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Chapter 6 The Normal Distribution Chapter 6 The Normal Distribution The Normal Distribution The Standard Normal Distribution Applications of Normal Distributions Sampling Distributions The Central Limit Theorem (CLT) Probability Distributions of Continuous random variables X 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 P(X) 0.000075339 0.000828733 0.004419907 0.015224123 0.038060308 0.073583262 0.114462852 0.147166524 0.159430401 0.147620742 0.118096593 0.082309747 0.050300401 0.027084831 0.012897539 0.005445627 0.002042110 0.000680703 0.000201690 0.000053076 0.000012384 0.000002556 0.000000465 0.000000074 0.000000010 0.000000001 0.000000000 0.000000000 0.000000000 0.000000000 0.000000000 0.000000000 0.000000000 0.000000000 Recall P( X 11) 0.08231 Guessing 33 Answers on a Test Data Sample size Probability of success 33 0.25 Statistics Mean Variance Standard deviation 8.25 6.1875 2.487469 Porbality Distribution of Total Number of Right Answers Probability Distribution of Total Number of Right Answers 0.180000000 0.160000000 0.140000000 P( X 11) .0823 1 area 0.120000000 P(X) 0.100000000 0.080000000 0.060000000 0.040000000 0.020000000 0.000000000 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 Number of Successes X 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 P(X) 0.000075339 0.000828733 0.004419907 0.015224123 0.038060308 0.073583262 0.114462852 0.147166524 0.159430401 0.147620742 0.118096593 0.082309747 0.050300401 0.027084831 0.012897539 0.005445627 0.002042110 0.000680703 0.000201690 0.000053076 0.000012384 0.000002556 0.000000465 0.000000074 0.000000010 0.000000001 0.000000000 0.000000000 0.000000000 0.000000000 0.000000000 0.000000000 0.000000000 0.000000000 Recall P(8 X 11) area four rectangles 0.50746 Guessing 33 Answers on a Test Data Sample size Probability of success 33 0.25 Statistics Mean Variance Standard deviation 8.25 6.1875 2.487469 Porbality Distribution of Total Number of Right Answers Probability Distribution of Total Number of Right Answers 0.180000000 P(8 X 11) area four rectangles 0.50746 0.160000000 0.140000000 0.120000000 P(X) 0.100000000 0.080000000 0.060000000 0.040000000 0.020000000 0.000000000 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 Number of Successes Continuous Probability Distribution A continuous random variable has an infinite number of possible values that can be represented by an interval on the number line. For example Average Grade in Math 1111 0 20 40 60 80 90 100 Average Grade can be any number between 0 and 100. The probability distribution of a continuous random variable is called a continuous probability distribution. Relative Frequency Histogram Average Final Grade for 60 students Relative Frequency Histogram Average Final Grade for 450 students Relative Frequency Histogram Average Final Grade for 1260 students Relative Frequency Histogram Average Final Grade for 1961 students Relative Frequency Histogram Average Final Grade for 1961 students Continuous Probability Distribution Average Final Grade for all students Data World Theory World Density Curve or Density Function Mathematical Model f ( x) x f ( x)dx ( x )2 f ( x)dx 2 It is mathematically easier to work with such a curve instead of a histogram Density Curve or Density Function A density curve is a curve that 1) is always on or above the horizontal axis. 2) has area exactly equal to 1 underneath it. A density curve describes the overall pattern of a continuous distribution. The area under the curve and above any range of values is the relative frequency of all observations that fall in that range (see picture in next slide). Density Curve or Density Function P(a X b) is equal to this area a b X Shaded area equals percentage of observations that fall between a and b Examples of Density Functions Normal Distributions and the Standard Distribution Properties of Normal Distributions The most important probability distribution in statistics is the normal distribution. Normal curve x A normal distribution is a continuous probability distribution for a random variable, X. The graph of a normal distribution is called the normal curve. 