Module 6: Probability and Probability Distributions
Module 6: Probability and Probability Distributions
Normal Random Variables (5 of 6)
Normal Random Variables (5 of 6)
Learning OUTCOMES
- Use a normal probability distribution to estimate probabilities and identify unusual events.
We now know that the empirical rule gives probabilities for values that lie exactly 1, 2, and 3 standard deviations away from the mean. But how do we determine the probability that a value lies some fraction of a standard deviation away from the mean? In this situation, we will use technology to find the probability.
For now, we use a simulation to find the probability based on a z-score. Statistical packages can also be used to find probabilities associated with a normal curve.
Example
The Distribution of z-Scores
Recall that we use the area under a normal density curve to find a probability. If we convert the x-values into z-scores, the distribution of z-scores is also a normal density curve. This curve is called the standard normal distribution.
Here we compare the normal density curve for the foot lengths to the standard normal curve:
The normal curve pictured on top is the model for the distribution of foot lengths. Note that the values on the axis are foot lengths. The distribution has a mean of 11 inches and a standard deviation of 1.5 inches. A foot length of 13 inches is marked. The shaded area is 0.0918. This is the probability that a randomly selected male will have a foot length greater than 13 inches: P(X > 13) = 0.0918.
The normal curve pictured on bottom is the standard normal distribution. This represents the distribution of z-scores. Note that the values on the axis are z-scores. The mean is 0 and the standard deviation is 1. The z-score corresponding to X = 13 inches is marked.
[latex]Z = \frac{x - \mu}{\sigma} = \frac{13 - 11}{1.5} = 1.33[/latex]
The shaded area here is the same, 0.0918. This is the probability that a z-score is greater than 1.33: P(Z > 1.33) = 0.0918.
Here is the main idea: Since the areas are the same, we use the standard normal curve to find the probabilities associated with any normal density curve.
Note: The standard normal distribution always has a mean = 0 and a standard deviation = 1. To understand this, recall that a z-score is the number of standard deviations X is above (or below) the mean.
- When the x-value is the mean, the z-score is 0.
- We can illustrate this for the foot lengths: If X is the mean, then
[latex]X = 11[/latex]
[latex]Z = \frac{x - \mu}{\sigma} = {11 - 11}{1.5} = \frac{0}{1.5} = 0[/latex]
- When X is 1 standard deviation above the mean, the z-score is 1.
- We can illustrate this for the foot lengths: If x-value is 1 standard deviation above the mean, X = 11 + 1.5 = 12.5.
[latex]Z = \frac{x - \mu}{\sigma} = {12.5 - 11}{1.5} = {1.5}{1.5} = 1[/latex]
- Similarly, the z-score for the x-value that is 1 standard deviation below the mean is −1.
Example
Using the Standard Normal Curve to Find Probabilities
We use a simulation based on the standard normal distribu