This tutorial will cover standard normal distribution. You will learn about:
Now, the standard normal distribution is a specific kind of normal distribution that uses standard scores. Standard scores are also known as z-scores.
Here is an example to show how to use standard normal distribution: Men's heights are normally distributed with a mean of 68 inches, which is five feet eight inches, and a standard deviation of 3 inches.
What percent of men are over six feet (or 72 inches) tall?
There will be several steps to take to answer this question.
First, recall the 68, 95, 99.7 rule, which says that 68% of the data points fall within one standard deviation of the mean. That means that 68% of men's heights will fall within three inches of 68. 95% will fall within two standard deviations, and 99.7% percent will fall within three standard deviations.
Here's the distribution of men's heights in inches, normally distributed with a mean of 68 and a standard deviation of 3.
Where is 72? It falls between the first and second standard deviation above the mean, and the goal was to find the amount of men above that height. So because the answer is not on an integer standard deviation, it becomes an issue. How do you find the answer?
To solve this challenge, take these heights and standardize them by turning these values into z-scores. Z-scores are how many standard deviations away from the mean an observation is. In the above distribution, 71 is one standard deviation above the mean, so in our new graph, it will be marked as +1. 74 is two standard deviations above the mean, so it will be marked as +2. The rest of the numbers will follow, resulting in a graph that looks like this:
In this graph, the 68 has been marked as 0, and all the other numbers into the integer number of standard deviations away from the mean that they were.
But remember, 72 wasn't an integer number of standard deviations. It was between one and two standard deviations away. If you calculate the z-score for that cutoff point of 72 inches, it's at 1.33 standard deviations above the mean.
So you want to find the percent of values that are above that.
To solve the problem you were addressing above, you need a standard normal table.
A standard normal table is a table of probabilities that lie below particular z-scores, a table that calculates the percent of values below a particular z-score. Z-tables will look like this:
With these tables, you can find the percent of values that fall at or below a particular z-score.
Negative z-score probability looks like this:
The image below is the negative z-score table. Notice that the values are always below the mean. The left side shows negative z-scores that fall to the left of the mean.
(For your reference, this table is included as a pdf below the tutorial.)
Positive z-score probability looks like this:
The image below is the positive z-score table. Notice that the values are always below the mean. The right side shows positive z-scores that fall to the right of the mean.
(For your reference, this table is included as a pdf below the tutorial.)
You're only going to need to use one of these tables at a time. Because your z-score was positive 1.33, you’re going to use the positive z-score table.
The column on the far left of the image represents the tenths place of your z-score. Your z-score was 1.33, so you’re going to find 1.3 as your tenths.
The row across the top is the hundredths place of your z-score. In this case, your tenths place was 1.3 and your hundredths place was also three: 1.33.
Look in the table for the value that corresponds to the row of 1.3 and the hundredths place a 0.03:
This is going to give you the amount of area below, meaning the percent of men shorter than 72 inches tall. 90.82% of men are shorter than six feet tall. Since the original question was what percent of men are taller than six feet, subtract from 100%. You should find that 9.18% of men that have height over 72 inches.
The same process can be used to:
What percent of men are between five foot six and five foot nine, or 66 and 69 inches tall?
The orange area in the above graph is what you’re looking for. The z-scores are negative 0.67 and positive 0.33.
Find the table values that correspond to positive 0.33, which is 0.6293 from the table. Then, find the table value for negative 0.67, which is 0.2514, and subtract.
In this graph, the orange area is equal to the combined yellow and orange minus the yellow. The combined yellow and orange is what the table value is for positive 0.33. The yellow is the table value for negative 0.67. When you subtract them, you end up with about 38%.
Therefore, the area contains about 38% (37.79%), which represents all men who fall in between those two heights.
It's possible to find the percent of values above or below a particular value using something other than the 68, 95, 99.7 rule using the z-scores on the standard normal distribution. The standard normal table helps you find the percent of values below a particular z-score in order to calculate a percentage of values above or below a particular value or between two values.
Thank you and good luck!
Source: THIS WORK IS ADAPTED FROM SOPHIA AUTHOR JONATHAN OSTERS
A normal distribution of z-scores. The mean is zero, and the standard deviation is one.
A table that calculates the percent of values below a particular z-score.