Chapter 5 The Normal Distribution Elementary Statistics Larson

Chapter 5 The Normal Distribution Elementary Statistics Larson Farber 1

Properties of a Normal Distribution Inflection point x • The mean, median, and mode are equal • Bell shaped and symmetric about the mean • The total area under the curve is one (1) or 100% • The curve approaches but never touches the x- axis as it extends farther and farther away from the mean in both directions. • The points at which the curvature changes are called inflection points. 2

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 3

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 4

Determining Intervals 3. 3 3 3. 62 3. 91 0 4. 2 1 4. 5 2 4. 8 3 5. 1 x 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 hours. 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. 5

The Standard Score The standard score, or z-score, represents the number of standard deviations a random variable x falls from the mean. The test scores for a civil service exam are normally distributed with a mean of 152 and standard deviation of 7. Find the standard z-score for a person with a score of: (a) 161 (b) 148 (b) (a) Larson/Farber Ch 5 (c) 152 (c) 6

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 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 7

The Standard Normal Distribution The standard normal distributionhas 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 Larson/Farber Ch 5 3 2 1 0 1 2 3 4 z 8

Cumulative Areas The total area under the curve is one. -3 -2 -1 0 1 2 3 z • 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. 9

Cumulative Areas Find the cumulative area for a z-score of -1. 25. 0. 1056 z -3 -2 -1 0 1 2 3 Read down the z column on the left to z = -1. 2 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. P ( z -1. 25) = 0. 1056 10

From Areas to z-scores Find the z-score corresponding to a cumulative area of 0. 9803 z 4 3 2 1 0 1 2 3 4 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 zscore is 2. 06. z = 2. 06 is roughly the 98 th percentile. 11

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. 24) -3 -2 -1 0 1 2 3 z Read down the z-column to -1. 2 and across to. 04. The cumulative area is 0. 1075. P ( z < 1. 24) = 0. 1075 12

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) Required area 0. 1075 0. 8925 -3 -2 -1 0 1 2 3 z 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 1. P(z < 1. 17) = 0. 8790 0 1 2 3 z 2. P(z < -1. 25) =0. 1056 3. P( -1. 25 < z < 1. 17) = 0. 8790 - 0. 1056 = 0. 7734 14

Summary To find the probability that z is less than a given value, read the corresponding cumulative area. z -3 -2 -1 0 1 2 3 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 To find the probability z is between two given values, find the cumulative areas for each and subtract the smaller area from the z -3 -2 -1 0 1 2 3 z 15

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 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. 16

Probabilities and Normal Distributions SAME Find P(x < 115) SAME Normal Distribution 100 115 Standard Normal Distribution Find P(z < 1) 0 1 P( z < 1) = 0. 8413, so P( x <115) = 0. 8413 Larson/Farber Ch 5 17

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. 18

Finding Percentiles 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? 90% 10% z Find the cumulative area in the table that is closest to 0. 9000 (the 90 th 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. $115. 36 is the smallest value for the top 10%. 19

Sampling Distributions A sampling distributionis the probability distribution of a sample statistic that is formed when samples of size n are repeatedly taken from a population. If the sample statistic is the sample mean, then the distribution is the sampling distribution of sample means. Sample Sample The sampling distribution consists of the values of the sample means, 20

The Central Limit Theorem If a sample n 30 is taken from a population with any type distribution that has a mean = and standard deviation = x the sample means will have a normal distribution with a mean and standard deviation 21

The Central Limit Theorem If a sample of any size is taken from a population with a normal distribution with mean = and standard deviation= the distribution of means of sample size n , will be normal with a mean x standard deviation Larson/Farber Ch 5 22

Application The mean height of American men (ages 20 -29) is inches. Random samples of 60 such men are selected. Find the mean and standard deviation (standard error) of the sampling distribution. 69. 2 Distribution of means of sample size 60 , will be normal. mean Standard deviation 23

