Download Section 6-2, 6-3

Section 6-2, 6-3 The Normal Distribution
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In this lesson, we’ll look at the “normal distribution”.
The normal distribution looks something like this. You see that it’s bell shaped, and also symmetric. In the
middle is mu, the mean, and there’s also sigma, the standard deviation. For a small sigma (or standard
deviation) we might get a distribution that has a pretty pronounced peak in the middle, and for a large standard
deviation we’ll get a distribution that’s more spread out.
The Standard normal distribution has a mean equal to 0 and standard deviation equal to 1. As with any
distribution the total area underneath the curve is equal to 1. Because of symmetry, if we divide the distribution
at the mean we have an area of 0.5 on either side.
If I know that I’ve cut my distribution in 2 parts and the area on 1 side is 0.05 then I know that the area on the
other side will be 0.95 and I get that by subtracting 0.05 from 1. And if I know that for example the area to the
right is say 0.025 I can again figure out the missing left area that will be 0.975 and I get that by again
subtracting from 1.
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Now let’s look at finding areas or probabilities from values.
We can do this using table ‘A2’, or I can use the built in calculator functions 2ndVARS – and then choosing
normalcdf, or I can use the program NORMAL83. I’ll focus on the 2 calculator functions while table A2 is
talked about in the book.
The built-in normalcdf function expects 4 things: the lower boundary; the upper boundry; the mean; and the
standard deviation. Remember for the ‘standard normal distribution’, mu, the mean is 0, and sigma the standard
deviation is equal to 1.
For the boundaries, focus on the horizontal axis. Remember this is just a number-line, and sometimes you’ll
need a value to the far far far far left. Since we can’t put infinity into the calculator we’re going to use some
extremely negative value, like – 999999. Then you’ve got your regular values coming up to possibly 0 in the
middle for standard normal distributions (the mean whatever it is), and then our far far far far right end we can
treat as some incredibly large positive number, so often we’ll use 999999.
So let’s consider the example - the probability that z is greater than -1.23. As soon as I see ‘z’, I think about the
‘standard normal distribution’, so I can draw a sketch with my mean 0 in the middle, my z score -1.23
somewhere to the left, and the fact that ‘z’ is greater than -1.23 tells me that I’m looking for the area to the right.
On the calculator I call up ‘normalcdf’ program or function, and the lower boundary - the left boundary is -1.23.
Then I’m moving out to the right from there, so my upper boundary is infinity, but I have to put in a number so I use this really big number 999,999, and then I have my mean of 0, and my standard deviation of 1. Put all
that in and press ‘enter’ and I get my answer. I get my answer, ‘the area is 0.8907’, which is reasonable because
I can see in the diagram in my sketch that it’s certainly more than half.
Now let’s try this example - the probability that ‘x’ is greater than 10, with the mean, mu = 5, and the standard
deviation, sigma = 3. The x tells me I’m no longer in the ‘standard normal distribution’, but I can still mark my
diagram with my mean of 5 in the middle, and that lets me realize – oh! ‘x’ being greater than 10 - where does
10 fall? 10’s going to be somewhere to the right, and then ‘x’ being greater than that means I’m looking for this
area here - to the right of 10.
I enter into the calculator ‘normalcdf’ – left boundary of 10, and then going up to positive infinity so essentially
99999 or I use 5 - 9’s here, 5 or 6 is ok; mean of 5; standard deviation of 3; press enter - and I get my area of
0.0478 - Which my diagram doesn’t quite look like that - it should be maybe even a little smaller of an area,
but at least it’s less than ½, that makes a lot of sense.
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The program ‘normal83’ is similar. It asks for the same pieces of information - it’s just you put them in 1 at a
time as prompted, so call up this program, and press enter to run it - and you see 3 choices. The first choice says
the probability that ‘x is greater than a’, and that would correspond to values being to the right of ‘a’, so you’d
see some sort of sketch like this. The 2nd choice - you’re looking for a probability that ‘x’ is less than ‘b’, and so
that’s an area to the left of some given value, and you have a 3rd choice here - the probability that ‘x’ is between
‘a’ and ‘b’, and that would be the area between the values ‘a’ and ‘b’. So just pick whichever choice is
appropriate for your problem.
