Module 6: Probability and Probability Distributions
Module 6: Probability and Probability Distributions
Normal Random Variables (6 of 6)
Normal Random Variables (6 of 6)
Learning OUTCOMES
- Use a normal probability distribution to estimate probabilities and identify unusual events.
Now we use the simulation and the standard normal curve to find the probabilities associated with any normal density curve.
Example
Length of Human Pregnancy
The length (in days) of a randomly chosen human pregnancy is a normal random variable with μ = 266, σ = 16. So X = length of pregnancy (in days)
(a) What is the probability that a randomly chosen pregnancy will last less than 246 days?
We want P(X < 246). To find this probability, we first convert X = 246 to a z-score:
[latex]Z = \frac{246 - 266}{16} = \frac{-20}{16} = -1.25[/latex]
Now we can use the simulation to find P(Z < −1.25). This is the area under the normal probability curve to the left of Z = −1.25.
The probability that a randomly chosen pregnancy lasts less than 246 days is 0.1056. In other words, there is an 11% chance that a randomly selected pregnancy will last less than 246 days.
(b) Suppose a pregnant woman’s husband has scheduled his business trips so that he will be in town between the 235th and 295th days of her pregnancy. What is the probability that the birth will take place during that time?
Compute the z-scores for each of these x-values:
[latex]Z = \frac{235 - 266}{16} = \frac{-31}{16} = -1.94[/latex]
and
[latex]Z = \frac{295 - 266}{16} = \frac{29}{16} = 1.81[/latex]
Use the simulation to find the area under the standard normal curve between these two z-scores.
So the desired probability is 0.9387.
[latex]P(235 < X < 295) = P(-1.94 < Z < 1.81) = 0.9387[/latex]
There is about a 94% probability that he will be home for the birth. Looks like he planned well.
Try It
The previous examples all followed the same general form: Given values of a normal random variable, we found an associated probability. The two basic steps in the solution process were as follows:
- Convert x-value to a z-score.
- Use the simulation to find associated probability.
The next example is a different type of problem: Given a probability, we will find the associated value of the normal random variable. The solution process will go in reverse order.
- Use a new simulation to convert statements about probabilities to statements about z-scores.
- Convert z-scores to x-values.
These types of problems are informally called “work-backwards” problems. We will use a new simulation for these types of problems. The new simulation requires us to enter a probability and then gives us the associated z-score. This is backwards from the simulation we worked with previously where we entered a z-score to find a probability. We will use this simulation in the next example.
Click here to open this simulation in its own window.
Example
Work Backwards to Find X
Foot length (in inches) of a randomly chosen adult male is a normal random variable with a mean of 11 and standard deviation of 1.5.