In the Exploratory Data Analysis sections of this course, we encountered data sets, such as lengths of human pregnancies, whose distributions naturally followed a symmetric unimodal bell shape, bulging in the middle and tapering off at the ends.
Many variables, such as pregnancy lengths, shoe sizes, foot lengths, and other human physical characteristics exhibit these properties: symmetry indicates that the variable is just as likely to take a value a certain distance below its mean as it is to take a value that same distance above its mean; the bell-shape indicates that values closer to the mean are more likely, and it becomes increasingly unlikely to take values far from the mean in either direction. The particular shape exhibited by these variables has been studied since the early part of the nineteenth century, when they were first called “normal” as a way of suggesting their depiction of a common, natural pattern.
Observations of Normal Distributions
There are many normal distributions. Even though all of them have the bell-shape, they vary in their center and spread.
More specifically, the center of the distribution is determined by its mean (μ) and the spread is determined by its standard deviation ( σ).
Some observations we can make as we look at this graph are:
- The black and the red normal curves have means or centers at μ = 10. However, the red curve is more spread out and thus has a larger standard deviation.As you look at these two normal curves, notice that as the red graph is squished down, the spread gets larger, thus allowing the area under the curve to remain the same.
- The black and the green normal curves have the same standard deviation or spread (the range of the black curve is 6.5-13.5, and the green curve’s range is 10.5-17.5).
Even more important than the fact that many variables themselves follow the normal curve is the role played by the normal curve in sampling theory, as we’ll see in the next module of probability. Understanding the normal distribution is an important step in the direction of our overall goal, which is to relate sample means or proportions to population means or proportions.
The Standard Deviation Rule for Normal Random Variables
We began to get a feel for normal distributions in the Exploratory Data Analysis (EDA) section, when we introduced the Standard Deviation Rule (or the 68-95-99.7 rule) for how values in a normally-shaped sample data set behave relative to their mean (x¯) and standard deviation (s). This is the same rule that dictates how the distribution of a normal random variable behaves relative to its mean μ and standard deviation σ. Now we use probability language and notation to describe the random variable’s behavior. For example, in the EDA section, we would have said “68% of pregnancies in our data set fall within 1 standard deviation (s) of their mean (x¯).” The analogous statement now would be “If X, the length of a randomly chosen pregnancy, is normal with mean (μ) and standard deviation (σ), then 0.68=P(μ-σ<X<μ+σ).”
In general, if X is a normal random variable, then the probability is
68% that X falls within 1 σ of μ , that is, in the interval μ±σ
95% that X falls within 2 σ of μ , that is, in the interval μ±2σ
99.7% that X falls within 3 σ of μ , that is, in the intervalμ±3σ
Using probability notation, we may write
Notice that the information from the rule can be interpreted from the perspective of the tails of the normal curve: since 0.68 is the probability of being within 1 standard deviation of the mean, (1 – .68) / 2 = 0.16 is the probability of being further than 1 standard deviation below the mean (or further than 1 standard deviation above the mean). Likewise, (1 – .95) / 2 = 0.025 is the probability of being more than 2 standard deviations below (or above) the mean; (1 – .997) / 2 = 0.0015 is the probability of being more than 3 standard deviations below (or above) the mean. The three figures below illustrate this.
Suppose that foot length of a randomly chosen adult male is a normal random variable with mean μ=11 and standard deviation σ=1.5. Then the Standard Deviation Rule lets us sketch the probability distribution of X as follows:
(a) What is the probability that a randomly chosen adult male will have a foot length between 8 and 14 inches? 0.95, or 95%.
(b) An adult male is almost guaranteed (.997 probability) to have a foot length between what two values? 6.5 and 15.5 inches.
(c) The probability is only 2.5% that an adult male will have a foot length greater than how many inches? 14.
Now you should try a few. (Use the figure that is just before part (a) to help you.)
