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The properties of normal distributions, including their symmetrical, bell-shaped, and unimodal nature. It also covers how to compute proportions of cases within given intervals using the 68-95-99.7% rule. An example of finding the percentage of 14-year-old kids with a cholesterol level above a certain threshold using both the original distribution and the standard normal distribution.
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NORMAL DISTRIBUTIONS Properties
They are described by giving the mean and standard deviation N (μ, 0 1 A 0)
This week we have learnt how to perform computations with the normal distribution. We are going to review now how to compute proportions of cases within given intervals of the normal distribution, and how to do some other computations with the standard normal distribution.
As we saw last week, sometimes when we have a lot of observations the distribution becomes regular and we can describe it with a smooth curve (the density curve). A very important class of density curves are symmetrical, with one peak and with the form of a bell. We call these normal curves and the distribution behind it normal distribution. This type of distribution is quite frequent in real data, such as the height of a group of people of the same sex.
All normal distributions share a set of properties. One of these properties is known as the 68-95-99.7% rule, and refers to the following:
In any normal distribution with mean and standard deviation ,
With this rule we can compute the proportion of cases for some intervals derived from the different parts of the rules.
For example, suppose that we know that the distribution of blood colesterol level in 14 year old kids has a mean equal to 170 mg/dl and a standard deviation of 30 mg/dl. Knowing that the distribution is approximately normal, what is the percentage of 14 year old kids who have more than 230 mg/dl of colesterol?
Draw the distribution and find the different points given by the 68-96-99.7% rule. Notice that 230 is exactly the mean plus two times the standard deviation. Therefore we can use the rule to answer to the following question.
What happens if we want to know the percentage of 14 year old kids with a cholesterol level larger than 240? This number does not coincide with any of the ones indicated by the 68-95-99.7% rule. To answer this question we need the standard normal distribution, and follow the steps that we studied in class:
1) State the problem in terms of the observed variable x.
x: blood cholesterol level
x has a normal distribution N(170,30) and we want to know which proportion has x>240 , in other words we are looking for the area under the curve to the right of 240
Week 3 and 4: Normal distribution
2) Standardize x to restate the problem in terms of the standardized variable z. Draw a picte to locate which area we are trying to figure out.
Since we do not have a table with proportions for the normal distribution with mean equal to 170 and standard deviation equal to 30, we have to standardize our variable x to look for the same proportions but using the standard normal distribution, that is the normal distribution with mean equal to 0 and standard deviation equal to 1.
We are looking for the percentage of observations with x > 240. Standardizing,
This means that the standardized value of 240 (x) is 2.
For the standardized variable z we do have a table (the table for the standard normal distribution). The area we are looking for iw the area to the right of 2.33 in a normal distribution with mean equal to 0 and standard deviation equal to 1. The table gives us only the area to the left of a given value.
This would be the area to the left of 2.33:
3) In the table we can find this area:
Therefore, 99% of the cases are to the left of 2.33 in a standard normal distribution. But we need the percentage to the left of 2.33. How would you compute it? Recall that the total area under the curve is 1, that is 100% of the observations. Therefore that area we are looking for has a proportion of 1-0.99 = 0.01 or 1%, this is the percentage of kids with more than 240 mg/dl of cholesterol.
We could be interested for instance in the maximum cholesterol level possible to be within the 20% lowest part of the distribution, that is to be part of this group, what is the maximum cholesterol level that a 14 year old kid may have? We proceed as follows:
1) State the problem (use a graph to understand better what are you trying to solve). 2) Use the table to find z.
Now we look for a value in the numbers inside of the table (a proportion of cases) and with the table we can figure out which z is associated with that proportion.
3) We undo the standardization to express our solution in terms of our original variable x and not in terms of z.
Apply this methodology to figure out which is the maximum blood cholesterol level to be in the lower 20% of the distribution, that is the 14 year old kids with less cholesterol.
Week 3 and 4: Normal distribution