Welcome to the video.
BLOCK 5 Probability and Statistics,
and important laws in Statistics
and common distributions.
My name is Lidia Montero and this is the companion
video to the theory We are going to see five problems
this is aligned
to the learning objectives that they have mentioned at the beginning,
two important laws in Statistics and common distributions.
Let's see
Problem 1. It's calculation of probabilities
based on a normal distribution.
We know that birth weights in a population are normally distributed
with a mean that is 109 ounces
and standard deviation that is 13 ounces.
So the model would be, as it's written on
the slide, normal with 'mu'
109 and sigma squared is 13
squared.
The question
addressing what's the chance of obtaining
an event defined
birth weight of 141 oz
or heavier? Then
the probability of this event
that let's call it A can be written as the probability over
the X variable that we have defined
to be greater or equal to 141.
So, this is not in any table because is probability greater than
So, we have to use the complement 1 minus
the complement that these 1 one minus the probability
of X less or equal 141.
Point probability is null in numeric
random variables, sorry, continuous random variables.
So, less or equal is fine.
Then I transform the X normal variable
by subtracting 'mu' and divided into 'sigma'.
And then what we get is the transformation that leads to the standard
normal distribution and the probability
in the standard normal distribution is
such that I have to look at the tables. It is a probability
1 minus the probability of zeta
to be less than 2.46.
If we take a look at the tables,
then we will see that in the tables the probability
for zeta to be less than 2.46 is 0.991.
So, the final outcome of baby
heavier than 141
has a probability of 0.69.
What's the chance of birthweight
less than 120 ounces?
This could be the probability of event B
that can be obtained by the normal model.
So, this can be stated as the probability of X random variable
modeling the process of birth weight
to be less than 120,
then what we do is just to transform
the probability based
on the nonstandard normal variable
into the probability in the standard normal variable.
So, this is the same information that we have seen before.
And now in this case we are going to take a look in detail at the tables.
Okay.
So, finally we have to obtain the probability
for a standard normal distribution to be less than 0.85.
So the distribution function at the 0.85.
Okay.
We will see that this is 0.80.
This are the tables
I'm interested in obtaining
the cumulative probability for 0.85.
I have to take a look at the row and selecting the row
corresponding to the first digit
and the first digit after the point.
So, 0.8 clearly is this.
Now, the second digit
after the point is 5.
So I'll have to select the column 0.05.
Here it is, now the cross.
And I see that the cumulative probability is 0.8023.
So, this is!
The probability of birth
weight to be less than 120 oz.
Let's continue.
Let's address the probability
for a normal distribution in this case,
we are lecturing, for example, a group of 1000 students,
and we ask them to select
between two today's for the re-sit.
So, we have two alternative days, 10th July and 19th July.
We assume that the positive outcome is 19th July.
Okay, so
assuming that
no favorite day is any of the two.
So in this case, the best
guess for how many students pick the 19th
July is just the expectation
of a binomial distribution, because each student is
a Bernoulli trial, either 10th or 19th, with probability one half.
So, this is a model that is a binomial model n equal
1000 and p = 0.5.
Okay, so the expectation is 500.
Okay?
If we have a bias
in the choice in the case
that 19th July is too late
for some Mediterranean students.
So, let's assume that the probability of 19th July is 0.1.
Okay.
Then in this case, which would be the number of expected
students choosing 19th July?
Again, it is a binomial n=1000 and p = 0.1.
So, the expectation is just the product of the two parameters
and in this case, it is 1000 by 0.1.
So 100.
And now let's focus on the
most interesting discussion,
Which would be the probability
of obtaining less than 90 students
for July 19th?
So, in this case I have to define
an event.
And this event is
X total number of students
in the group
less than 90.
Okay.
So, the probability of this event
is a probability that can be obtained
based on the binomial distribution that works,
that fits this process, in the modeling process.
So, technically, the binomial distribution as can be seen in
the slide, it is binomial with n=1000
p=0.1 and
based on the rule of thumb, when
the expectation is greater than five.
