Download Z-Scores Exercises

Z-Scores1 Exercises
Interpreting Z-scores: A Z-score tells us “how many standard deviations above/below2 the
mean.” Every time you obtain a Z-score, interpret it. Use “above” or “below” in your phrasing.
1) The yields of apples trees in an orchard have Normal distribution with mean 42.4 pounds and
standard deviation 6.1 pounds.
a) Identify the variable and units.
b) What does the Empirical Rule (page 106) say about yields of trees?
c) Compute the Z-score for a tree yielding 35 pounds of apples. Interpret this Z-score.
d) Suppose a tree has a Z-score of 1.85. Interpret this Z-score. What is the yield of this tree?
2) Consider all the trees in the apple orchard. The yield of these trees has mean 42.4 pounds
with standard deviation 6.1 pounds. Suppose each tree’s yield is converted to a Z-score. (The
“new” – derived – variable is “Z-score for weight”; the units are still the apple trees.) For the
data set of all Z-scores: What is the shape? The mean? The standard deviation?
3) Chris’ web page viewing varies from day to day with mean 280 pages and standard deviation
260 days.
a) Identify the variable and units.
b) Compute the Z-score for a day in which Chris views 500 web pages. Interpret this Zscore.
c) Suppose a day has a Z-score of -0.85. Interpret this Z-score. How many web pages did
Chris view on this day?
d) Could a day have a Z-score of -1.5? How many web pages would Chris view on such a
day? Use your answer to argue that the distribution of web pages viewed is not Normal.
4) Magnus takes the SAT Math and scores 730. He is applying to the University of Southern
North Dakota (USND), where the mean score on the SAT Math is 585 with standard
deviation 95.
a) Identify the variable and units (of which Magnus is one).
b) Compute the Z-score for Magnus’s SAT Math score. Interpret this Z-score.
c) Magnus is thinking about taking the ACT. From the admissions office at USND he learns
that the mean score on the ACT Math component is 24.50 with a standard deviation of
4.75. What ACT score should Magnus anticipate? (Assume that this ACT score would
have the same Z-score as that for his SAT.)
5) Consider SAT Math scores at UNSD for which the mean is 585 with standard deviation 95.
The distribution is slightly left skewed. Suppose each student’s SAT Math is converted to a
Z-score. (The “new” – derived – variable is “Z-score for SAT Math”; the units are still the
students.) For the distribution of all Z-scores: What is the shape? The mean? The standard
deviation?
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Z-scores are also referred to as “standardized values.”
Negative Z-scores go with data below the mean.
6) Among females, Becky’s weight has a Z-score of -2.20. Weights are approximately Normal
in shape. What does this tell you? Is Becky heavier or lighter than most women? By a lot?
Use the Empirical Rule to judge. What can you say about Becky’s percentile rank among
women?
Solutions
1)
a) Variable = weight (or yield); units = apple trees
b) 68% have weight between 36.3 and 48.5 pounds; 95% between 30.2 and 54.6; almost all
between 24.1 and 60.7 pounds.
c) Z = (35 – 42.4) / 6.1 =-7.4/6.1 = -1.21. This tree’s yield is 1.21 standard deviations below
the mean yield.
d) This tree’s yield is 1.85 standard deviations above the mean yield. X = 42.4 + 1.85(6.1) =
53.685 pounds.
2) The shape is Normal (Z-scores have the same shape distribution as does the data); the mean
Z-score is 0 (always) the standard deviation is 1 (always).
3) Chris’ web page viewing varies from day to day with mean 280 pages and standard deviation
260 days.
a) The variable is number of pages viewed; the units are days.
b) (500 – 280) / 260 = 220 / 260 = 0.85. 500 web pages viewed is 0.85 standard deviations
above the mean.
c) This value is 0.85 standard deviations below the mean. X = 280 – 0.85(260) = 280 – 221
= 59 pages. This value is as far below (220) the mean of 280 as 500 is above (this slight
difference in published answers is due to rounding error). If Z-scores are opposites, the
corresponding data values are equally far from the mean – one above the other below.
d) X = 280 – 1.5(260) = 280 – 390 = -110. This is impossible. Since a Normal distribution
allows for a Z-score of -1.5 (a value 1.5 standard deviations below the mean), the
distribution of web page views cannot be Normal. Not even close, actually.
4)
a) The variable is SAT Math score; the units are students at USND.
b) Z = (730 – 585) / 95 = 1.53. Magnus’ score is 1.53 standard deviations above the mean.
c) X = 24.50 + 1.53(4.75) = 31.75 is the score Magnus can anticipate.
5) The distribution of Z-scores is [always the same] slightly left skewed with mean [always for
Z-scores] 0 and standard deviation [always for Z-scores] 1.
6) Becky’s weight falls 2.2 standard deviations below the mean. She is rather light. Only 2.5%
of women are more than 2 standard deviations below the mean. Becky’s percentile rank is
less than 2.5! (Her weight falls at about the 1.39 percentile, with only 1.4% of women lighter
and (about 1 in every 72) and 98.61% heavier.