletic scholarship, but not to compete, as a combined SAT score is at least 720. What proportion of all students who take the SAT would be partial qualifiers? That is, what proportion have scores between 720 and 820? area between = area left of 820 - area left of 720 720 and 820 = 0.1611 - 0.0721 ≈ 9% About 9% of all students who take the SAT have scores between 720 and 820.

  30. Ex. Gestation time in malnourished mothers What improvement did we get by adding better food? What is the effects of better maternal care on gestation time and premies? The goal is to obtain pregnancies 240 days (8 months) or longer. • 266 s 15 • 250 s 20

  31. Under each treatment, what percent of mothers failed to carry their babies at least 240 days? Vitamins Only m=250, s=20, x=240 Vitamins only: 30.85% of women would be expected to have gestation times shorter than 240 days.

  32. Vitamins and better food m=266, s=15, x=240 Vitamins and better food: 4.18% of women would be expected to have gestation times shorter than 240 days. Compared to vitamin supplements alone, vitamins and better food resulted in a much smaller percentage of women with pregnancy terms below 8 months (4% vs. 31%).

  33. 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.

  34. Normal Distributions • Assessing Normality • The Normal distributions provide good models for some distributions of real data. Many statistical inference procedures are based on the assumption that the population is approximately Normally distributed. Consequently, we need a strategy for assessing Normality. • Plot the data. • Make a dotplot, stemplot, or histogram and see if the graph is approximately symmetric and bell-shaped. • Check whether the data follow the 68-95-99.7 rule. • Count how many observations fall within one, two, and three standard deviations of the mean and check to see if these percents are close to the 68%, 95%, and 99.7% targets for a Normal distribution.

  35. Normal Distributions • Normal Probability Plots • Most software packages can construct Normal probability plots. These plots are constructed by plotting each observation in a data set against its corresponding percentile’s z-score. Interpreting Normal Probability Plots If the points on a Normal probability plot lie close to a straight line, the plot indicates that the data are Normal. Systematic deviations from a straight line indicate a non-Normal distribution. Outliers appear as points that are far away from the overall pattern of the plot.

  36. Describing Location in a Distribution Learning Objectives After this section, you should be able to… • INTERPRET cumulative relative frequency graphs • MEASURE position using z-scores • TRANSFORM data • DEFINE and DESCRIBE density curves

  37. Example: Using the Empirical Rule 37 In a survey conducted by the National Center for Health Statistics, the sample mean height of women in the United States (ages 20-29) was 64 inches, with a sample standard deviation of 2.71 inches. Estimate the percent of the women whose heights are between 64 inches and 69.42 inches.

  38. Solution: Using the Empirical Rule • Because the distribution is bell-shaped, you can use the Empirical Rule. 34% 13.5% 55.87 58.58 61.29 64 66.71 69.42 72.13 34% + 13.5% = 47.5% of women are between 64 and 69.42 inches tall. 38

  39. Example: Comparing z-Scores from Different Data Sets 39 In 2007, Forest Whitaker won the Best Actor Oscar at age 45 for his role in the movie The Last King of Scotland. Helen Mirren won the Best Actress Oscar at age 61 for her role in The Queen. The mean age of all best actor winners is 43.7, with a standard deviation of 8.8. The mean age of all best actress winners is 36, with a standard deviation of 11.5. Find the z-score that corresponds to the age for each actor or actress. Then compare your results.

  40. Solution: Comparing z-Scores from Different Data Sets 0.15 standard deviations above the mean • Helen Mirren 2.17 standard deviations above the mean 40 Forest Whitaker

  41. Solution: Comparing z-Scores from Different Data Sets z = 0.15 z = 2.17 The z-score corresponding to the age of Helen Mirren is more than two standard deviations from the mean, so it is considered unusual. Compared to other Best Actress winners, she is relatively older, whereas the age of Forest Whitaker is only slightly higher than the average age of other Best Actor winners. 41

  42. Describing Location in a Distribution Describing Density Curves Our measures of center and spread apply to density curves as well as to actual sets of observations. Distinguishing the Median and Mean of a Density Curve The median of a density curve is the equal-areas point, the point that divides the area under the curve in half. The mean of a density curve is the balance point, at which the curve would balance if made of solid material. The median and the mean are the same for a symmetric density curve. They both lie at the center of the curve. The mean of a skewed curve is pulled away from the median in the direction of the long tail.

  43. Normal Distributions Summary In this section, we learned that… • The Normal Distributions are described by a special family of bell-shaped, symmetric density curves called Normal curves. The mean µ and standard deviation σ completely specify a Normal distribution N(µ,σ). The mean is the center of the curve, and σ is the distance from µ to the change-of-curvature points on either side. • All Normal distributions obey the 68-95-99.7 Rule, which describes what percent of observations lie within one, two, and three standard deviations of the mean.

  44. Normal Distributions In this section, we learned that… • All Normal distributions are the same when measurements are standardized. The standard Normal distribution has mean µ=0 and standard deviation σ=1. • Table A gives percentiles for the standard Normal curve. By standardizing, we can use Table A to determine the percentile for a given z-score or the z-score corresponding to a given percentile in any Normal distribution. • To assess Normality for a given set of data, we first observe its shape. We then check how well the data fits the 68-95-99.7 rule. We can also construct and interpret a Normal probability plot.

  45. For an area to the left of 1.25 % (0.0125), the z-value is -2.24 Inverse normal calculations For an area to the left of 1.25 % (0.0125), We may also want to find the observed range of values that correspond to a given proportion/ area under the curve. For that, we use Table A backward: • we first find the desired area/ proportion in the body of the table, • we then read the corresponding z-value from the left column and top row.

  46. Summary the z-value is -2.24 Vitamins and better food How long are the longest 75% of pregnancies when malnutritioned mothers are given vitamins and better food? m=266, s=15, upper area 75% upper 75% ? Remember that Table A gives the area to the left of z. Thus we need to search for the lower 25% in Table A in order to get z.  The 75% longest pregnancies in this group are about 256 days or longer.

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