Chapter 6 Evan Johnson & Mike Bi, Period 1. Z-Scores ●The easiest way to compare two dissimilar values is to compare their standard deviations. ●Z-scores.

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Presentation transcript:

Chapter 6 Evan Johnson & Mike Bi, Period 1

Z-Scores ●The easiest way to compare two dissimilar values is to compare their standard deviations. ●Z-scores are used to determine how many standard deviations you are above the mean. ●We can shift the data (change the values on the x-axis) or rescale the data to provide a more accurate and readable representation of the data.

Z-Scores ●When a z-score is positive, the value is above the mean. ●When a z-score is negative, the value is below the mean. ●The absolute value of the z-score determines its extremity.

●The normal model is distributed into areas under the curve for each z- score. ●About 66% of the values fall within one standard deviation of the mean. ●About 95% of the values fall within two standard deviations of the mean. ●About 99.7% of the values fall within three standard deviations of the mean.

The Normal Model ●The Normal Model can show us how extreme a value is based off of its standard deviation from the mean. ●Find the area between two z-score points using the calculator function normalcdf(zleft, zright). ●Find the z-scores of a certain percentile with invNorm().

Homework Problem #41 Assume the cholesterol levels of adult American women can be described by a Normal model with a mean of 188 mg/dL and a standard deviation of 24. a.Draw and label the Normal model. Cholesterol (mg/dL)

Homework Problem #41 b.What percent of adult women do you expect to have cholesterol levels over 200 mg/dL? normalcdf(200, ∞) = > 30.85% c.What percent of adult women do you expect to have cholesterol levels between 150 and 170 mg/dL? normalcdf(150, 170) =.17 -> 17% d.Estimate the interquartile range of the cholesterol levels. invNorm(.75) - invNorm(.25) = 32 mg/dL e.Above what value are the highest 15% of women’s cholesterol levels? invNorm(.85) = 213 mg/dL

Homework Problem #43 Companies who design furniture for elementary school classrooms produce a variety of sizes for kids of different ages. Suppose the heights of kindergarten children can be described by a Normal model with a mean of 38.2 inches and a standard deviation of 1.8 inches. a.What fraction of kindergarten kids should the company expect to be less than 3 feet tall? normalcdf(-∞, 36) =.11 -> 11.1% b.In what height interval should the company expect to find the middle 80% of kindergartners? invNorm(.1) = 35.89, invNorm(40.5) In between and c.At least how tall are the biggest 10% of kindergartners? invNorm(.9) = 40.5 inches tall