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An important application of the
central limit theorem is

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in the approximate calculation
of the binomial probabilities.

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Here is what is involved.

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We start with random
variables--

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Xi--

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that are independent.

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And they have the same
distribution.

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They're all Bernoulli
with parameter p.

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We add n of those random
variables, and the resulting

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random variable, Sn, we know
that it has a binomial PNF

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with parameters n and p.

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We also know its mean, and
we do know its variance.

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What the central limit theorem
tells us, in this case, since

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we're dealing with the sum of
independent identically

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distributed random variables,
is the following.

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If we take this random variable
here that we have

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been denoting by Zn, which is a
standardized version of Sn--

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we subtract the mean of Sn and
divide by the standard

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deviation--

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this random variable has a CDF
that approaches as n goes to

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infinity, the CDF of
a standard normal.

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So let us use what we now
know to calculate some

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probabilities.

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Let us fix some parameters.

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n is 36.

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p is 0.5.

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And we wish to calculate the
probability that Sn is less

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than or equal to 21.

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Now, in this case, we can
calculate it exactly using the

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binomial formula.

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The probability of being less
than or equal to 21 is the sum

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of the probabilities of all
the numbers from 0 to 21.

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And this is the probability
of obtaining a number k.

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And by calculating this
expression, we obtain this

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number, which is the
exact answer.

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Now, let us proceed using the
central limit theorem.

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We are interested in this
probability, but we will use

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the fact about the CDF of this
related random variable.

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So the first step is
to calculate n

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times p, which is 18.

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The second step is to calculate
this denominator

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here, which in our case
evaluates to 3.

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Now, since we know something
about the CDF of this random

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variable, what we need to do
is to take this event and

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rewrite it in terms of
this random variable.

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So we have the event of
interest, which is that Sn is

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less than or equal to 21.

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This is the same as the event
that Sn minus 18 is less than

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or equal to 21 minus 18.

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And it's the same as this event
here, where we divide

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both sides by 3.

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Now, what we have here is the
probability that this random

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variable Zn is less than
or equal to 1.

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But now, Zn is approximately a
standard normal, so we can use

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here the CDF of the standard
normal distribution,

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which is Phi of 1.

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And at this point, we look at
the tables for the normal

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distribution.

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We'll find this entry here.

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And this gives us an
answer of 0.8413.

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This is a pretty good
approximation of the exact

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answer, which is 0.8785.

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But it is not a great
approximation.

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It is off by about four
percentage points.

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Can we do better than that?

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It turns out that we can get
a better approximation.

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And let us see how
this can be done.

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Recall that we approximated
this probability using the

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central limit theorem and found
this numerical value.

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But we make an observation that
this probability is equal

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to this probability here.

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Why is that?

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Sn is an integer random
variable.

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Therefore, if I tell you that
it is strictly less than 22,

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I'm also telling you that
it is 21 or less.

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Therefore, this event here is
the same as that event here.

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And therefore, their
probabilities are the same.

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So instead of using the central
limit approximation to

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calculate this probability,
let us follow the same

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procedure but try to calculate
this probability here.

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And this probability here
is equal to the

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probability that Sn minus--

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we subtract the mean, divide
by the standard

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deviation of Sn--

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is strictly less than 22 minus
18 divided by 3, which is the

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probability that the random
variable that we denote by Zn,

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which is this expression here,
is strictly less than 22

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minus 18 over 3.

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And this is 1.33.

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Now, at this point, we pretend
that Zn is a standard normal

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random variable--

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the probability that the
standard normal is less than a

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certain number.

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This is the standard normal CDF
evaluated at that number.

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And then we look up at the
normal tables at 1.33 and we

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find this value of 0.9082.

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Now, we compare this value
with the exact

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answer for this problem.

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And we see that we
again missed it.

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Using this approximation to
this quantity gave us an

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underestimate of this number.

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Now, we obtained an
overestimate.

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The true value is somewhere
in the middle.

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So this suggests that we may
want to do something that

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combines these two alternative
choices here.