In this case, then I can use an approximation
based on the standard normal distribution that it is easier to calculate.
And in this case, I have the the expectation
of the actually normal variable would be 100.
Okay.
And the variance would be 90.
But this is 9.49 squared
Okay.
So, if I address the calculus
of the probability of less than 90 students,
then what they have to do is, it is state
the event, then write the probability of the event
depending on the values of the random variable modeling the process
then transform this
probability calculus
using the approximation, the normal approximation,
and directly transform the calculus
on the normal approximation variable to the standard
normal distribution.
So, I performed directly ninety
since as you'll see in the slide, we have that zeta
has to be 90 minus
the expectation that it's 100, divided into the standard
deviation of the approximation
to the binomial model distribution.
So, this leads an equivalent probability of, sorry,
this leads to a value in the transformed
standard normal distribution that it is to
be less than -1.05.
Okay.
We do not have negative values in the table,
so, we have to use the property of symmetry.
So, being less
than -1.05 is equivalent
to be greater than the positive value.
1.05.
Since the normal distribution is symmetric.
With this, I use the complement.
This is 1- minus the probability of the standard normal
to be one less than 1.05.
And then take a look at the tables
and we will finally see that
the probability of less than 90 students
selecting July 19th
is 0.85.
Okay,
Now let's see these examples.
These are basic concepts
I move forward to
Problem 3. Basic concepts of random variables
In the a) example I give you a density function.
In fact, it is a probability function for a discrete variable
having values 9, 10, 11 and 12 each one with a probability value
of 0.25. So, if I write the table,
I can see that adding up the probabilities leads to
a sum that it's 1, so it's perfectly defined.
And for each of the values, what they
have it's a probability that ranges between 0 to 1.
So, this is fine.
This is a
well written
probability function
for a discrete random variable.
Now, let's use the second example that it's b).
I have a function and the values for the random variable are 1,
2, 3 ande 4. It's discrete, but the probability is given by the formula.
If I write the four values in the table
and then you look at the function returning the probability,
we will see that
at 4 the probability is negative.
I don't care that adding up
the probabilities, I get 1.
But it's not possible to have a probability for a value
that is less than 0.
So, this is not a probability
function properly defined. It's not.
Okay.
And then the final example.
In the final example, what we have is, again, very fast.
We have the table with the four values
0 to 3, and the probabilities that are given
by the formula that is included on the slide.
We see, directly, that adding up all the probabilities,
the sum of the probabilities is
24/25, less than 1.
So again, this it's not a probability
function well defined.
Problem 4. Lottery.
Just highlight that
in the examples of lottery
we have, for example, the case of betting.
This is just to play a little bit, but
let's assume that we have to pay
just $1 or €1 and the case of lottery
where you have to select 6 numbers
ranging between 1 and 49.
Okay, the order is not relevant.
So, you have to pick these 6 numbers
and if you win, then you win 2 million.
So, this is great.
You have to pay $1. Each time you play.
So taking into account the combination of all
the possible values for these
49 units and taking 6 without repetition,
then what we see is that the possible numbers is huge.
And in fact the probability of selecting the positive outcome.
So, winning the lottery is 7.2 by 7.2e-8.
Okay, so in this case,
if I have to write the probability function in this case,
what I say 'possible
values for my pocket money, what they win'
- 1 and the probability of
losing is 1 minus winning.
Okay, so it's 0.99999.
And the second value for the random variable would be 2 million positive.
This is if I win, but all in all probability,
which is the expected value?
So the quantity, the expectation value
that they will be winning in one trial, one week.
Then, I have this table and I calculate
the expectation that it is probability by value
and after some calculations what we see is that is
$-0.86 (euros).
The currency that you would like
so it's negative! If you repeat the exercise during ten years,
then you are expecting to lose
$447.2 (euros).
So, this is a lot!
PROBLEM 5.
I have problem 5 that it's related to 'The batting example'.
So, consider a baseball player that his probability of getting a hit each
time at bat is 0.29.
