Basic HTML Version
Q
uantitative
R
easoning &
P
roblem
S
olving
308
© 2014 Pacific Crest
Step
Explanation
6.
Compute standard
deviation
Compute the square root of the variance in order to determine the standard
deviation.
2
65.1 8.068
σ
σ
= =
=
7.
Validate
Chebyshev’s theorem states that more than 75% of the data is within two
standard deviations of the mean. Check to see if the results satisfy this
condition.
The mean is 81 and the standard deviation is about 8. Two standard deviations would
be 16. 81 – 16 = 65 and 81 + 16 = 97. The lowest test score was 68 and the highest
test score was 92, so 100% of the data was within two standard deviations of the mean.
This result agrees with Chebyshev’s Theorem because 100% is more than 75%.
Note: In later statistics courses, students study two different types of variances. In the other case the
denominator is
n
when the variance is of an entire population. In most cases, the formula for
σ
2
that appears in this methodology is the appropriate one so that is what will be used in this
book. There are also tools available online that will perform the same calculations. See the
companion website for an example; try entering the test scores and then press “Calculate” to
validate the results above.
YOUR
TURN!
Scenario:
Mark Buehrle, of the Toronto Blue
Jays, has had a fairly consistent
career. The following table lists the
number of wins for each year from
2001 to 2013.
Compute the variance and standard
deviation for the number of wins.
Year
Wins
Year
Wins
2001
16
2008
15
2002
19
2009
13
2003
14
2010
13
2004
16
2011
13
2005
16
2012
13
2006
12
2013
12
2007
10
O
ops
! A
voiding
C
ommon
E
rrors
●
Substituting Variance for Standard Deviation
Example
: The calculation of the variance for the final exam was a mean of 80 and a variance
of 50. Based upon Chebyshev’s Theorem, 75% of the exam scores should then fall
between 80 ± (2 × 50) which we know is true because all the exam scores are between
0 and 100.
Why?
The standard deviation is the square root of 50, which is just over 7. The two standard
deviations would be 80 ± (2 × 7) or approximately 66 to 94 which is of course more
reasonable and expected.
●
Not connecting Units with Measure of Spread
Example
: A consumer advocacy agency was conducting a comparison of the gas consumption
among the 2014 model year fleet of cars and found that 34 different automobiles had
an average between 14.4 m.p.g. and 45.3 m.p.g. The standard deviation was 6.5.