Interpreting the Central Limit Theorem The mean height of American men (ages 20 -29) is = 69. 2”. If a random sample of 60 men in this age group is selected, what is the probability the mean height for the sample is greater than 70”? Assume the standard deviation is 2. 9”. Since n > 30 the sampling distribution of will be normal mean standard deviation Find the z-score for a sample mean of 70: 24

Interpreting the Central Limit Theorem P( > 70) = P (z > 2. 14) = 1 - 0. 9838 z 2. 14 = 0. 0162 There is a 0. 0162 probability that a sample of 60 men will have a mean height greater than 70”. 25

Application Central Limit Theorem During a certain week the mean price of gasoline in California was $1. 164 per gallon. What is the probability that the mean price for the sample of 38 gas stations in California is between $1. 169 and $1. 179? Assume the standard deviation = $0. 049. Since n > 30 the sampling distribution of will be normal mean standard deviation Calculate the standard z-score for sample values of $1. 169 and $1. 179. 26

Application Central Limit Theorem P( 0. 63 < z < 1. 90) = 0. 9713 - 0. 7357 = 0. 2356 z. 63 1. 90 The probability is 0. 2356 that the mean for the sample is between $1. 169 and $1. 179. 27

Normal Approximations to the Binomial • • • Characteristics of a Binomial Experiment There a fixed number of trials. (n) The n trials are independent and repeated under identical conditions Each trial has 2 outcomes, Success or Failure. The probability of success on a single trial is p and the probability of failure is q. p+q=1 The central problem is to find the probability of x successes out of n trials. Where x = 0 or 1 or 2 … n. x is a count of the number of successes in n trials. 28

Application 34% of Americans have type A+ blood. If 500 Americans are sampled at random, what is the probability at least 300 have type A+ blood? Using techniques of chapter 4 you could calculate the probability that exactly 300, exactly 301…exactly 500 Americans have A+ blood type and add the probabilities. Or…you could use the normal curve probabilities to approximate the binomial probabilities. If np 5 and nq 5, the binomial random variable x is approximately normally distributed with mean and 29

Why do we require np 5 and nq 5? 0 1 2 3 4 4 5 n=5 p = 0. 25, q =. 75 np =1. 25 nq = 3. 75 n = 20 p = 0. 25 np = 5 nq = 15 4 n = 50 p = 0. 25 np = 12. 5 nq = 37. 5 0 10 20 30 40 50 30

Binomial Probabilities The binomial distribution is discrete with a probability histogram graph. The probability that a specific value of x will occur is equal to the area of the rectangle with midpoint at x. If n = 50 and p = 0. 25 find P (14 x 16) Add the areas of the rectangles with midpoints at x = 14, x = 15, x = 16. 0. 111 + 0. 089 + 0. 065 = 0. 265 0. 111 14 Larson/Farber Ch 5 0. 089 15 0. 065 16 P (14 x 16) = 0. 265 31

Correction for Continuity Use the normal approximation to the binomial to find P(14 x 16) if n = 50 and p = 0. 25 Check that np= 12. 5 5 and nq= 37. 5 5. 14 15 16 The interval of values under the normal curve is 13. 5 x 16. 5. To ensure the boundaries of each rectangle are included in the interval, subtract 0. 5 from a left-hand boundary and add 0. 5 to a right-hand boundary. 32

Normal Approximation to the Binomial Use the normal approximation to the binomial to find P(14 x 16) if n = 50 and p = 0. 25 Find the mean and standard deviation using binomial distribution formulas. Adjust the endpoints to correct for continuity P(13. 5 x 16. 5) Convert each endpoint to a standard score P(0. 33 z 1. 31) = 0. 9049 - 0. 6293 = 0. 2756 Larson/Farber Ch 5 33

Application A survey of Internet users found that 75% favored government regulations on “junk” e-mail. If 200 Internet users are randomly selected, find the probability that fewer than 140 are in favor of government regulation. Since np=150 5 and nq = 50 5 use the normal approximation to the binomial. Use the correction for continuity P(x < 139. 5) P(z < -1. 71) = 0. 0436 The probability that fewer than 140 are in favor of government 34 regulation is 0. 0436