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The Standard Normal Distribution Section 2.2.1
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Starter Weights of adult male Norwegian Elkhounds are N(42, 2) pounds. What weight would represent the 16 th percentile?
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Today’s Objectives State mean and standard deviation for The Standard Normal Distribution. Given a raw score from a normal distribution, find the standardized “z- score”. Use the Table of Standard Normal Probabilities to find: 1) the area between given borders of the Standard Normal curve. 2)The raw score (x value) that leads to a given area.
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Standardizing Observations All normal distributions have fundamentally the same shape. If we measure the x axis in units of size σ about a center of 0, then they are all exactly the same curve. This is called the Standard Normal Curve To standardize observations, we change from x values (the raw observations) to z values (the standardized observations) by the formula:
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The Standard Normal Distribution Notice that the z-score formula always subtracts μ from each observation. –So the mean is always shifted to zero Also notice that the shifted values are divided by σ, the standard deviation. –So the units along the z-axis represent numbers of standard deviations Thus the Standard Normal Distribution is always N(0,1).
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Example 2.3 (p 84) The heights of young women are: N(64.5, 2.5) Use the formula to find the z-score of a woman 68 inches tall. A woman’s standardized height is the number of standard deviations by which her height differs from the mean height of all young women.
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Normal Distribution Calculations What proportion of all young women are less than 68 inches tall? (Examples 2.4 & 2.5, p 86) –Notice that this does not fall conveniently on one of the σ borders, so the empirical rule does not help! We already found that 68 inches corresponds to a z-score of 1.4 So what proportion of all standardized observations fall to the left of z = 1.4? Since the area under the Standard Normal Curve is always 1, we can ask instead what is the area under the curve and to the left of z=1.4 –For that, we need a table.
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The Standard Normal Table Find Table A on pages 13 & 14 of the handout –It is also in your textbook inside the front cover Z-scores (to the nearest tenth) are in the left column –The other 10 columns round z to the nearest hundredth Find z = 1.4 in the table and read the area –You should find area to the left =.9192 So the proportion of observations less than z = 1.4 is about 92% –Now put the answer in context: “About 92% of all young women are 68 inches tall or less.”
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What about area above a value? Still using the N(64.5, 2.5) distribution, what proportion of young women have a height of 61.5 inches or taller? Z = (61.5 – 64.5)/2.5 = -1.2 From Table A, area to the left of -1.2 =.1151 –So area to the right = 1 -.1151 =.8849 So about 88% of young women are 61.5” tall or taller.
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What about area between two values? What proportion of young women are between 61.5” and 68” tall? We already know 68” gives z = 1.4 and area to the left of.9192 We also know 61.5” gives z = -1.2 and area to the left of.1151 So just subtract:.9192 -.1151 =.8041 So about 80% of young women are between 61.5” and 68” tall –Remember to write your answer IN CONTEXT!!!
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Given a proportion, find the observation x (Ex. 2.8, p 90) SAT Verbal scores are N(505, 110). How high must you score to be in the top 10%? If you are in the top 10%, there must be 90% below you (to the left). Find.90 (or close to it) in the body of Table A. What is the z-score? –You should have found z = 1.28 Now solve the z definition equation for x So you need a score of at least 646 to be in the top 10%.
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Today’s Objectives State mean and standard deviation for The Standard Normal Distribution. Given a raw score from a normal distribution, find the standardized “z- score”. Use the Table of Standard Normal Probabilities to find: 1) the area between given borders of the Standard Normal curve. 2)The raw score (x value) that leads to a given area.
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Homework Read pages 83 – 91 Do problems 22 – 25
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