WEBVTT

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>> Hey, everyone.

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Thanks for joining
us in a discussion

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of binomial distributions.

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As you can see, the first
objective is to determine

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if we've got an experiment
that is a binomial experiment.

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First, for an experiment
to be considered binomial,

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four things must hold.

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You see the four
things there listed.

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The experiment that is
performed needs to be

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for a fixed number of trials.

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Secondly, the trials
need to be independent.

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Third, there are only two
mutually exclusive outcomes

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for each trial.

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This is the biggest piece of
evidence in my opinion in terms

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of being able to
recognize, and of course,

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we'll discuss this
in the examples.

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And finally, the probability
of success needs to be the same

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for each trial where we let
P denote the probability

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of success.

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So, let's go ahead into
our first example here

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where they're asking us

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to determine whether the
following experiments are

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binomial experiments and we
have to explain the reasoning.

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Taking a look at part A then,
according to a recent study,

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33% of Americans 23 years
or older have been arrested.

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A random sample of
500 Americans 23 years

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or older are asked whether or
not they have been arrested.

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Remember we're trying

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to determine whether this
is a binomial experiment,

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so first just to make it a bit
clearer, we're going to go ahead

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and state that this is
a binomial experiment.

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But we're going to
explain why, of course.

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It's pulling in the
four criteria.

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The first one, of course, is
that the experiment is performed

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for a fixed number of
trials where they told us

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that we have a random
sample of 500 Americans.

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Next, we need to recognize that
the trials are independent.

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That is, whether or not a
randomly selected person has

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been arrested does
not affect whether

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or not another randomly selected
person has been arrested.

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Therefore, the trials
are independent Third,

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and the pivotal one, there
are only two outcomes,

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either they have
been arrested or not.

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There's no other outcome,
so we have the situation

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that is only two outcomes.

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And finally, the probability of
success, being arrested this is,

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is the same for each trial
or P is 33%, or 0.33.

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Let's jump over to part B,
then, where we're told Mary is

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at the fair playing pop
the balloon with six darts.

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There are twenty balloons
total, fifteen which is say

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"lose" and five which "win".

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Now, this particular example
is not a binomial experiment

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since the trials are dependent
and the probability of success

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of winning changes
with each trial.

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And what we mean is when
we pop the first balloon,

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now the total number of balloons
goes from twenty to nineteen,

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and whether we win or lose

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on that first one changes
the probability regardless.

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To explain a little more for
some elaboration, we can set it

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up like this where
we're saying, okay,

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what's the probability
of winning?

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Well, the probability of winning
says that we have five balloons

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that say win out of
twenty total balloons.

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Therefore, the probability
that we will win

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on the first dart would
be a five out of twenty.

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Now, if we went on the
first dart and that means

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for the second dart, the
probability of we win changes

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because now there are only
nineteen balloons available

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and with only nineteen balloons
available we have only now four

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of them that would allow us to
win since we already first one

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of the balloons that was a win.

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Now, even if the
first one was a lose,

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it would still change
the probabilities

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because we would go from
twenty to nineteen balloons.

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So, we have dependent trials.

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Let's now look at part
C where Bob flips a coin

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until the coin lands on heads.

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Now, we have the idea that
he's looking for heads.

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There is the idea that is
only two possible outcomes

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where we have heads or tails.

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However, the key piece here,

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this is not a binomial
experiment

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since there is not a
fixed number of trials.

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They use the keyword until,
meaning he's going to keep

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on flipping until he gets a
heads, so he may need one trial

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or two trials or three trials
until the coin lands on heads.

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Pause the video and
try these problems.

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That takes us over to
our second objective

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where we will be
calculating probabilities

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for a binomial distribution.

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This can be a bit tough based
on the calculator input.

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We have a table here to help
us calculate probabilities

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for these binomial
distributions.

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It's all based on the phrase
that is be said to us,

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for example, when the phrase is
"exactly" or "equals" or "is",

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what's happening is
they are telling us

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to find the probability
of a specific outcome

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with our math symbol being P of
X being equal to some value X.

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In the calculator, this is where
we use the binom PDF function.

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This is very specific, again,
with a specific result.

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As we continue the comparison,
you see the other options

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where they are set up
with both the phrase

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and then the corresponding
math symbols.

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You do have the option of
when they tell us between A

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and B inclusive there,
we have the option

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of adding these together
with the binom PDF function.

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Or we can use a combination
of the binom CDF

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where we find the probability of
something happening and so track

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that from the probability
of outcomes that are not

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in the event we're discussing.