You’ll enter the mean, and standard deviation as prompted, pressing ‘enter’ after each. Then you’ll also enter
‘A’ and/or ‘B’ as prompted. After you get your answer you’re given the option to look at tthe graph which will
be a sketch as above which is also kind of fun.
Let’s try this example the probabillty that ‘z’ is less than 1.03. Remember that ‘z’ scores generally refer to the
standard normal distribution so the mean is going to be 0, and the standard deviation is going to be 1 when
you’re prompted for those.
So I’ll call up the program, and press ‘enter’ to run it - and choose option # 2, the probability that ‘x’ or ‘z’ in
this case is less than some number. I enter the information as prompted, and I get an answer that the area is
equal to 0.8485.
Now let’s try this problem - the probability that ‘x’ is between 10 and 25 with a mean of 12, and a standard
deviation of 7. Here’s a sketch - we’ll want to be between 10 and 25, and the mean of 12 is also in the middle of
that. We’ll call up the program ‘normal83’ and choose option number 3 - the probability that ‘x’ is between ‘a’
and ‘b’. I get an answer of the area is equal to 0.5808.
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Now we’ll look at finding values from areas, or probabilities. Again we have 3 choices. We have table ‘A-2’
which is talked about in the book - and then we also have on the calculator a built in function ‘2nd VARS’ and
then ‘invNorm’, and we have a corresponding program ‘INVNORM83’. The function invnorm() needs 3 pieces
of information. It needs the left area - the area to the left of whatever value we’re looking for, and that’s
important - and then it also needs the mean, and the standard deviation.
Let’s try this example - supposing that I know the area to the left of my mystery value is 0.38, my goal is to find
this value right here - And first lets assume that we’re in the standard standard normal distribution.
So were gonna find our ‘z’ value that creates this left area of 0.38. I call up the invNorm(), enter that left area of
0.38, the mean of 0, the standard deviation of 1 to get an answer of -0.31, and that negative makes sense,
because my area of 0.38 is less than ½ of the total distribution, and so it makes sense that my ‘z’ value would be
somewhere to the left of 0.
Now let’s try this example. This time we’re given the right area of 0.45, and we’re also given a mean of 10 and
a standard deviation of 3. So the first thing we need to do is translate this into a left area - what’s to the left of
this mystery value that I’m looking for right here. Well, I can take 1 minus 0.45 to get my left area of 0.55. So,
into the calculator, into the invNorm(), I enter 0.55, 10, 3, and I get an answer of 10.38 - and this answer makes
sense because 0.45 is almost ½ of the distribution and so 10.38 is just a little bit to the right.
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The program ‘inverseNorm83’ is a little bit more flexible. The first option is if you are given a left area, as we
saw for inverseNorm(). The 2nd option is if your given a right area - a right tail, and if you’re looking for a
value to create that right tail - and you even get a 3rd option if you have a situation where you have 2 equal tails,
and what you get there is some kind of positive and negative answer because of symmetry those two tails being
equal will give a positive and negative value for ‘z’.
The program 1st gives a ‘z’ score, which remember is associated with the standard normal distribution. Then
you have the option to choose to continue if you’ve got some other normal distribution with a mean that’s not 0,
and a standard deviation that’s not equal to 1.
So lets’ try this example where we have the area in the right tail is 0.25 - and we want to find the value that
creates this tail. So we call up the program invNorm83, and choose option 2 for the right tail. I get an answer, z
= 0.67, and we’ll stop there.
I didn’t say it explicitly, but notice that I did label my mean as equal to 0 there, and we’ll just assume that the
standard deviation was 1 for this example.
Now let’s try this example where I have 2 equal tails - each tail is 0.01, and I’ve given you a mean of 5, and a
standard deviation of 2. So were actually going to get 2 values here.
So I call up the program invNorm83 and choose option3, 2 equal tails. And at the prompt I need to enter the
area of the tails, and notice I enter 0.02 - the total area of both tails together. I get my ‘z’ values plus and minus
2.33, but now I have to continue to translate these into values in terms of my other mean being 5, and standard
deviation of 2. So I choose to continue - enter the mean of five, and the standard deviation of 2 as prompted I
get the x values of .35 and 9.65.