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Chapter 6 Normal Probability Distributions Overview The Standard Normal Distribution Normal Distributions: Finding Probabilities Normal Distributions: Finding Values The Central Limit Theorem 1 Overview Continuous random variable Normal distribution Curve is bell shaped and symmetric µ Score y= e 1 2 x-µ 2 ( ) 2p 2 Definitions Density Curve (or probability density function) the graph of a continuous probability distribution 1. The total area under the curve must equal 1. 2. Every point on the curve must have a vertical height that is 0 or greater. 3 Because the total area under the density curve is equal to 1, there is a correspondence between area and probability. 4 Heights of Adult Men and Women Women: µ = 63.6 = 2.5 Men: µ = 69.0 = 2.8 63.6 69.0 Height (inches) 5 Definition Standard Normal Deviation a normal probability distribution that has a mean of 0 and a standard deviation of 1 6 The Empirical Rule Standard Normal Distribution: µ = 0 and = 1 99.7% of data are within 3 standard deviations of the mean 95% within 2 standard deviations 68% within 1 standard deviation 34% 34% 2.4% 2.4% 0.1% 0.1% 13.5% x - 3s x - 2s 13.5% x-s x x+s x + 2s x + 3s 7 Probability of Half of a Distribution 0.5 0 8 Notation P(a < z < b) denotes the probability that the z score is between a and b P(z > a) denotes the probability that the z score is greater than a P (z < a) denotes the probability that the z score is less than a 9 Finding the Area to the Right of z = 1.27 This area is 0.1020 0 z = 1.27 10 To find: z Score the distance along horizontal scale of the standard normal distribution; refer to the leftmost column and top row of theTable Area the region under the curve; refer to the values in the body of the Table 11 Table E Standard Normal Distribution =1 µ=0 0 x z 12 Definition Standard Normal Deviation a normal probability distribution that has a mean of 0 and a standard deviation of 1 Area found in Table A2 Area = 0.3413 0.4429 -3 -2 -1 0 1 2 3 0 z = 1.58 Score (z ) 13 Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees. 14 Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees. Area = 0.4429 P ( 0 < x < 1.58 ) = 0.4429 0 1.58 The probability that the chosen thermometer will measure freezing water between 0 and 1.58 degrees is 0.4429. 15 Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water, and if one thermometer is randomly selected, find the probability that it reads freezing water between -2.43 degrees and 0 degrees. Area = 0.4925 P ( -2.43 < x < 0 ) = 0.4925 -2.43 0 The probability that the chosen thermometer will measure freezing water between -2.43 and 0 degrees is 0.4925. 16 Finding the Area Between z = 1.20 and z = 2.30 Area A is 0.1044 A 0 z = 1.20 z = 2.30 17 Other Normal Distributions 0 1 If or (or both), we will convert values to standard scores using Formula 5-2, then procedures for working with all normal distributions are the same as those for the standard normal distribution. z= x-µ 18 Converting to Standard Normal Distribution P (a) x 19 Converting to Standard Normal Distribution x- z= P P (a) x (b) 0 z 20 Probability of Weight between 143 pounds and 201 pounds z= x = 143 s = 29 143 201 201 - 143 29 = 2.00 Weight z 0 2.00 21 Probability of Weight between 143 pounds and 201 pounds OR - 47.72% of women have weights between 143 lb and 201 lb. x = 143 s = 29 143 201 Weight z 0 2.00 22 Finding a probability when given a z-score using the TI83 • • • • 2nd Distributions 2:normalcdf (lower bound, upper bound, mean, s.d.) 23 Finding a z - score when given a probability Using the Table 1. Draw a bell-shaped curve, draw the centerline, and identify the region under the curve that corresponds to the given probability. If that region is not bounded by the centerline, work with a known region that is bounded by the centerline. 2. Using the probability representing the area bounded by the centerline, locate the closest probability in the body of Table E and identify the corresponding z score. 3. If the z score is positioned to the left of the centerline, make it a negative. 24 Finding z Scores when Given Probabilities 95% 5% 5% or 0.05 0.45 0.50 0 1.645 (z score will be positive) Finding the 95th Percentile 25 Finding z Scores when Given Probabilities 90% 10% Bottom 10% 0.10 0.40 -1.28 0 (z score will be negative) Finding the 10th Percentile 26 Finding z Scores when Given Probabilities Using the TI83 • • • • 2nd Distributions 3:Invnorm (%,mean,s.d) 27 Cautions to keep in mind 1. Don’t confuse z scores and areas. Z scores are distances along the horizontal scale, but areas are regions under the normal curve. Table A-2 lists z scores in the left column and across the top row, but areas are found in the body of the table. 