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Of course, we will refer back
to this as we use our examples,

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so let's go ahead
and jump into those.

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Let's first note, to get
to either the binom PDF

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or binom CDF functions
in your calculator,

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you press the second button, and
then the vars [phonetic] button

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and scroll up until you
find either the binom PDF

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or the binom CDF.

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With our first example,
we're told a study was done

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which stated that 41% of
Americans only have a cell phone

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in their house, no
landline that is.

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What is the probability
that in a random sample

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of fifty American households

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that exactly twenty
only have a cell phone?

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The first keyword that we're
told is that the probability

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that we're looking

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for is exactly twenty
only have a cell phone.

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Translating this to our
symbols, have a little bit

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of language here, we're
looking for the probability

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that exactly twenty, which
would be the probability that X,

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the variable we're looking
for, would be twenty.

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Or X represents the
number of households

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that have only a cell phone.

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Now, referring to
the previous table,

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remember since we have the
value that is exactly twenty,

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this tells us to use
the binom PDF function.

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Now, with the binom
PDF function,

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remember when you
did input N, P,

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and then X with commas
separating those values,

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which means we have to
address each of those

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from the given information
in the problem.

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N being the sample and
they tell us up here

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that we've got a random sample
of fifty American households.

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So, that means that N
is equal to our fifty.

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Next, we're looking for a P
representing the probability

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of success.

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The study stated that 41% of
Americans have only a cell phone

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in their house, therefore the
41% represents the probability

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of finding that success.

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We, of course, write this
as a decimal, as 0.41.

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And lastly, X would be
the number we're looking

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for for the number of successes,

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which we previously stated was
twenty because they tell us

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that we want exactly
twenty in our problem.

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If we input this into a TI-83,
our screen ends up looking

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like this with the
binom PDF function.

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On a TI-84 screen, it will
end up looking like this.

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If you have something called
the stat wizard mode on,

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or you can turn that off
using the mode function,

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with each calculator we should
obtain approximately the same

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result, getting a
probability of 0.11.

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And you're all set.

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On to the next example were told
according to a recent article,

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38% of busses in
Chicago arrive on time.

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A random sample of thirty
Chicago busses is taken.

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In a random sample of
thirty Chicago busses,

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what is the probability that
less than 10 arrive on time?

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Which means, we're looking for
the probability of less than 10.

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Referring back to the
table, we can translate this

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as a math symbol and to
X being less than 10.

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In order to lead that to the
correct input, now depending

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on your approach or your
instructor's approach,

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we are able to put this directly
to the binom CDF function,

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referring to the table where
we have strictly less than 10.

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That is one route we can
take is going to that table

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where we say we've
got a binom CDF now,

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making sure to create
some emphasis

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on the fact that it's CDF.

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And since it's strictly less
than 10, that table tells us

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to do N, P, and then more
specifically this is an X minus

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1, because what we are doing
is we are excluding the value

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of ten itself, which changes
that to be a binom CDF.

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And referring to
the problem said

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that we had thirty
Chicago busses.

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Therefore, we've got a value
of 30 for N. The problem said

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that 38% of busses
arrive on time.

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Therefore, our 38%
represents the probability

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of success, or P, being at 0.38.

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And finally, making sure
to be careful about it,

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X minus 1 over X was
set to be less than 10.

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Therefore, we convert that to
be a 10 minus 1, giving us 9.

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Now, another way to
think of this though is

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to translate the
original statement.

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It all depends on
your preference.

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A lot of instructors will say
instead of thinking of this

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as P being less than 10 we
can convert that statement.

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We know that since
we are working

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with a discrete variable,
if X is less than 10,

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we can call that X being
less than or equal to 9.

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If we've got less
than or equal to 9,

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according to that table this
would be equal to a binom CDF.

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And on our table we see that it
says N, P, and then just X. So,

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the idea is whatever
you're comfortable with.

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We can think of it
as less than 10.

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If we think of it as less than
10, then we use X minus 1.

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If we think of it as X less
than or equal to 9 by converting

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that to a less than or equal
to, we've included the 9,

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which allows us to just
use X as our key number,

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which means we end
up with a binom CDF.

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Our N is still 30, 0.38 is still
P, and now our value for X is 9.

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The result should be the
same for each of them,

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giving us a probability of 0.24,
which means the probability

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that less than 10 of the 30
Chicago busses will arrive

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on time is about 24%.

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Over to part B in
our example then.

