Transcript Chapter 6

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.
66-95-99.7
● The normal model is distributed into areas under the curve for each zscore.
● 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
Homework Problem #41
b. What percent of adult women do you expect to have cholesterol levels
over 200 mg/dL?
normalcdf(200, ∞) = .3085 -> 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 35.89 and 40.5.
c.
At least how tall are the biggest 10% of kindergartners?
invNorm(.9) = 40.5 inches tall