2. Choose the correct (right/left) side of the graph. 3. A z score must be negative whenever it is located to the left of the centerline of 0. 28 Finding z Scores when Given Probabilities 95% 5% 5% or 0.05 0.45 0.50 0 1.645 (z score will be positive) Finding the 95th Percentile 29 Finding z Scores when Given Probabilities 90% 10% Bottom 10% 0.10 0.40 -1.28 0 (z score will be negative) Finding the 10th Percentile 30 Procedure for Finding Values Using the Table and By Formula 1. Sketch a normal distribution curve, enter the given probability or percentage in the appropriate region of the graph, and identify the value(s) being sought. x 2. Use Table E to find the z score corresponding to the region bounded by x and the centerline of 0. Cautions: Refer to the BODY of Table E to find the closest area, then identify the corresponding z score. Make the z score negative if it is located to the left of the centerline. 3. Using the Formula, enter the values for µ, , and the z score found in step 2, then solve for x. x = µ + (z • ) 4. Refer to the sketch of the curve to verify that the solution makes sense in the context of the graph and the context of the problem. 31 Finding P10 for Weights of Women 10% 90% 40% x=? 143 50% Weight 32 Finding P10 for Weights of Women The weight of 106 lb (rounded) separates the lowest 10% from the highest 90%. 0.10 0.40 x = 106 -1.28 0.50 143 Weight 0 33 Forgot to make z score negative??? UNREASONABLE ANSWER! 0.10 0.40 x = 180 1.28 0.50 143 Weight 0 34 REMEMBER! Make the z score negative if the value is located to the left (below) the mean. Otherwise, the z score will be positive. 35 Definition Sampling Distribution of the mean the probability distribution of sample means, with all samples having the same sample size n. 36 Central Limit Theorem Given: 1. The random variable x has a distribution (which may or may not be normal) with mean µ and standard deviation . 2. Samples all of the same size n are randomly selected from the population of x values. 37 Central Limit Theorem Conclusions: 1. The distribution of sample x will, as the sample size increases, approach a normal distribution. 2. The mean of the sample means will be the population mean µ. 3. The standard deviation of the sample means n will approach 38 Practical Rules Commonly Used: 1. For samples of size n larger than 30, the distribution of the sample means can be approximated reasonably well by a normal distribution. The approximation gets better as the sample size n becomes larger. 2. If the original population is itself normally distributed, then the sample means will be normally distributed for any sample size n (not just the values of n larger than 30). 39 Notation the mean of the sample means µx = µ the standard deviation of sample mean x = n (often called standard error of the mean) 40 Distribution of 200 digits from Social Security Numbers Frequency (Last 4 digits from 50 students) 20 10 0 0 1 2 3 4 5 6 7 8 9 Distribution of 200 digits 41 x SSN digits 1 5 9 5 9 4 7 9 5 7 2 6 2 2 5 0 2 7 8 5 8 3 8 1 3 2 7 1 3 3 7 7 3 4 4 4 5 1 3 6 6 3 8 2 3 6 1 5 3 4 6 7 3 7 3 3 8 3 7 6 4 6 8 5 5 2 6 4 9 4.75 4.25 8.25 3.25 5.00 3.50 5.25 4.75 5.00 2 6 1 9 5 7 8 6 4 0 7 4.00 5.25 4.25 4.50 4.75 3.75 5.25 3.75 4.50 6.00 42 Frequency Distribution of 50 Sample Means for 50 Students 15 10 5 0 0 1 2 3 4 5 6 7 8 9 43 As the sample size increases, the sampling distribution of sample means approaches a normal distribution. 44 Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, a.) if one woman is randomly selected, find the probability that her weight is greater than 150 lb. b.) if 36 different women are randomly selected, find the probability that their mean weight is greater than 150 lb. 45 Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, a.) if one woman is randomly selected, the probability that her weight is greater than 150 lb. is 0.4052. 0.5 - 0.0948 = 0.4052 0.0948 = 143 = 29 0 150 0.24 46 Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, b.) if 36 different women are randomly selected, the probability that their mean weight is greater than 150 lb is 0.0735. z = 150-143 = 1.45 29 36 0.5 - 0.4265 = 0.0735 0.4265 x = 143 x = 4.83333 0 150 1.45 47 Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, a.) if one woman is randomly selected, find the probability that her weight is greater than 150 lb. P(x > 150) = 0.4052 b.) if 36 different women are randomly selected, their mean weight is greater than 150 lb. P(x > 150) = 0.0735 It is much easier for an individual to deviate from the mean than it is for a group of 36 to deviate from the mean. 48 Finding a z - score for a group Using TI83 • • • • 2nd Distributions 2:normalcdf (lower bound, upper bound, mean, s.e) – Replace s.d. with standard error 49