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In a random sample of
thirty Chicago busses,

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what is the probability that
exactly 17 arrive on time,

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which means again we're
referring to a keyword

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that is saying exactly 17.

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When they tell us exactly
17, that means we are trying

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to find the probability that X,

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the variable representing the
number of busses that arrive

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on time being specifically 17.

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We know that since it's a
specific value like that,

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we need to use a binom PDF
function as opposed to a CDF,

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where N is still 30, P is
still a 0.38 from previous,

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and now X being the value
17 that we're looking for.

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And put this in our calculator
and we got ourselves a 0.02,

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that is the probability
that exactly 17

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of the 30 Chicago busses will
arrive on time is about 2%.

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In part C of the example,
we've got a random sample

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of our 30 Chicago busses still.

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What is the probability that
at least 12 arrive on time?

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The key phrase we're
highlighting now is at least 12.

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And we're looking for
the probability of.

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Make sure we're careful
with this translation

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into our math symbols because
at least 12 includes the 12.

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Therefore, we have the
probability that X is greater

00:14:15.976 --> 00:14:19.346 A:middle
than or equal to the 12,
meaning it's including that 12.

00:14:19.936 --> 00:14:22.676 A:middle
This again provides a
couple opportunities

00:14:22.676 --> 00:14:24.166 A:middle
in order to translate this.

00:14:25.516 --> 00:14:28.676 A:middle
Referring back to the
table, we can use the option

00:14:28.676 --> 00:14:34.426 A:middle
that is 1 minus the
binom CDF function.

00:14:34.796 --> 00:14:40.086 A:middle
And as we see in the table,
our input would be N, P,

00:14:40.306 --> 00:14:45.366 A:middle
and now an X minus 1 because
it's including the 12.

00:14:45.556 --> 00:14:48.096 A:middle
The 1 minus of course means
the complement of that.

00:14:48.526 --> 00:14:51.946 A:middle
If we use this, then I'm
going 1 minus my binom CDF--

00:14:56.856 --> 00:15:03.336 A:middle
Where N is still 30, P was
our 0.38, and X minus 1 needs

00:15:03.336 --> 00:15:06.476 A:middle
to be 12 minus our
1, getting 11.

00:15:07.756 --> 00:15:10.196 A:middle
Similar to the previous
discussion, we can think of this

00:15:10.196 --> 00:15:12.966 A:middle
in terms of sample
space and outcomes.

00:15:13.596 --> 00:15:16.066 A:middle
So, that means we can
translate the probability

00:15:16.066 --> 00:15:21.616 A:middle
of X being greater than or equal
to 12 into the complement of,

00:15:21.836 --> 00:15:24.316 A:middle
which would be our 1
minus, the probability

00:15:24.316 --> 00:15:26.906 A:middle
that X is less than
or equal to 11.

00:15:27.286 --> 00:15:33.046 A:middle
If we do that, then that means
we've got 1 minus a binom CDF--

00:15:35.626 --> 00:15:41.676 A:middle
Where N is still the number of
trials, P, and since it's less

00:15:41.756 --> 00:15:45.916 A:middle
than or equal to that just
stays X. And this leads us

00:15:45.916 --> 00:15:48.166 A:middle
to the same calculator
input that we had

00:15:48.446 --> 00:15:49.646 A:middle
in the initial translation.

00:15:50.156 --> 00:15:52.536 A:middle
Either route we take leads
us to the same result,

00:15:52.656 --> 00:15:54.176 A:middle
which would be a 0.48.

00:15:55.426 --> 00:15:58.056 A:middle
That is the probability
that at least 12

00:15:58.056 --> 00:16:02.426 A:middle
of the 30 Chicago busses
will arrive on time is 48%.

00:16:03.636 --> 00:16:06.446 A:middle
Finally, in part D, we
have our random sample

00:16:06.446 --> 00:16:07.806 A:middle
of 30 Chicago busses still.

00:16:08.336 --> 00:16:11.876 A:middle
Question's asking us what is
the probability that between 5

00:16:11.876 --> 00:16:14.536 A:middle
and 7 inclusive arrive on time?

00:16:15.066 --> 00:16:17.986 A:middle
Of course, that being our key
phrase again, we're looking

00:16:17.986 --> 00:16:22.506 A:middle
for the probability that between
5 and 7 inclusive are on time.

