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We will now go through a
sequence of examples that

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illustrate the different types
of questions that we usually

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answer using a normal
approximation based on the

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central limit theorem.

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In general, one uses these
approximations to make

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statements of this type.

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That the probability of
the sum of n, i.i.d.

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random variables being less than
a certain number, that

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this probability is
approximately equal to some

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other number.

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Notice that this statement
involves three parameters, a,

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b, and n, and you can imagine
problems where you are given

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two of these parameters,
and you're

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asked to find the third.

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And this gives us the different
variations of the

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questions that we might
be able to answer.

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So we will go through examples
of each one of these

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variations.

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Our setting will
be as follows.

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We have a container, and the
container receives packages.

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Each package has a random
weight, which is an

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independent random variable
that's drawn from an

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exponential distribution
with a parameter 1/2.

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And we load the container
with 100 packages.

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We would like to calculate the
probability that the total

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weight of the 100 packages
exceeds 210.

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For example, 210 might be the
capacity of the container.

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Since we will be using the
central limit theorem, we will

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have to work with the
standardized version of Sn in

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which we subtract the mean of Sn
and divide by the standard

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deviation of Sn.

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And to do that, we will need
to know the mean and the

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standard deviation.

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Now for an exponential, the mean
is the inverse of lambda

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and the standard deviation is
also the inverse of lambda and

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so we know what these
quantities are.

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Then, the next step is to take
this event here and rewrite it

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in a way that involves
this random variable.

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So what we would do is that we
take the original description

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of the event, subtract from both
sides of this inequality

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this number n times mu.

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In this case, n is
100, mu is 2.

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So we subtract 200, divide by
this quantity, square root of

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100 is 10 times sigma,
this gives us 20.

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And we do the same on the other
side of the inequality.

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This is just an equivalent
representation of the original

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event, but we have here is
the probability that this

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standardized version of Sn is
larger than or equal to this

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number, which is 0.5.

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00:03:01,580 --> 00:03:04,880
And at this point, we can use
the central limit theorem

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approximation to say that this
probability is approximately

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the same if we use a standard
normal instead of Zn.

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Now for a standard normal, we
can calculate probabilities in

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terms of the CDF that's
given in the table.

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But here, we have the
probability that Z is larger

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than something, not smaller
than something.

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The CDF gives us the probability
that Z is less

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than something.

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This is easy to handle.

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This probability is 1 minus the
probability that Z is less

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than 0.5, which is 1 minus the
CDF of the standard normal

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evaluated at 0.5.

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And at this point, we look up
the normal table, the standard

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normal table, and value for
an argument of 0.5.

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The corresponding value is this
one, so we obtain 1 minus

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0.6915, which evaluates
to 0.3085.

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And this is the answer to
this particular problem.

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In the next example, we ask a
somewhat different question.

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We fix again the number of
packages to be 100, but we're

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given some probabilistic
tolerance.

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We allow the packages, their
total weight, to exceed the

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capacity of the container.

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But we don't want that to happen
too often, we want to

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have only 5% probability of
exceeding that capacity.

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How should we choose the
capacity of the container if

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we want to have this kind
of a specification?

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So we proceed as follows.

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We want this number, 0.05, to be
approximately equal to this

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probability.

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But now, we take this event and
rewrite it in terms of the

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standardized random variable.

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That is, we start from both
sides of the inequality and

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subtract n times mu, which is
200, and then divide by the

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standard deviation of Sn, which
is this quantity and

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which is 20, exactly as in
the previous example.

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And now, this random variable,
Zn, is approximately a

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standard normal.

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So we're asking for the
probability of the standard

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normal is larger than or equal
to something which, using the

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argument as in the previous
example, is 1 minus the CDF of

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the standard normal, evaluated
at this particular value.

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Now, what this tells us is that
this quantity here, the

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value of the CDF, should be
equal to 1 minus 0.05.

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So this quantity here
should be 0.95.

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What does this tell us about
the argument of the CDF?

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We can look at the table
and try to find

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somewhere an entry of 0.95.

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And we find it either
here or there.

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We could choose either one, or
we might decide to split the

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difference and say that we get
the value of 0.95 when the

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argument is 1.645.

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And so we conclude that in order
for this to be 0.95, we

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need a minus 200 divided by
20 to be equal to 1.645.

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And then we solve for a and we
find that a should be 232.9.

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And we can choose the capacity
of our container this way.

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Our next example is a little
more challenging.

