1. 5.1 Normal Probability Distributions Normal distribution A continuous probability distribution for a continuous random variable, x. The most important continuous probability distribution in statistics. The graph of a normal distribution is called the normal curve. x 1 Larson/Farber

2. Properties of Normal Distributions 1.The mean, median, and mode are equal. 2.The normal curve is bell-shaped and symmetric about the mean. 3.The total area under the curve is equal to one. 4.The normal curve approaches, but never touches the x-axis as it extends farther and farther away from the mean. Total area = 1 2 5.Between μ – σ and μ + σ (in the center of the curve), the graph curves downward. The graph curves upward to the left of μ – σ and to the right of μ + σ. The points at which the curve changes from curving upward to curving downward are called the inflection points. μ  3σ μ + σ μ  2σμ  σμ  σ μμ + 2σμ + 3σ x Inflection points

3. Means and Standard Deviations A normal distribution can have any mean and any positive standard deviation. The mean gives the location of the line of symmetry. The standard deviation describes the spread of the data. μ = 3.5 σ = 1.5 μ = 3.5 σ = 0.7 μ = 1.5 σ = 0.7 3 Larson/Farber

4. Example: Understanding Mean and Standard Deviation Which curve has the greater mean and Standard Deviation? Curve A has the greater mean (The line of symmetry of curve A occurs at x = 15. The line of symmetry of curve B occurs at x = 12.) 4 Larson/Farber Curve B has the greater standard deviation (Curve B is more spread out than curve A.)

5. Interpreting Graphs The heights of fully grown white oak trees are normally distributed. The curve represents the distribution. μ = 90 (A normal curve is symmetric about the mean) σ = 3.5 (The inflection points are one standard deviation away from the mean) 5 Larson/Farber 012-233Z-Scores:

6. The Standard Normal Distribution Standard normal distribution A normal distribution with a mean of 0 and a standard deviation of 1. 33 1 22 11 023 z Area = 1 Any x-value can be transformed into a z-score by using the formula 6 Larson/Farber

7. The Standard Normal Distribution If each data value of a normally distributed random variable x is transformed into a z-score, the result will be the standard normal distribution. Normal Distribution x     z Standard Normal Distribution Use Standard Normal Table or Calculator function: NormalCdf() to find cumulative area under the standard normal curve. 7 Larson/Farber

8. Properties of the Standard Normal Distribution 1.The cumulative area is close to 0 for z-scores close to z =  3.49. 2.The cumulative area increases as the z-scores increase. 3.The cumulative area for z = 0 is 0.5000. 4.The cumulative area is close to 1 for z-scores close to z = 3.49. z =  3.49 Area is close to 0 z 33 1 22 11 023 8 Larson/Farber z = 3.49 Area is close to 1 Z = 0 Area = 0.500 ** Note: Cumulative Area = Area under the curve to the ‘LEFT’ of the z-score.

9. Using The Standard Normal Table Find the cumulative area that corresponds to a z-score of 1.15 and –0.24 The area to the left of z = 1.15 is 0.8749 9 Larson/Farber The area to the left of z = 0.24 is 0.4052 TI-83/84 2-NormalCdf Normalcdf(-10000, 1.15) Why use –10000 in normal cdf()?

10. Finding Areas Under the Standard Normal Curve #1 To find the area to the left of z, find the area that corresponds to z in the standard normal table or Ti83/84 normalcdf(-10000, zscore) The area to the left of z = 1.23 is 0.8907 10 Larson/Farber Area to the left of z = 1.23 is 0.8907. #2 To find area to the right of z, find area corresponding to z in table & subtract from 1 (total area) or Ti83/84 normalcdf (zscore, 10000) Area to the right is 1 – 0.8907 = 0.1093 #3 To find the area between two z-scores, find both z-scores in table & subtract smaller from larger area, or Ti83/84 normalcdf(low-zscore, hi-zscore) The area to the left of z =  0.75 is 0.2266. Subtract to find the area of the region between the two zscores: 0.8907  0.2266 = 0.6641.

11. Practice #1: Find the area under the standard normal curve to the left of z = -0.99  0.99 0 z Larson/Farber Answer: 0.1611 #2: Find the area under the standard normal curve to the right of z = 1.06 Answer: 0.1446 1.06 0 z #3: Find the area under the standard normal curve between z=-1.5 and 1.25 Answer: 0.8276 1.25 0 z  1.50

12. μ = 500 σ = 100 600 μ =500 x 5.2 Probability and Normal Distributions If a random variable x is normally distributed, you can find the probability that x will fall in a given interval by calculating the area under the normal curve for that interval. P(x < 600) = Area 12 Larson/Farber 600μ =500 P(x < 600) Normal Distribution μ = 500 σ = 100 x 1 μ = 0 Standard Normal Distribution μ = 0 σ = 1 z P(z < 1) Same Area P(x < 600) = P(z < 1) Example:

13. Example1: Finding Probabilities for Normal Distributions A survey indicates that people use their computers an average of 2.4 years before upgrading to a new machine. The standard deviation is 0.5 year. A computer owner is selected at random. Find the probability that he or she will use it for fewer than 2 years before upgrading. Assume that the variable x is normally distributed. 13 Larson/Farber 22.4 P(x < 2) Normal Distribution μ = 2.4 σ = 0.5 x -0.80 0 Standard Normal Distribution μ = 0 σ = 1 z P(z < -0.80) P(x < 2) = P(z < -0.80) = 0.2119 Find the probability using your Ti-84/83 Calculator___________________ 0.2119