What's the
probability of no hit in the next
five at bats?
This is a probability that can be obtained
based on a binomial distribution.
A binomial distribution with n = 5.
So five times at bat, and the probability of a positive outcome
that is getting a hit in this case is p=0.29.
So, which is the probability of zero
hits? It tells the probability for this binomial distribution
to take the value zero, thus the probability of value zero.
If you take the formulas that we have seen in the companion video, then
it's possible to see that in this case
the probability of 0 hits is 0.178.
You have the details on the slide.
What's the probability
of two hits in the next 8 at bats?
Then again,
we have a binomial distribution to consider
n=8 times at bats
and the probability, that this is exactly the same.
And in this case, what I'm asking is exactly two hits.
So, a probability for the binomial distribution
that is the model that fits to the situation.
Then, it's the probability for X = 2.
We substitute and we see that it's 0.3.
And finally, which is the expected number of hits in the next 10 at bats?
But binomial distribution in this case n=10.
So, the expectation is np
so 10x0.29
So, 2.92
hits in the next ten at bats.
It is the expectation.
And finally,
I have an example that it's about
the Chi-square test for independence.
This example is a little bit long to explain
but this is a case of a contingency table where you have to
check whether
rows and columns are independent.
So, in the case I have on rows gender
and I have on columns, the political party they prefer.
Okay, I have a sample of 1000 units.
and I state one null hypothesis.
Stating the hypothesis, in this case, is
gender and voting preferences are independent.
And the alternative hypothesis is 'they are not'.
I choose a significant level.
Let's say alpha=0.05,
and now what we have to do it's
the chi square test
takes into account degrees of freedom.
I will explain the meaning in the case square distribution.
This is the degrees of freedom
is number of rows minus one
by number of columns minus one.
So, in this case, we don't rely on the totals
because these are artificially added.
So what we have is (2-1) by (3-1).
So, 2.
Then under the null hypothesis,
So, independence between rows and columns,
which would be the number of expected
observations in each cell of the table?
This can be obtained
by the product of the marginal proportions,
and this leads to the possibility
of calculating the chi square statistic
that is defined for each of the cells
as the squared difference between the observation in the cell
minus this expected value under the null hypothesis,
squared and divided into the expected value in the cell.
This is the famous Chi square test.
Here you have in the new table that has appeared
which could be the number of expected number of observations
if the null hypothesis is true,
then we'll obtain the symmetry.
The squared discrepancy, observation and expectation
values squared divided into the expected values.
And this leads to a quantity that it is chi square
statistic, 16.2
This is the test statistic and
it can be
proved that it follows a chi squared distribution with
2 degrees of freedom.
The P value, the probability
for the null hypothesis,
is the probability of a chi square
with 2 degrees of freedom to be greater than 16.2.
And taking a look at the tables
that in this case are not the same tables
as in the normal situation, we will see that
this probability is very low, 0.003,
less than the significant level that we have set at the beginning, at 0.05.
So, it means that we cannot not accept
the null hypothesis.
Thus, there is a relationship between gender and voting preferences.
Okay, Let's take a look at the chi-square tables.
And in this case, what we see it is a table
that is not exactly the same as in the normal distribution,
because what we say we have in roles is the degrees of freedom.
Remember that in our case, we have a chi squared, two degrees
of freedom based on the product of the
rows by columns, and in this case we have set
the significance level to 0.05
and for the chi squared distribution
we have a tabulation of which are the values
to obtain a cumulative
probability of 0.95.
In fact we have 0.95, 0.99, etc.
So, it depends on the significance level.
So, if I take a look then based on the degrees of freedom
and the significance level, I have 5.99
So, since my chi squared
value is 16.2,
the probability for a chi squared to be greater than
16.2 will be a probability
that it is less than 0.05
because the threshold is defined at 5.99.
So greater than 16,
so directly it means that, what we have
is rejection of the null hypotheses, as we have seen in the previous slide.
Thank you very much for your attention.