The first step to assessing a probability associated with a normal value is to determine the relative value with respect to all the other values taken by that normal variable. This is accomplished by determining how many standard deviations below or above the mean that value is.
Example: Foot Length
How many standard deviations below or above the mean male foot length is 13 inches? Since the mean is 11 inches, 13 inches is 2 inches above the mean. Since a standard deviation is 1.5 inches, this would be 2 / 1.5 = 1.33 standard deviations above the mean. Combining these two steps, we could write:
(13 in. – 11 in.) / (1.5 inches per standard deviation) = (13 – 11) / 1.5 standard deviations = +1.33 standard deviations.
In the language of statistics, we have just found the z-score for a male foot length of 13 inches to be z = +1.33. Or, to put it another way, we have standardized the value of 13. In general, the standardized value z tells how many standard deviations below or above the mean the original value is, and is calculated as follows:
z-score = (value – mean)/standard deviation
The convention is to denote a value of our normal random variable X with the letter “x.” Since the mean is written μ and the standard deviation σ, we may write the standardized value asz=x-μσ
Notice that since σ is always positive, for values of x above the mean (μ), z will be positive; for values of x below μ, z will be negative.
Example: Standardizing Foot Measurements
Let’s go back to our foot length example, and answer some more questions.
(a) What is the standardized value for a male foot length of 8.5 inches? How does this foot length relate to the mean?
z = (8.5 – 11) / 1.5 = -1.67. This foot length is 1.67 standard deviations below the mean.
(b) A man’s standardized foot length is +2.5. What is his actual foot length in inches? If z = +2.5, then his foot length is 2.5 standard deviations above the mean. Since the mean is 11, and each standard deviation is 1.5, we get that the man’s foot length is: 11 + 2.5(1.5) = 14.75 inches.
z-scores also allow us to compare values of different normal random variables. Here is an example:
(c) In general, women’s foot length is shorter than men’s. Assume that women’s foot length follows a normal distribution with a mean of 9.5 inches and standard deviation of 1.2. Ross’ foot length is 13.25 inches, and Candace’s foot length is only 11.6 inches. Which of the two has a longer foot relative to his or her gender group?
To answer this question, let’s find the z-score of each of these two normal values, bearing in mind that each of the values comes from a different normal distribution.
Ross: z-score = (13.25 – 11) / 1.5 = 1.5 (Ross’ foot length is 1.5 standard deviations above the mean foot length for men).
Candace: z-score = (11.6 – 9.5) / 1.2 = 1.75 (Candace’s foot length is 1.75 standard deviations above the mean foot length for women).
Note that even though Ross’ foot is longer than Candace’s, Candace’s foot is longer relative to their respective genders.
TO SUM UP…
Part (c) illustrates how z-scores become crucial when you want to compare distributions.
Finding Probabilities with the Normal Table
Now that you have learned to assess the relative value of any normal value by standardizing, the next step is to evaluate probabilities. In other contexts, as mentioned before, we will first take the conventional approach of referring to a normal table, which tells the probability of a normal variable taking a value less than any standardized score z.
Since normal curves are symmetric about their mean, it follows that the curve of z scores must be symmetric about 0. Since the total area under any normal curve is 1, it follows that the areas on either side of z = 0 are both 0.5. Also, according to the Standard Deviation Rule, most of the area under the standardized curve falls between z = -3 and z = +3.
The normal table outlines the precise behavior of the standard normal random variable Z, the number of standard deviations a normal value x is below or above its mean. The normal table provides probabilities that a standardized normal random variable Z would take a value less than or equal to a particular value z*.
These particular values are listed in the form *.* in rows along the left margins of the table, specifying the ones and tenths. The columns fine-tune these values to hundredths, allowing us to look up the probability of being below any standardized value z of the form *.**. Here is part of the table.
By construction, the probability P(Z < z*) equals the area under the z curve to the left of that particular value z*.
A quick sketch is often the key to solving normal problems easily and correctly.