14. Example2: Finding Probabilities for Normal Distributions A survey indicates that for each trip to the supermarket, a shopper spends an average of 45 minutes with a standard deviation of 12 minutes in the store. The length of time spent in the store is normally distributed and is represented by the variable x. A shopper enters the store. Find the probability that the shopper will be in the store for between 24 and 54 minutes. P(24 < x < 54) = P(-1.75 < z < 0.75) = 0.7734 – 0.0401 = 0.7333 Find the probability using your Ti-84/83 Calculator_________________ 2445 P(24 < x < 54) x Normal Distribution μ = 45 σ = 12 0.0401 54 -1.75 z Standard Normal Distribution μ = 0 σ = 1 0 P(-1.75 < z < 0.75) 0.75 0.7734

15. Example3: Finding Probabilities for Normal Distributions Find the probability that the shopper will be in the store more than 39 minutes. (Recall μ = 45 minutes and σ = 12 minutes) P(x > 39) = P(z > -0.50) = 1– 0.3085 = 0.6915 Find the probability using your Ti-84/83 Calculator__________________ 3945 P(x > 39) x Normal Distribution μ = 45 σ = 12 Standard Normal Distribution μ = 0 σ = 1 0.3085 0 P(z > -0.50) z -0.50 15 If 200 shoppers enter the store, how many shoppers would you expect to be in the store more than 39 minutes? 200(.6915) = 138.3 (approx. 138 shoppers)

16. You Try this one! Assume that cholesterol levels of men in the United States are normally distributed, with a mean of 215 milligrams per deciliter and a standard deviation of 25 milligrams per deciliter. You randomly select a man from the United States. Draw pictures, use Z-scores and use your calculator to answer the questions below: 1. A lower risk of heart attack is associated with a cholesterol level below 200. What is the probability that the man’s cholesterol level is less than 200? 2. A moderate risk of heart attack is associated with a cholesterol level between 200 and 239. What is the probability that the man’s cholesterol level is between 200 and 239? 3. A higher risk of heart attack is associated with cholesterol levels above 239. What is the probability that the man’s cholesterol level is above 239? (How would you do this using a complement?) 16 Answers: 1).2743 2).5572 3).1685

17. 5.3 Finding a z-Score Given an Area Example1: Find the z-score that corresponds to a cumulative area of 0.3632. z 0 z 0.3632 17 Larson/Farber 4th ed The z-score is -0.35. TI 83/84 3:InvNorm(.3632)

18. Finding a z-Score Given an Area Example2: Find the z-score that has 10.75% of the distribution’s area to its right. z0 z 0.1075 1 – 0.1075 = 0.8925 Because the area to the right is 0.1075, the cumulative area is 0.8925. Locate.8925 in the Standard Normal Table. Or InvNorm (.8925) Answer: Z-score = 1.24 18 Larson/Farber 4th ed

19. Finding a z-Score Given a Percentile Find the z-score that corresponds to P 5 (or the 5 th Percentile) (Recall: 5 th percentile refers to a value that is higher than 5% of the data) The z-score for P 5 is the same z-score that corresponds to an area of 0.05. The areas closest to 0.05 in the table are 0.0495 (z = -1.65) and 0.0505 (z = - 1.64). Because 0.05 is halfway between the two areas in the table, use the z- score that is halfway between -1.64 and -1.65. The z-score is -1.645. OR InvNorm (.05) = -1.645 z 0 z 0.05 19 Larson/Farber 4th ed

20. Transforming a z-Score to an x-Score To transform a standard z-score to a data value x in a given population, use the formula : x = μ + zσ 20 Larson/Farber 4th ed The speeds of vehicles along a stretch of highway are normally distributed, with a mean of 67 miles per hour and a standard deviation of 4 miles per hour. Find the speeds x corresponding to z-sores of 1.96, -2.33, and 0. z = 1.96:x = 67 + 1.96(4) = 74.84 miles per hour (Above the mean) z = -2.33:x = 67 + (-2.33)(4) = 57.68 miles per hour (Below the mean) z = 0: x = 67 + 0(4) = 67 miles per hour (Equal to the mean)

21. Finding a Specific Data Value Scores for a civil service exam are normally distributed, with a mean of 75 and a standard deviation of 6.5. To be eligible for civil service employment, you must score in the top 5%. What is the lowest score you can earn and still be eligible for employment? ? 0 z 5% ? 75 x 1 – 0.05 = 0.95 An exam score in the top 5% is any score above the 95 th percentile. Find the z-score that corresponds to a cumulative area of 0.95. From the Standard Normal Table, the areas closest to 0.95 are 0.9495 (z = 1.64) and 0.9505 (z = 1.65). Because 0.95 is halfway between the two areas in the table, use the z-score that is halfway between 1.64 and 1.65. That is, z = 1.645 OR Using the equation x = μ + zσ : x = 75 + 1.645(6.5) ≈ 85.69 The lowest score you can earn and still be eligible for employment is 86. invNorm (.95)