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Here, we will fix a and
b and we will ask for

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the value of n.

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Here's a type of question
that has this flavor.

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We are given the capacity
of our container.

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We want to have small
probability of

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exceeding that capacity.

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How many packages should
you try to load?

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What is the value of
n for which this

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relation will be true?

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So we proceed, as usual, by
taking this event and

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rewriting it in a way that
involves the standardized

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version of Sn.

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So we need to subtract n
times mu, which in this

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problem is 2 times n.

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We subtract it from both sides
of the inequality.

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And then we divide by the square
root of n times sigma,

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which is 2.

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So we divide both sides of the
inequality by this number.

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Once more, this event here is
identical to the original

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event, but now it is expressed
in terms of the standardized

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version of Sn.

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This is a random variable
that's approximately a

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standard normal, so once more,
we're talking about the

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probability that the standard
normal exceeds a certain

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value, and by the central
limit theorem, this is

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approximately equal to 1 minus
the standard normal CDF,

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evaluated at this particular
value here.

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Now we want this quantity to
be approximately equal to

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0.05, which, once more, means
that this quantity should be

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0.95 and arguing as before that
we try to find 0.95 in

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the standard normal table.

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And this tells us that the
argument of the normal CDF

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should be equal to 1.645.

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Here, we get an equation
for n.

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Unfortunately, it is a
quadratic equation.

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However, we can solve it.

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And after you solve it,
numerically or using the

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formula for the solution of
quadratic equations, you find

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the value of n that's somewhere
between 89 and 90.

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Now, n is an integer, so you
could choose either 89 or 90.

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If you want to be conservative,
then you would

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set n to the smaller value of
the two and set n to be 89.

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Our last example is going to
be a little different.

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Here's what happens.

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We start loading the container,
and the container

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has a capacity of 210.

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Once we load the package and
we see that the weight has

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exceeded 210, we stop.

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Let N be the number of packages
that have been

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loaded, and this number
is random.

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If you're unlucky and you happen
to get lots of heavy

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packages, then you will
stop earlier.

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We would like to calculate,
approximately, the probability

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that the number of packages
that have been loaded is

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larger than 100.

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Now, this problem feels
a little different.

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The reason is that N is not the
sum of independent random

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variables and so we do not have
a version of the central

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limit theorem that we could
apply to N. What can we do?

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Well, we try to take this event
and express it in terms

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of the Xi's.

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And here's how we go about it.

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What does it mean that we loaded
more than 100 packages?

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This means that at the time
we were loading the 100th

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package, we didn't stop.

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And this means that at that
time, after we loaded the

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100th package, the weight
had not exceeded 210.

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So the event that we're dealing
with here is the same

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as the event that the first
100 packages have a total

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weight which is less than
or equal to 210.

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But now we're back into
a problem that

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we know how to solve.

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And the way to solve it is to
take this random variable,

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standardize it--

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this actually is essentially the
same calculation as in our

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very first example--

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00:12:40,720 --> 00:12:46,450
and we will get the standard
normal CDF evaluated at 210

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minus the mean of this random
variable, which is 200,

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00:12:51,120 --> 00:12:53,610
divided by the standard
deviation of this random

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00:12:53,610 --> 00:12:55,920
variable, which is 20.

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So we're looking at phi of 0.5,
which we look up at the

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00:13:04,400 --> 00:13:14,900
standard normal table, and
has a value of 0.6915.

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00:13:14,900 --> 00:13:19,480
So this was our last example,
and these four examples that

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00:13:19,480 --> 00:13:24,330
we worked through cover pretty
much all of the types of

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00:13:24,330 --> 00:13:27,410
problems that you
might encounter.

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00:13:27,410 --> 00:13:31,110
Of course, sometimes it might
not be entirely obvious what

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00:13:31,110 --> 00:13:33,370
kind of problem you
are dealing with.

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00:13:33,370 --> 00:13:36,310
You may have to do some
translation from a problem

195
00:13:36,310 --> 00:13:40,100
statement to bring it in the
form that we dealt with at

196
00:13:40,100 --> 00:13:41,080
this point.

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00:13:41,080 --> 00:13:46,120
But once you bring it into a
form where you can get close

198
00:13:46,120 --> 00:13:49,480
to applying the central limit
theorem, then the steps are

199
00:13:49,480 --> 00:13:53,340
pretty much routine, as long
as you carry them out in a

200
00:13:53,340 --> 00:13:54,990
systematic and organized
manner.