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Think about this…. If Jenny gets an 86% on her first statistics test, should she be satisfied or disappointed? Could the scores of the other students in the class affect her feelings? In this chapter, we will investigate the individual observations (data values) in comparison to the entire distribution.
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Empirical Rule: For data with a symmetric, bell shaped distribution, the standard deviation has the following characteristics:
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The distribution of heights of adult American males is approximately normal with a mean of 69 inches and a standard deviation of 2.5 inches. 1) What percent of men are taller than 74 inches? 2)Between what heights do the middle 95% of men fall? 3)What percent of men are shorter than 66.5 inches? 4) A height of 71.5 inches corresponds to what percentile of adult American heights?
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If a distribution is fairly symmetric and can be “idealized” by a normal distribution curve, the distribution can be symbolized by the function, N( ). Standard Scores (z-scores) are based upon the Standard Normal Distribution of N(0,1) where the mean is 0 and a standard deviation of 1. This scale shows the z- score values (# of standard deviations from the mean)
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Properties of a Density Curve Always on or above the horizontal axis Has an area of exactly 1 under the curve Describes the overall pattern of a distribution Because a density curve is an “idealized” description of the distribution, the density curve can be used to generalize from a sample to a population The mean of a density curve is symbolized and the standard deviation is symbolized
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Here is the list of the 25 students in Jenny’s class: 79 81 80 77 73 83 74 93 78 80 75 67 73 77 83 86 90 79 85 83 89 84 82 77 72 If we wanted to compare Jenny’s score (86) to the class, what could we do? observation (value) mean standard deviation
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Putting the z-scores on a standard normal distribution together with the fact that the area under the curve represents the percent of data that falls within a particular interval… we can use z-scores to calculate exact percentiles of an observation. To find these percentiles, we can use a table. This is near the front cover of your textbook. This shaded area would equal the percentile of that z-score What if the value is not EXACTLY at a whole standard deviation?
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What would be the percentile for a z-score of 1.15? This means that an observation with a z-score of 1.15, would fall in the 87.49 th percentile.
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2nd VARS (Distr), #2 (normalcdf) normalcdf(lowest x value, highest x-value, , ) Use -9999 for - and 9999 for + If you given a z-score instead of x values, use = 0 & = 1. Finding Percentiles of Normal Distributions with a Calculator For Jenny’s stat test, you would enter: Normalcdf (0, 86, 80, 6.07)
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Using a standard normal distribution, we can relate an observations standard score with its percentile. This shaded area would equal the percentile of that z-score If you know the percentile, how would you find the value of the observation?
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What would be the z-score for an observation that had a percentile value of approximately 92% ? Once you know the z-score, you can use the formula to find the actual observation value…but you must know the original data’s mean and standard deviation Find the percentile that is CLOSEST to the desired percent
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Example: The mean score on a test was 76 with a standard deviation of 10.4. If a teacher wishes to give awards to those students in approximately the top 8%, what is cutoff score for those students? top 8% = 92% to the left of the cutoff score
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Finding a Value given a Percentile for a Normal Distribution with a Calculator 2nd VARS (Distr), #3 (invnorm) Invnorm(percentile or area to the left, , ) For the previous example you would enter: InvNorm(.08, 76, 10.4)
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