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But before doing that, it's
good to understand what

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exactly have we be
doing all along.

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What we're doing is
the following.

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We have the PMF of the binomial
centered at 18, which

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is the mean.

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It's a discrete random
variable.

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But when we use the central
limit theorem, we pretend that

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the binomial is normal, but
while we keep the same mean

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and variance.

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Now, when we calculate
probabilities, if we want to

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find the discrete probability
that Sn is less than or equal

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to 21, which is the sum of these
probabilities, what we

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do is we look at the area
under the normal

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PDF from 21 and below.

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In the alternative approach,
when we use the central limit

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theorem to approximate the
probability of this event, we

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go to 22, and we look at the
event of falling below 22.

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This means that we're looking at
the area from 22 and lower.

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So in one approach, this
particular region is not used

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in the calculation.

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That's what we did here.

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But in the second approach, it
was used in the calculation.

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Should it be used or not?

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It makes more sense to use only
part of this solid region

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and assign it to the calculation
of the probability

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of being at 21 or less.

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Namely, we can take the mid
point here, where the mid

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point is at 21.5, and calculate
the area under the

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normal PDF only going
up to 21.5.

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What this amounts to is looking
at this particular

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event here.

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Now, this event is, of course,
identical to this event that

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we have been considering,
because again, Sn is a

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discrete random variable that
takes integer values.

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But when we approximate it by
a normal, it does make a

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difference whether we
write the event

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this way or that way.

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So here, we're going to obtain
the probability that the

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standardized version
of Zn is less than.

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We follow the same calculation,
but now we have

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21.5 minus 18 divided by 3.

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And this number here is 1.17.

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And using the central limit
theorem calculation, this is

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the CDF of the standard normal
evaluated at 1.17, which we

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can go and look up in the
normal table to find

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the value of 0.8790.

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And now, we notice that this
value is remarkably close to

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the true value.

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It is much better as an
approximation that what we

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obtained using either this
choice or that choice.

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And since this approximation
is so good, we may consider

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even using it to approximate
individual probabilities of

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the binomial PMF.

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Let's see what that takes.

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Let us try to approximate, as
an example, the probability

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that Sn takes a value
of exactly 19.

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So what we will do will be to
write the event that Sn is

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equal to 19 as the event that Sn
lies between 18.5 and 19.5.

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In terms of the picture that
we were discussing before,

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what we are doing, essentially,
is to take the

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area under the normal PDF that
extends from 18.5 to 19.5 and

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declare that this area
corresponds to the discrete

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event that our binomial random
variable takes a value of 19.

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Similarly, if we wanted to
calculate approximately the

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value of the probability that
Sn takes a value of 21, we

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would consider the area
under the normal PDF

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from 20.5 to 21.5.

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So let us now continue
with this approach.

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We do the usual calculations,
which is to express this event

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in terms of standardized
values.

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That is, we subtract throughout
the mean of Sn and

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divide by standard deviation.

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So what we obtain here is the
standardized version of Sn.

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And that has to be, now, less
than or equal to 19.5 minus 18

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divided by 3, which is the
probability that our

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standardized random variable
lies between 0.17 and 0.5.

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And now, if we pretend that Zn
is a standard normal random

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variable, which is what the
central limit theorem

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suggests, this is going to be
equal to the probability that

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the standard normal is less than
or equal to 0.5 minus the

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probability that it
is less than 0.17.

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And if we look up those entries
in the normal tables,

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what we find is an answer of
0.6915 minus this number,

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which evaluates to 0.124.

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And what is the exact answer if
we were to use the binomial

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probability formulas?

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The exact answer is remarkably
close to what we obtained in

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our approximation.

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This example illustrates a more
general fact that this

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approach of calculating
individual entries of the

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binomial PMF gives very
accurate answers.

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And in fact, there are
theorems, there are

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theoretical results to this
effect, that tell us that this

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way of approximating--

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asymptotically, as n goes
to infinity and

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in a certain regime--

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does give us very accurate
approximations.