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Advanced Math Topics
3.6 Chebyshev’s Theorem:
Applying the Standard Deviation
We learned how to calculate the standard deviation (s) and we learned that
a high value of s means the data is spread far from the mean and a low
value of s means the data is close to the mean. This next theorem gives us more
concrete information of this idea.
Chebyshev’s Theorem
In any population or sample…
Let k be any number equal to or greater than 1,
Then the proportion of any distribution that lies
Within k standard deviations of the mean is at least…
1-
1
k2
What does this theorem mean?
Computing some values of k…
1
k
1-
(Std. deviations)
What does this look like?
Remember k = the standard deviation = s
k2
1
1
1-
12
= 0 = 0%
At least 75%
of the data
will lie in
this interval.
1
2
1-
22
= 3/4 = .75 = 75%
1
3
1-
32
= 8/9 = .8889 = 88.89%
x – 2s
x
x + 2s
At least 88.89%
of the data
will lie in
this interval.
x – 3s
x
x + 3s
What if we had values for the mean and standard deviations?
x = 50
s = 18
Let’s use k = 2.
75
Then at least ______%
of the data
will lie within 2 standard
deviations of the mean .
75% of the data or more lie within
the interval x – 2s and x + 2s
50 – 2(18) and 50 + 2(18)
14 and
14
50
86
Chebyshev’s Theorem states that at least 75% of the data will lie in this interval.
86
1) In one month, the average file claimed with an insurance
company was $831.
The standard deviation was $150.
Find the interval that will contain all values that are within
2.5 standard deviations of the mean.
831 – 2.5(150)
and
$456
and
831 + 2.5(150)
$1206
At least what % of the data will lie between $456 and $1206?
1
1= .84 = 84%
2
(2.5)
At least 84% of the claims will be in the interval $456-$1206.
2) All G.E. light bulbs have a mean life of 1600 hours with a standard
deviation of 100 hours. At least what % of the bulbs will have a
mean life between 1450 and 1750 hours?
How far is the upper and lower limit of the interval from the mean?
1750 – 1600 = 150
and
1600 – 1450 = 150
How many standard
deviations from the
mean is this?
We know that k standard deviations is 150 thus k(100) = 150.
k = 1.5
1
1= .5556 = 55.56%
2
(1.5)
At least 55.56% of the bulbs will have a life in the interval of 1450-1750 hours.
Pafnuty Lvovich Chebyshev
Born
May 16 [O.S. May 4] 1821
Borovsk, Kaluga, Russia
Died
December 8 [O.S. November 26] 1894
St Petersburg, Russia
Residence
Russia
Nationality
Russian
Field
Institutions
Mathematician
St Petersburg University
HW
 P. 139 #1-5
Let’s do #1 together.
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