00:16:22.696 --> 00:16:26.516 A:middle
We can translate that into
saying that we are looking

00:16:26.516 --> 00:16:29.916 A:middle
for the probability
between 5 and 7 inclusive,

00:16:31.006 --> 00:16:36.236 A:middle
or that is X is greater than or
equal to 5 while also being less

00:16:36.236 --> 00:16:39.806 A:middle
than or equal two 7, which
means remember we have two

00:16:39.986 --> 00:16:41.696 A:middle
possibilities for
this approach as well.

00:16:42.006 --> 00:16:45.236 A:middle
We can piece this together
as discrete inputs.

00:16:45.706 --> 00:16:47.086 A:middle
If we're piecing
this together then

00:16:47.086 --> 00:16:48.976 A:middle
that means this would be
equal to a binom PDF--

00:16:53.526 --> 00:16:58.666 A:middle
Where N is 30, P
is still at 0.38,

00:16:59.176 --> 00:17:02.246 A:middle
and what we're doing is
we're using the three options

00:17:02.246 --> 00:17:05.896 A:middle
that we're looking for, 5,
6, or 7, because remember,

00:17:06.026 --> 00:17:07.966 A:middle
they're discrete variables.

00:17:09.056 --> 00:17:12.316 A:middle
So, the first possible outcome
we're looking for would be 5.

00:17:12.896 --> 00:17:16.306 A:middle
We add that to the second
outcome we're looking for,

00:17:16.306 --> 00:17:18.536 A:middle
which would be 6 busses on time.

00:17:19.156 --> 00:17:23.306 A:middle
And we add that to the third
option, that would be 7 busses

00:17:23.306 --> 00:17:27.536 A:middle
on time, leading us to
a probability of a 0.07.

00:17:27.876 --> 00:17:30.536 A:middle
In your calculator you can
put all three of those at once

00:17:30.536 --> 00:17:31.986 A:middle
or you can add them together.

00:17:32.706 --> 00:17:36.016 A:middle
The other route we can go
conceptually would be to think

00:17:36.016 --> 00:17:39.386 A:middle
of this in terms of binom CDF
function where we're looking

00:17:39.386 --> 00:17:40.726 A:middle
for the probability
that we want.

00:17:40.726 --> 00:17:42.616 A:middle
And we're going to subtract
that by the probability

00:17:42.616 --> 00:17:47.216 A:middle
that we do not want, which means
our first input would be a binom

00:17:47.216 --> 00:17:51.196 A:middle
CDF now of 30 with P being 0.38

00:17:51.656 --> 00:17:53.826 A:middle
and that first value
for X being 7.

00:17:54.186 --> 00:17:57.246 A:middle
What this would do would
be the probability of 7

00:17:57.246 --> 00:17:58.956 A:middle
or less arriving on time.

00:17:59.106 --> 00:18:00.866 A:middle
But we don't want 7 or less.

00:18:00.866 --> 00:18:02.306 A:middle
We want between 5 and 7,

00:18:02.856 --> 00:18:06.386 A:middle
which means we subtract the
probability that we don't want,

00:18:06.636 --> 00:18:07.976 A:middle
which tells us to
do a binom CDF--

00:18:12.196 --> 00:18:16.456 A:middle
N still being 30, of course,
0.38 for our probability

00:18:16.456 --> 00:18:20.166 A:middle
of success, and now
we want 7, 6, and 5.

00:18:20.166 --> 00:18:23.006 A:middle
Therefore, the first output

00:18:23.006 --> 00:18:26.576 A:middle
that we don't want would
be 4 busses on time.

00:18:27.216 --> 00:18:28.686 A:middle
So, we're subtracting that out

00:18:28.686 --> 00:18:30.266 A:middle
from the probability
we're looking for.

00:18:30.516 --> 00:18:34.596 A:middle
We can input that entire line
all tighter in our calculator

00:18:34.596 --> 00:18:36.776 A:middle
if you want or piece by piece.

00:18:36.776 --> 00:18:40.146 A:middle
And we end up with the
same result of a 0.07,

00:18:41.146 --> 00:18:46.676 A:middle
that is the probability that
between 5 and 7 inclusive

00:18:46.706 --> 00:18:49.336 A:middle
of the 30 Chicago
busses will arrive

00:18:49.336 --> 00:18:52.016 A:middle
on time is approximately 7%.

00:18:52.746 --> 00:18:54.976 A:middle
Pause the video and
try these problems.

00:19:17.136 --> 00:19:19.266 A:middle
And thanks for joining
us in a discussion

00:19:19.266 --> 00:19:20.856 A:middle
of binomial distributions.