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Describing Location in a Distribution 2.2 Normal Distributions YMS3e 2.2 Objectives Identify the main properties if the Normal curve as a particular density curve. List three reasons why Normal Distributions are important in statistics. Explain the 68-95-99.7 rule (the empirical rule). Explain the notation N(μ,σ). Define the standardized Normal distribution. Use a table of values for the standard Normal curve (Table A) to compute the proportion of observations that are Less than a given z-score. Greater than a given z-score. Between two given z-scores. 2.2 Objectives Use a table of values for the standard Normal curve to find the proportion of observations in any region given any Normal distribution (i.e., given raw data rather than z-scores) Use a table of values for the standard Normal curve to find a value with a given proportion of observations above or below it (inverse Normal) Identify at least two graphical techniques for assessing Normality. Explain what is meant by a Normal probability plot, use it to help assess the Normality of a given data set. Use technology to perform Normal distribution calculations and to make Normal probability plots. Properties of a Normal Distribution x • The mean, median, and mode are equal • Bell shaped and is symmetric about the mean • The total area that lies under the curve is one or 100% Properties of a Normal Distribution Inflection point Inflection point x • As the curve extends farther and farther away from the mean, it gets closer and closer to the x-axis but never touches it. • The points at which the curvature changes are called inflection points. The graph curves downward between the inflection points and curves upward past the inflection points to the left and to the right. Means and Standard Deviations Curves with different means, same standard deviation 10 11 12 13 14 15 16 17 18 19 20 Curves with different means, different standard deviations 9 10 11 12 13 14 15 16 17 18 19 20 21 22 Empirical Rule 68% About 68% of the area lies within 1 standard deviation of the mean About 95% of the area lies within 2 standard deviations About 99.7% of the area lies within 3 standard deviations of the mean Determining Intervals x 3.3 3.6 3.9 4.2 4.5 4.8 5.1 An instruction manual claims that the assembly time for a product is normally distributed with a mean of 4.2 hours and standard deviation 0.3 hour. Determine the interval in which 95% of the assembly times fall. 95% of the data will fall within 2 standard deviations of the mean. 4.2 – 2 (0.3) = 3.6 and 4.2 + 2 (0.3) = 4.8. 95% of the assembly times will be between 3.6 and 4.8 hrs. The Standard Normal Distribution The standard normal distribution has a mean of 0 and a standard deviation of 1. Using z-scores any normal distribution can be transformed into the standard normal distribution. –4 –3 –2 –1 0 1 Larson/Farber Ch 5 2 3 4 z 9 Cumulative Areas The total area under the curve is one. z –3 –2 –1 0 1 2 3 • The cumulative area is close to 0 for z-scores close to –3.49. • The cumulative area for z = 0 is 0.5000. • The cumulative area is close to 1 for z-scores close to 3.49. Cumulative Areas Find the cumulative area for a z-score of –1.25. 0.1056 –3 –2 –1 0 1 2 3 z Read down the z column on the left to z = –1.25 and across to the column under .05. The value in the cell is 0.1056, the cumulative area. The probability that z is at most –1.25 is 0.1056. Finding Probabilities To find the probability that z is less than a given value, read the cumulative area in the table corresponding to that z-score. Find P(z < –1.45). P (z < –1.45) = 0.0735 –3 –2 –1 0 1 2 3 z Read down the z-column to –1.4 and across to .05. The cumulative area is 0.0735. Finding Probabilities To find the probability that z is greater than a given value, subtract the cumulative area in the table from 1. Find P(z > –1.24). 