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. 5) Between μ σ and μ + σ the graph curves downward. The graph curves upward to the left of μ σ and to the right of μ + σ. Properties of Normal Distributions Inflection points Total area = 1 μ 3σ μ 2σ μσ μ μ+σ μ + 2σ μ + 3σ x If X is a continuous random variable having a normal distribution with mean μ and standard deviation σ, you can graph a normal curve with the equation 1 -(x - μ )2 2σ 2 y= e σ 2 Example of Normal Distribution The following table displays a frequency and a relativefrequency distribution for the heights of the 3264 female students who attend a Midwestern college. The figure displays a relative-frequency histogram for the same data. Included is a shaded area between 67” and 68”. This area is P(67 Height 68) = 0.0735 We now included a normal curve that approximates the overall shape of the distribution. The area under the curve. between 67 and 68 is P(67 Height 68) = 0.0735 Means and Standard Deviations A normal distribution can have any mean and any positive standard deviation. Inflection points 1 2 3 4 5 6 Inflection points The mean gives the location of the line of symmetry. x 1 2 3 4 5 6 7 8 9 10 11 Mean: μ = 3.5 Mean: μ = 6 Standard deviation: σ 1.3 Standard deviation: σ 1.9 The standard deviation describes the spread of the data. x Means and Standard Deviations Example: 1. Which curve has the greater mean? 2. Which curve has the greater standard deviation? B A 1 3 5 7 9 11 13 x The line of symmetry of curve A occurs at x = 5. The line of symmetry of curve B occurs at x = 9. Curve B has the greater mean. Curve B is more spread out than curve A, so curve B has the greater standard deviation. The Standard Normal Distribution The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. The horizontal scale corresponds to z-scores. 3 2 1 0 1 2 3 z If a variable X has a normal distribution with mean μ and standard deviation σ, then the standardized variable X -μ Z = σ has the standard normal distribution 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 Area is close to 1. Area is close to 0. z = 3.49 3 z 2 1 0 z=0 1 Area is 0.5000. 2 3 z = 3.49 Areas Under the Standard Normal Curve The Standard Normal Table P(Z z) = The Standard Normal Table Example - Find P(Z 1.40) P(Z 1.40)=0.9192 Finding the Area to the Left of a Specified z-Score Finding P(Z 1.23), that is, finding the area under the standard normal curve to the left of 1.23 Always draw the curve! Finding the Area to the Left of a Specified z-Score Finding P(Z 1.23), that is, finding the area under the standard normal curve to the left of 1.23 Finding the Area to the Right of a Specified z-Score Finding P(Z ≥ 0.76), that is, finding the area under the standard normal curve to the right of 0.76 Always draw the curve! Finding the Area to the Right of a Specified z-Score Finding P(Z ≥ 0.76), that is, finding the area under the standard normal curve to the right of 0.76 Find the Area Between Two Specified z-Scores Finding P(-0.68 Z 1.82), that is, finding the area under the standard normal curve that lies between –0.68 and 1.82 Always draw the curve! Find the Area Between Two Specified z-Scores Finding P(-0.68 Z 1.82), that is, finding the area under the standard normal curve that lies between –0.68 and 1.82 Summary Using Table II to find the area under the standard normal curve that lies (a) to the left of a specified z-score, (b) to the right of a specified z-score, (c) between two specified z-scores Find the z-Score Having a Specified Area to Its Left Finding the z-score having area 0.04 to its left, that is, finding z, so that P(Z z)=0.04 Always draw the curve! Second decimal place for z-score z score Inside the table, look for the best approximation to the given area Find the z-Score Having a Specified Area to Its Left Finding the z-score having