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An Introduction to Statistics
Odd Solution for Statistics Lesson 7
- Graduating Math and Science Center students have a mean ACT score
of 29. Calculate the z-score for their mean
relative to the national mean of 21.0 and standard deviation of 4.7.
(29-21)/4.7 = 1.70
- Given the fact that 50% of a normally distributed data set is within
0.675 standard deviations of the mean, estimate Q1,
Q3, and the interquartile range for Center Senior ACT
scores, given also a mean of 29 and standard deviation of 3.0.
Would an ACT score of 36 be unusual for a Center student?
z=±0.675 = (x - 29)/3.0
x = 26.975 or 31.025
Q1=27.0 Q3=31.0
Q3 - Q1 = 31.0 - 27.0 = 4.0
29 + 2s = 29 + 6.0 = 35.0 < 36, so yes.
New info: mean = 28.3; s=2.6; 3 35's known.
- Calculate the z-score for the largest value in the above data set.
Is it an ordinary score? Is it an outlier?
Which definition works best?
z = (360,000,000 - 66400)/373000 = 965.00!
It is VERY unusual data value (outlier).
The hinge and quartile definitions fail!
- Using the data set: {0, 2, 4, 5, 6, 3, 6, 1, 1, 50}, as given in
the lesson, calculate its 5-number summary, using the quartiles.
minX=0 Q1=1
Median=3.5 Q3=6
maxX=50
- Using the fifty 1999 class of 2003 Algebra Diagnostic Test scores:
140, 122, 119, 99, 92, 90, 90, 88, 85, 82,
82, 81, 80, 80, 77, 74, 74, 73, 72, 71,
70, 70, 69, 69, 69, 68, 68, 68, 67, 66,
64, 64, 62, 60, 59, 59, 58, 58, 56, 56,
56, 56, 55, 54, 53, 53, 50, 47, 35, 32,
find P10, P90
and the 10-90 percentile range. Show all your work.
L10 = (10/100)(50) = 5 (53 + 53)/2 = 53 = P10
L90 = (90/100)(50) = 45 (90+92)/2 = 91.0 = P90