0.1075 0.8925 z –3 –2 –1 0 1 2 3 The cumulative area (area to the left) is 0.1075. So the area to the right is 1 – 0.1075 = 0.8925. P(z > –1.24) = 0.8925 Larson/Farber Ch 5 13 Finding Probabilities To find the probability z is between two given values, find the cumulative areas for each and subtract the smaller area from the larger. Find P(–1.25 < z < 1.17). –3 –2 –1 0 1 2 1. P(z < 1.17) = 0.8790 3 z 2. P(z < –1.25) = 0.1056 3. P(–1.25 < z < 1.17) = 0.8790 – 0.1056 = 0.7734 Summary To find the probability that z is less than a given value, read the corresponding cumulative area. -3 -2 -1 0 1 2 3 z To find the probability is greater than a given value, subtract the cumulative area in the table from 1. -3 -2 -1 0 1 2 3 z To find the probability z is between two given values, find the cumulative areas for each and subtract the smaller area from the larger. -3 -2 -1 0 1 2 3 z Examples a. Find the area under the curve to the left of a z-score of –2.19. b. Find the area under the curve to the left of a z-score of 2.17. Examples c. Find the area under the standard normal curve to the left of z = 2.13. Examples d. Find the area under the standard normal curve to the right of z = -2.16. Probabilities and Normal Distributions If a random variable, x is normally distributed, the probability that x will fall within an interval is equal to the area under the curve in the interval. IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. Find the probability that a person selected at random will have an IQ score less than 115. 100 115 To find the area in this interval, first find the standard score equivalent to x = 115. Z = 115 – 100 = 1 15 Probabilities and Normal Distributions Normal Distribution Standard Normal Distribution 100 115 SAME SAME Find P(x < 115). Find P(z < 1). 0 1 P(z < 1) = 0.8413, so P(xCh <115) = 0.8413 Larson/Farber 5 20 Application Monthly utility bills in a certain city are normally distributed with a mean of $100 and a standard deviation of $12. A utility bill is randomly selected. Find the probability it is between $80 and $115. Normal Distribution P(80 < x < 115) P(–1.67 < z < 1.25) 0.8944 – 0.0475 = 0.8469 The probability a utility bill is between $80 and $115 is 0.8469. From Areas to z-Scores Find the z-score corresponding to a cumulative area of 0.9803. z = 2.06 corresponds roughly to the 98th percentile. 0.9803 –4 –3 –2 –1 0 1 2 3 4 z Locate 0.9803 in the area portion of the table. Read the values at the beginning of the corresponding row and at the top of the column. The z-score is 2.06. Finding z-Scores from Areas Find the z-score corresponding to the 90th percentile. .90 0 z The closest table area is .8997. The row heading is 1.2 and column heading is .08. This corresponds to z = 1.28. A z-score of 1.28 corresponds to the 90th percentile. Finding z-Scores from Areas Find the z-score with an area of .60 falling to its right. .40 .60 z 0 z With .60 to the right, cumulative area is .40. The closest area is .4013. The row heading is 0.2 and column heading is .05. The z-score is 0.25. A z-score of 0.25 has an area of .60 to its right. It also corresponds to the 40th percentile Finding z-Scores from Areas Find the z-score such that 45% of the area under the curve falls between –z and z. .275 .275 .45 –z 0 z The area remaining in the tails is .55. Half this area is in each tail, so since .55/2 = .275 is the cumulative area for the negative z value and .275 + .45 = .725 is the cumulative area for the positive z. The closest table area is .2743 and the z-score is 0.60. The positive z score is 0.60. From z-Scores to Raw Scores To find the data value, x when given a standard score, z: The test scores for a civil service exam are normally distributed with a mean of 152 and a standard deviation of 7. Find the test score for a person with a standard score of: (a) 2.33 (b) –1.75 (c) 0 (a) x = 152 + (2.33)(7) = 168.31 (b) x = 152 + (–1.75)(7) = 139.75 (c) x = 152 + (0)(7) = 152 Larson/Farber Ch 5 26 Finding Percentiles or Cut-off Values Monthly utility bills in a certain city are normally distributed with a mean of $100 and a standard deviation of $12. What is the smallest utility bill that can be in the top 10% of the bills? $115.36 is the smallest value for the top 10%. 90% 10% z Find the cumulative area in the table that is closest to 0.9000 (the 90th percentile.) The area 0.8997 corresponds to a z-score of 1.28. To find the corresponding x-value, use x = 100 + 1.28(12) = 115.36.