Module 6: Probability and Probability Distributions

# Normal Random Variables (3 of 6)

## Normal Random Variables (3 of 6)

### Learning OUTCOMES

- Use a normal probability distribution to estimate probabilities and identify unusual events.

### Example

## The Empirical Rule in a Context

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 empirical 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?**Answer:**0.95, or 95%**(b)**An adult male is almost guaranteed (0.997 probability) to have a foot length between what two values?**Answer:**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?**Answer:**14 inches

Ninety-five percent of the area is within 2 standard deviations of the mean, so 2.5% of the area is in the tail above 2 standard deviations. The *x*-value 2 standard deviations above the mean is 14 inches.

Now you should try a few: questions (d), (e), and (f) are presented in the Try It activity. Use the figure preceding question (a) to help you.

### Try It

**Comment**

Notice that there are two types of problems we may want to solve: those like (a) and, from the Try It activity, (d) and (e), in which a particular interval of values of a normal random variable is given and we are asked to find a probability; and those like (b), (c), and, from the Try It, (f), in which a probability is given and we are asked to identify values of the normal random variable.

- Concepts in Statistics.
**Provided by**: Open Learning Initiative.**Located at**: http://oli.cmu.edu.**License**:*CC BY: Attribution*

Feedback for interactive question

(d) How likely or unlikely is a male’s foot length to be smaller than 9.5 inches? Not too unlikely, since the probability of being smaller than 9.5 is 0.16, which is not a particularly low probability.

Feedback: Indeed, the probability that foot length is between 9.5 and 12.5 is 0.68, and therefore the remaining two tails together have probability 1 − 0.68 = 0.32. We conclude, then, that: P(X 15.5) = 0.003/2 = 0.0015.

(e) How likely or unlikely is a foot length longer than 15.5 inches? Extremely unlikely, since the probability of being longer than 15.5 is only 0.0015.

Feedback: Indeed, the probability that foot length is between 6.5 and 15.5 is 0.997, and therefore the remaining two tails together have probability 1 − 0.997 = 0.003. We conclude, then, that P(X > 15.5) = 0.003/2 = 0.0015.

(f) There is probability of 0.5 that a male’s foot is shorter than 11.

Feedback: The value that divides the area under the curve into two halves is 11, so that P(X 11) = 0.5.