1. Key Concepts, continued To determine the probability of an outcome using continuous data, we use the proportion of the area under the normal curve associated with the distribution of that data. A normal curve is a symmetrical curve representing the normal distribution. 1 1.2.1: Normal Distributions and the 68–95–99.7 Rule

2. Key Concepts, continued Continuous Normal Distributions Another type of a continuous distribution is a normal distribution. A normal distribution is a set of values that are continuous, are symmetric to the mean, and have higher frequencies in intervals close to the mean than equal-sized intervals away from the mean. When graphed, data following a normal distribution forms a normal curve. Normal distributions are symmetric to the mean. This means that 50% of the data is to the right of the mean and 50% of the data is to the left of the mean. 2 1.2.1: Normal Distributions and the 68–95–99.7 Rule

3. Key Concepts, continued T he mean is a measure of center in a set of numerical data, computed by adding the values in a data set and then dividing the sum by the number of values in the data set. The population mean is denoted by the Greek lowercase letter mu, μ, whereas the sample mean is denoted by 3 1.2.1: Normal Distributions and the 68–95–99.7 Rule

4. Key Concepts, continued A population is made up of all of the people, objects, or phenomena of interest in an investigation. A sample is a subset of the population—that is, a smaller portion that represents the whole population. The standard deviation is a measure of average variation about a mean. 4 1.2.1: Normal Distributions and the 68–95–99.7 Rule

5. Key Concepts, continued Approximately 68% of the values in a normal distribution are within one standard deviation of the mean. Written as an equation, this is In other words, the mean, μ, plus or minus the standard deviation σ times 1 is approximately equal to 68% of the values in the distribution. In the graph that follows, the shading represents these 68% of values that fall within one standard deviation of the mean. 5 1.2.1: Normal Distributions and the 68–95–99.7 Rule

6. Key Concepts, continued Data Within One Standard Deviation of the Mean 6 1.2.1: Normal Distributions and the 68–95–99.7 Rule

7. Key Concepts, continued Approximately 95% of the values in a normal distribution are within two standard deviations of the mean, as shown by the shading in the graph on the next slide. 7 1.2.1: Normal Distributions and the 68–95–99.7 Rule

8. Key Concepts, continued Data Within Two Standard Deviations of the Mean 8 1.2.1: Normal Distributions and the 68–95–99.7 Rule

9. Key Concepts, continued Approximately 99.7% of the values in a normal distribution are within three standard deviations of the mean, as shaded in the graph on the next slide. 9 1.2.1: Normal Distributions and the 68–95–99.7 Rule

10. Key Concepts, continued Data Within Three Standard Deviations of the Mean 10 1.2.1: Normal Distributions and the 68–95–99.7 Rule

11. Key Concepts, continued These percentages of data under the normal curve follow what is called the 68–95–99.7 rule. This rule is also known as the Empirical Rule. 11 1.2.1: Normal Distributions and the 68–95–99.7 Rule

12. Key Concepts, continued The standard normal distribution has a mean of 0 and a standard deviation of 1. A normal curve is often referred to as a bell curve, since its shape resembles the shape of a bell. Normal distribution curves are a common tool for teachers who want to analyze how their students performed on a test. If a test is “fair,” you can expect a handful of students to do very well or very poorly, with most scores being near average—a normal curve. If the curve is shifted strongly toward the lower or higher ends of the scores, then the test was too hard or too easy. 12 1.2.1: Normal Distributions and the 68–95–99.7 Rule

13. Common Errors/Misconceptions applying the 68–95–99.7 rule to distributions that are not normally distributed assuming that all normal distributions have a mean of 0 and/or a standard deviation of 1 not applying symmetry in a normal distribution to calculate probabilities 13 1.2.1: Normal Distributions and the 68–95–99.7 Rule

14. Guided Practice Example 4 The scores of a particular college admission test are normally distributed, with a mean score of 30 and a standard deviation of 2. Erin scored a 34 on her test. If possible, determine the percent of test-takers whom Erin outperformed on the test. 14 1.2.1: Normal Distributions and the 68–95–99.7 Rule

