Hey guys, in this video I wanted to do a quick tutorial on how you can convert
any normal distribution that has mean mu and variance sigma squared into the standard normal
distribution, which is normally distributed with mean 0 and variance 1. The main reason
you're going to want to do this conversion from a normal to a standard normal is usually if you're
taking a statistics class where you have to use a statistics table. I have a sample statistics table
here and I will put a link in the description box below for this table, but this is one way
of getting the probability or the Z-score. From any normal distribution, you usually will use a table
like this, which has the values for a standard normal distribution. You might be given a problem
where you're given information about a normal distribution and then to calculate probabilities
or to calculate Z-scores, you're going to need to convert it first into a standard normal
distribution. I'm going to go over the formula to do that in this video as well as an example
problem. The formula for converting from a normal distribution to a standard normal distribution
is very simple and it relies on the concept of a Z-score. For a regular normal distribution, this
is a continuous function and at any point you can find the area under the curve of a certain value
on the function, which is usually denoted x or sometimes it's also denoted y but it's usually
denoted some kind of letter of the alphabet. On a standard normal distribution, these same
types of values or locations on the function are usually denoted as z for Z-score and these are
scores or values that tell you how far away you are from the central mean of the distribution.
The formula for converting from a normal distribution to a standard normal distribution is
going to basically use the formula for calculating a Z-score, which is z is equal to x minus mu
divided by the standard deviation. So the mean and the standard deviation in this formula comes from
the normal distribution and you plug it into this formula to convert it to a standard normal. Let's
go over an example of how you would do this type of conversion using a practice problem. I'm going
to go over this problem here and I'm going to use the statistic table that I showed in the beginning
of this video to solve this problem. This problem says, the marks on a statistics test are normally
distributed with a mean of 62 and a variance of 225. If the instructor wishes to assign B's
or higher to the top 30 percent of students in the class, what mark is required to get a B or higher.
In this problem, we're given that the marks on a statistics test are normally distributed with
a mean equal to 62 and a variance equal to 225 and we want to know what value to assign people as the
cutoff for having a B and a B cutoff--the x that we want to find--is going to be the value such that
30% of people who get at least this score are going to be graded as a B or an A. So we want to find
x and we're given mu and sigma squared. If we write out what we want to find as an equation,
we want the probability that x is greater than or equal to some number to be equal to 0.3 or 30
percent and this is going to be a probability from the normal distribution. Instead of x,
I'm going to plug in the Z-score formula. So z is going to be equal to x minus mu divided by
sigma and we want this value to be greater than or equal to some number and that should be equal to
0.3. I'm plugging in the formula for the Z-score so we can find this value on a standard normal
table and then I'm going to plug into the value mu and sigma. We're given that mu is equal to 62
and we're given that sigma squared is 225, so that means the square root of 225 is going to give you
sigma and the square root of 225 is 15. Into this formula we can plug in z is equal to x minus 62
divided by 15 has to be greater than or equal to some number and that should be equal to 0.3. If I
go back and look at the table that I'm given, instead of having the probability that Z is
greater than or equal to some value, I have scores for the probability that Z is less than or equal
to some value, so I need to use the property of symmetry in the normal distribution to be able to
find the values on this table. Going back to this drawing of the normal distribution, instead of
finding the probability that Z is greater than some number to equal 0.3, because my standard
normal statistics table has Z has to be less than or equal to some number, this equation is equal
to finding the Z-score that's less than or equal to some number being equal to 0.7. This value--if
this is a Z-score--we want that any value greater than or equal to this Z-score would give you 30%,
which is equivalent to saying that anything less than or equal to the Z-score would be 70%
because the entire distribution has to be equal to one hundred percent. And I'm making
this distinction because this is the value I have to look up in the standard normal table that I'm
given. I'm going to change this equation to the probability that z is equal to x minus 62 divided
by 15 has to be less than or equal to some number, which should be equal to the probability of 0.7.
I'm trying to solve for x, so I need to find some number on the standard normal Z-score table that
will give me the probability equal to 0.7. I'm going to go back to the table and the way to read
this table is the Z-scores are given in the first column and the first row. For example, this would
be the probability that a Z-score of -3.50 is equal to the probability of .0002.
This would be for example the probability that Z--a Z score of -3.39 and that would be
equal to 0.0003 and so on. What it gives you in the body of the table are the probabilities and
these are the corresponding Z-scores at the cutoff for a probability that Z is less than or equal to
some value. I want to find the probability closest to 0.7. I don't get exactly 0.7 in the table
but the closest value I'm going to get is a Z-score of about 0.5--I would say 0.525. This
gives me a probability of about 0.7. I'm going to plug this Z-score value back into the equation.
I have the probability of x minus 62 over 15 has to be less than or equal to the critical
value, which I found to be 0.525 is equal to 0.7 and then I'm going to solve this inequality
within the parentheses. I'm going to multiply both sides by 15 and then I'm going to add a 62
and that's going to give me the final value, which is that x has to be less than or equal to 69.875
for this probability to hold true. The final solution for this problem is that the cutoff
value should be approximately 69.875 and any score that is at least 69.875 should have a B so that
30 percent of the students get at B or higher in the class and that's it for this problem.
*chiptune music*