5 1 Normal Probability Distributions Normal distribution A

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 Larson/Farber 1

Properties of Normal Distributions 1. 2. 3. 4. The mean, median, and mode are equal. The normal curve is bell-shaped and symmetric about the mean. The total area under the curve is equal to one. The normal curve approaches, but never touches the x-axis as it extends farther and farther away from the mean. 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. Inflection points Total area = 1 μ 3σ μ 2σ μ μ+σ μ + 2σ μ + 3σ x 2

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 Larson/Farber μ = 3. 5 σ = 0. 7 μ = 1. 5 σ = 0. 7 3

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. ) Curve B has the greater standard deviation (Curve B is more spread out than curve A. ) Larson/Farber 4

Interpreting Graphs The heights of fully grown white oak trees are normally distributed. The curve represents the distribution. σ = 3. 5 (The inflection points are one standard deviation away from the mean) μ = 90 (A normal curve is symmetric about the mean) Z-Scores: Larson/Farber 3 -2 -1 0 1 2 3 5

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

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 s Standard Normal Distribution s=1 m x m=0 z • Use Standard Normal Table or Calculator function: Normal. Cdf() to find cumulative area under the standard normal curve. Larson/Farber 7

Properties of the Standard Normal Distribution 1. 2. 3. 4. The cumulative area is close to 0 for z-scores close to z = 3. 49. The cumulative area increases as the z-scores increase. The cumulative area for z = 0 is 0. 5000. The cumulative area is close to 1 for z-scores close to z = 3. 49. ** Note: Cumulative Area = Area under the curve to the ‘LEFT’ of the z-score. Area is close to 0 z = 3. 49 Larson/Farber Area is close to 1 z 3 2 1 0 Z=0 Area = 0. 500 1 2 3 z = 3. 49 8

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 TI-83/84 <2 nd><DISTR> 2 -Normal. Cdf Normalcdf(-10000, 1. 15) The area to the left of z = 0. 24 is 0. 4052 Larson/Farber Why use – 10000 in normal cdf()? 9

Finding Areas Under the Standard Normal Curve #1 To find the area to the left of z, find #2 To find area to the right of z, find the area that corresponds to z in area corresponding to z in table & the standard normal table or subtract from 1 (total area) or Ti 83/84 normalcdf(-10000, zscore) Ti 83/84 normalcdf (zscore, 10000) The area to the left of z = 1. 23 is 0. 8907 Area to the left of z = 1. 23 is 0. 8907. 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 Ti 83/84 normalcdf(low-zscore, hi-zscore) The area to the left of z = 0. 75 is 0. 2266. Larson/Farber Subtract to find the area of the region between the two zscores: 0. 8907 0. 2266 = 0. 6641. 10

Practice #1: Find the area under the standard normal curve to the left of z = -0. 99 Answer: 0. 1611 0. 99 z 0 #2: Find the area under the standard normal curve to the right of z = 1. 06 Answer: 0. 1446 0 1. 06 z #3: Find the area under the standard normal curve between z=-1. 5 and 1. 25 Answer: 0. 8276 Larson/Farber 1. 50 0 1. 25 z

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 μ = 500 σ = 100 x μ =500 600 Example: Normal Distribution μ = 500 σ = 100 Standard Normal Distribution μ=0 σ=1 P(x < 600) P(z < 1) x μ =500 600 Larson/Farber Same Area P(x < 600) = P(z < 1) z μ=0 1 12

Example 1: 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. Normal Distribution μ = 2. 4 σ = 0. 5 Standard Normal Distribution μ=0 σ=1 P(z < -0. 80) P(x < 2) 0. 2119 z x 2 2. 4 -0. 80 0 P(x < 2) = P(z < -0. 80) = 0. 2119 Find the probability using your Ti-84/83 Calculator__________ Larson/Farber 13

Example 2: 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. Normal Distribution μ = 45 σ = 12 Standard Normal Distribution μ=0 σ=1 P(24 < x < 54) P(-1. 75 < z < 0. 75) 0. 7734 0. 0401 x 24 45 54 z -1. 75 0 0. 75 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_________

Example 3: 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) Normal Distribution μ = 45 σ = 12 Standard Normal Distribution μ=0 σ=1 P(z > -0. 50) P(x > 39) 0. 3085 z x 39 45 -0. 50 0 P(x > 39) = P(z > -0. 50) = 1– 0. 3085 = 0. 6915 Find the probability using your Ti-84/83 Calculator_________ 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) 15

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? ) Answers: 1). 2743 2). 5572 3). 1685 16

5. 3 Finding a z-Score Given an Area Example 1: Find the z-score that corresponds to a cumulative area of 0. 3632 z 0 z TI 83/84 <2 nd><DISTR> 3: Inv. Norm(. 3632) The z-score is -0. 35. Larson/Farber 4 th ed 17

Finding a z-Score Given an Area Example 2: Find the z-score that has 10. 75% of the distribution’s area to its right. 1 – 0. 1075 = 0. 8925 0 0. 1075 z z Because the area to the right is 0. 1075, the cumulative area is 0. 8925. Locate. 8925 in the Standard Normal Table. Or Inv. Norm (. 8925) Answer: Z-score = 1. 24 Larson/Farber 4 th ed 18

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 z 0 z 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 zscore that is halfway between -1. 64 and -1. 65. The z-score is -1. 645. OR Inv. Norm (. 05) = -1. 645 Larson/Farber 4 th ed 19

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 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) Larson/Farber 4 th ed 20

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? 5% 1 – 0. 05 = 0. 95 0 ? z x inv. Norm (. 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. 75 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.