15. Guided Practice: Example 4, continued 1.Sketch a normal curve and shade the area of the interval of interest. To sketch the normal curve, start by drawing a number line with a range of values that includes Erin’s score, 34. The mean of the scores is 30, so the middle number on the line will be 30. A range of 24 to 36 will give us a number line that has the mean, 30, in the middle, and an even number of data points on either side of the mean. 15 1.2.1: Normal Distributions and the 68–95–99.7 Rule

16. Guided Practice: Example 4, continued We want to know how many test-takers had scores lower than Erin’s. Erin scored a 34; therefore, the area of interest is the area to the left of 34. 16 1.2.1: Normal Distributions and the 68–95–99.7 Rule

17. Guided Practice: Example 4, continued 2.Determine how many standard deviations away from the mean Erin’s score is. From the problem statement, we know that Erin scored a 34, the mean is 30, and the standard deviation is 2. Erin’s score is greater than the mean. Also, we can determine that Erin scored two standard deviations above the mean. μ + 1 = 30 + 1(2) = 32 μ + 2σ = 30 + 2(2) = 34Erin’s score μ + 3 = 30 + 2(3) = 36 17 1.2.1: Normal Distributions and the 68–95–99.7 Rule

18. Guided Practice: Example 4, continued 3.Use symmetry and the 68–95–99.7 rule to determine the area of interest. We know that the data in a normal curve is symmetrical about the mean. Since the area under the curve is equal to 1, the area to the left of the mean is 0.5, as shaded in the graph on the next slide. 18 1.2.1: Normal Distributions and the 68–95–99.7 Rule

19. Guided Practice: Example 4, continued Erin’s score is above the mean; therefore, we need to determine the area between the mean and Erin’s score and add it to the area below the mean to find the total area of interest. 19 1.2.1: Normal Distributions and the 68–95–99.7 Rule

20. Guided Practice: Example 4, continued Recall that the 68–95–99.7 rule states the percentages of data under the normal curve are as follows: We know that We have already accounted for the area to the left of the mean, which includes from the mean down to –2σ. Since we found that Erin’s score is two standard deviations from the mean, we need to determine the area from the mean up to +2σ. 20 1.2.1: Normal Distributions and the 68–95–99.7 Rule

21. Guided Practice: Example 4, continued Since data is symmetric about the mean, we know that half of the area encompassed between μ ± 2 is above the mean. Therefore, divide 0.95 by 2. The following graph shows the shaded area of interest to the right of the mean up until Erin’s score of 34. 21 1.2.1: Normal Distributions and the 68–95–99.7 Rule

22. Guided Practice: Example 4, continued Add the two areas together to get the total area below 2σ, which is equal to Erin’s score of 34. 0.50 + 0.475 = 0.975 The total area of interest for this data is 0.975. 22 1.2.1: Normal Distributions and the 68–95–99.7 Rule

23. Guided Practice: Example 4, continued A graphing calculator can also be used to calculate the area of interest. On a TI-83/84: Step 1: Press [2ND][VARS] to bring up the distribution menu. Step 2: Arrow down to 2: normalcdf. Press [ENTER]. Step 3: Enter the following values for the lower bound, upper bound, mean (μ), and standard deviation (σ). Press [ENTER] after typing each value to navigate between fields. Lower: [(–)][99]; upper: [34]; μ: [30]; σ: [2]. Step 4: Press [ENTER] twice to calculate the area of interest. 23 1.2.1: Normal Distributions and the 68–95–99.7 Rule

24. Guided Practice: Example 4, continued 4.Interpret the proportion in terms of the context of the problem. Convert the area of interest to a percent. 0.975 = 97.5% Erin outperformed 97.5% of the students who also took the exam. 24 1.2.1: Normal Distributions and the 68–95–99.7 Rule ✔