The Law of Large Numbers
The nut does not fall far from the tree
 but in fact, how far away does it fall?
Suppose I roll a die 100 times, and find the percent of times I rolled a 4.
I expect about
of the rolls to be 4s.
1
6
Of course, in any particular series of 100 rolls, I don’t expect the percent to be
exactly
.
(In fact, it can’t be, because
of 100 rolls is not a whole number.)
But
1
6
1
6
how far away will my percent be?
To answer this question we calculate the standard
deviation of the percent.
But first, some technical details:
The jargon:
Repeating some action such as rolling a die and counting the number of times
you get a particular outcome is called “Bernouilli trials.”
(The trials must be
independent.)
The outcome you are counting is called a “success”  even if you are
counting failures!
The number of successes is a random variable called the “Binomial
random variable.”
The notation:
the number of trials (rolls) is n
the chance of a “success” on any one trial (roll) is p
the actual percent of “successes” in an actual series of trials (rolls) is
$
p
Then
is a random variable (It will come out differently for different actual series.)
$
p
(
29
μ
σ
$
$
p
p
p
p
p
n
=
=

1
The formulas for
and
have been found using algebra.
If you are interested, I
μ
$
p
σ
$
p
can show you the details.
Notice that the number of trials (n) appears in the denominator of the formula for the
standard deviation.
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 Spring '08
 Staff
 Statistics, Law Of Large Numbers, Probability, Standard Deviation, highest percent, keno

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