We are going to introduce some basic notation here for the normal distribution.
The first is, this cursive "N" character.
It says the variable x comes from the normal distribution with mean
of mu and variance of sigma squared.
Notice the second term is variance, not standard deviation.
If I wanted to say the height of people in Australia is normally distributed with a mean
of 180 cm and a variance of 36cm squared - in other words, a standard deviation of 6 cm -
I would write it this way, mathematically.
The next idea is to take the normal distribution and integrate it
to get the probability distribution that we saw before.
As we integrate from left to right, the total area tends to 1.0 units.
Most times we will stop before this.
We will find the cumulative integral from minus infinity up to a particular value of z.
For example, what is the cumulative area up to a value of z=0?
It is 50%, because we stopped halfway, and we know the distribution is symmetric.
Now people have calculated the integral for us and presented them as tables.
This is useful when we don't have a computer, or when we need it in a test or exam.
What is the area from minus infinity to z=2?
That's right, you can read it from the table it is 97.7% You can also use the pnorm(...)
function in R, by saying pnorm(2), and you get the same answer.
De can do this backwards as well, for the case when you know the probability or area,
but want to find the corresponding z-value.
Use the qnorm(...)
function in R, or use these inverse tables.
Let's try it out: what is the value of z when you have an area of 50%?
Both from the table or by typing qnorm(0.5) in R, you will get the result of z as zero.
Here's another one; what is the value of z, when you have an area of 2.5% under the left tail?
2.5% is a very small probability: so we expect a negative value of z. In fact, from the table,
we see it is z=-1.96, and R also confirms this for us.
Notice that you cannot find the z-value when the area
under the curve is exactly zero percent or exactly 1.
It is undefined for those two values.
Let's go in the forwards direction, but change the question slightly.
What is the area under the curve between z=-1 and z=+1?
That's right, we have to subtract the areas: the cumulative area up to minus 1 is 15.87%,
and the cumulative area up to +1 is 84.13%.
So there's a small overlap piece and the area then between -1
and +1 should subtract these two numbers, to avoid double counting.
That gives an area of 84.13 - 15.87 = 68.3%.
That's about 70%, which was the rough number I had given you earlier as the area
between -1 and +1 standard deviations.
Let's have two more examples.
What is the probability of observing a biological activity from our process,
when the biological activity values are normally distributed, with a mean of about 27,
and a variance of about 9 units squared.
You should be able to do this without a calculator.
First, let's standardize.
We want a probability of x<=30.
In standard variables, we have to say 30, minus the mean, divided by the standard deviation.
Thats 30 minus 27 divided by the square root of nine.
Our question then becomes what is the probability
of a z value smaller than or equal to 1?
That's easy: it is about 84%, and that's the answer to the question.
Here's another one to try without a calculator.
Historically we have observed the yield values from our batch process to be 85 g/L,
with a variance of 16 units squared.
What is the probability of yield values between 77 and 93 g/L.
You should have found the standardized z-values for 77 and 93 respectively, relative to a mean
of 85 g/L and a variance of 16 units squared.
These z-values are -2 and +2 respectively.
The area from minus infinity to +2 is 97.73.
But we have to subtract the small area at the start, the cumulative area up to minus 2.
So thats 97.73 minus 2.27%, in other words, and area of 95.46%.
So that's our answer: there's a probability of around 95%
of observing yield values between 77 and 93 g/L.
Your should make sure you can follow those calculations.
They wi;l get quicker and quicker as you become more comfortable.