area 0.04 to its left, that is, finding z, so that P(Z z)=0.04 Useful Notation Example-Find z0.025 Example-Find z0.025 Finding the z-Scores for a Specified Area Find the two z-scores that divide the area under the standard normal curve into a middle 0.95 area and two outside 0.025 areas, as depicted in the figure. Finding the z-Scores for a Specified Area Find the two z-scores that divide the area under the standard normal curve into a middle 0.95 area and two outside 0.025 areas, as depicted in the figure. Normal distribution calculations Three nonstandard normal distributions Standardizing three nonstandard normal distributions Normal Distribution Calculations The process of Standardizing preserves the areas in the sense indicated in above. Normal Distribution Calculations and areas are probabilities, that is Normal Distribution Calculations P ( a X b) P ( a X b ) a X b P b a P Z Normal Distribution Calculations P ( a X b) P ( a X b ) a X b P b a P Z Normal Distribution Calculations P ( a X b) P ( a X b ) a X b P b a P Z Normal Distribution Calculations P ( a X b) P ( a X b ) a X b P b a P Z Normal Distribution Calculations b a P ( a X b) P Z z-score for a z-score for b Percentage of people having IQs between 115 and 140 Percentage of people having IQs between 115 and 140 Percentage of people having IQs between 115 and 140 P(115 X 140) P 0.94 Z 2.50 0.1674 Conclusion: The percentage of people having IQs between 115 and 140 is 16.74% More Examples on Finding Probabilities Probability and Normal Distributions Example: The average on a statistics test was 78 with a standard deviation of 8. If the test scores are normally distributed, find the probability that a student receives a test score less than 90. μ = 78 σ=8 z x σ- μ = 90-78 1.5 8 P(X < 90) μ =78 90 μ =0 ? 1.5 P(X < 90) = P(Z < 1.5) = 0.9332 x z The probability that a student receives a test score less than 90 is 0.9332. Probability and Normal Distributions Example: The average on a statistics test was 78 with a standard deviation of 8. If the test scores are normally distributed, find the probability that a student receives a test score greater than than 85. z = x σ- μ = 85-78 0.88 8 μ = 78 σ=8 P(X > 85) μ =78 85 μ =0 0.88 ? x z The probability that a student receives a test score greater than 85 is 0.1894. P(X > 85) = P(Z > 0.88) = 1 P(Z < 0.88) = 1 0.8106 = 0.1894 Probability and Normal Distributions Example: The average on a statistics test was 78 with a standard deviation of 8. If the test scores are normally distributed, find the probability that a student receives a test score between 60 and 80. z1 = x σ- μ = 608- 78 2.25 P(60 < X < 80) z 2 x σ- μ = 808- 78 0.25 μ = 78 σ=8 60 μ =78 80 2.25 μ =0 0.25 ? ? x z The probability that a student receives a test score between 60 and 80 is 0.5865. P(60 < X < 80) = P(2.25 < Z < 0.25) = P(Z < 0.25) P(Z < 2.25) = 0.5987 0.0122 = 0.5865 Normal Distributions: Finding Values Finding a z-Score Given a Percentile Example: Find the z-score that corresponds to P75. That is P75 = 0.67 Area = 0.75 μ =0 ? 0.67 z The z-score that corresponds to P75 is the same z-score that corresponds to an area of 0.75. 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 + z Example: The monthly electric bills in a city are normally distributed with a mean of $120 and a standard deviation of $16. Find the x-value corresponding to a z-score of 1.60. x μ + z = 120 +1.60(16) = 145.6 We can conclude that an electric bill of $145.60 is 1.6 standard deviations above the mean. Finding a Specific Data Value Example: The weights of bags of chips for a vending machine are normally distributed with a mean of 1.25 ounces and a standard deviation of 0.1 ounce. Bags that have weights in the lower 8% are too light and will not work in the machine. What is the least a bag of chips can weigh and still work in the machine? P(Z < ?) = 0.08 8% P(Z < 1.41) = 0.08 ? 1.41 z 0 x ? 1.25 1.11 x μ + z 1.25 (1.41)0.1 1.11 The least a bag can weigh and still work in the machine is 1.11 ounces.