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Ch. 5: Continuous Random Variables

Module by: Susan Dean, Barbara Illowsky, Ph.D.. E-mail the authors

Summary: This module is the complementary teacher's guide for the Continuous Random Variables chapter of the Collaborative Statistics collection (col10522) by Barbara Illowsky and Susan Dean.

This chapter is a good introduction to continuous types of probability distributions (the most famous of all is the normal). Two continuous distributions are covered – the uniform (or rectangular) and the exponential. For the uniform, probability is just the area of a rectangle. This distribution easily gets across the concept that probability is equal to area under a "curve" (a function). The exponential, which is used in industry and models decay, is a nice lead-in to the normal. The uniform and exponential distributions are also nice distributions to start with when you teach the Central Limit Theorem. It is interesting to note that the amount of money spent in one trip to the supermarket follows an exponential distribution. Several of our students discovered this idea when they chose data for their second project.

Compare Binomial v. Continuous Distribution

Begin this chapter by a comparison of a binomial (discrete) distribution and a continuous distribution. Using the normal for this comparison works well because the students are already familiar with it. The binomial graph has probability = height and the normal graph has probability = area. Tell the students that the discovery of probability = area in the continuous graph comes from calculus (which most of them have not studied). Draw the two graphs to make these ideas clear.

Introduce Uniform Distribution

Introduce the uniform distribution using the following example: The amount of time a student waits in line at the college cafeteria is uniformly distributed in the interval from 0 to 5 minutes (the students must wait in line from 0 to 5 minutes - each time in this interval is equally likely). Note: all the times cannot be listed. This is different from the discrete distributions.

Example 1

Let XX= the amount of time (in minutes) a student waits in line at the college cafeteria. The notation for the distribution is XX~ U(a,b) U(a,b) where a=0a=0 and b=5b=5. The function is f(x)= 15f(x)=15 where 0<x<50<x<5. The pattern is f(x)= 1b−af(x)=1b−a where a<x<ba<x<b.

In this example a=0a=0 and b=5b=5. The function f(x)f(x) where 0<x<50<x<5 graphs as a horizontal line segment.

Figure 1: Because 0<x<5 0<x<5, the maximum area = (15)(5)=1, the largest probability possible.
Figure 1 (graph.PNG)

Example 2

Problem 1

Find the probability that a student must wait less than 3 minutes. Draw the picture and write the probability statement.

Solution

Figure 2: Probability statement: P(X<3)=(3 −0)(15 )=35P(X<3)=(3−0)(15)=35.
Figure 2 (graph3.PNG)
The probability is the shaded area (the area of a rectangle with base = b−a=3− 0=3base=b−a=3−0=3 and height = 15height=15. The probability a student must wait in the cafeteria line less than 3 minutes is 3535.

Example 3

Problem 1

Find the average wait time.

Solution

μ=a+b2=0+52 =2.5 minutesμ=a+b2=0+52=2.5minutes

If the students take the time to draw the picture and write the probability statement, the problem becomes much easier.

Example 4

Problem 1

Find the 75th percentile of waiting times. A time is being asked for here. Percentiles often confuse students. They see "75th" and think they need to find a probability. Draw a picture and write a probability statement. Let kk = the 75th percentile.

Solution

Figure 3
Figure 3 (graph4.PNG)
  • Probability statement: P(X<k)=0.75P(X<k)=0.75
  • Area: (k−0)(15) =0.75k=3.75(k−0)(15)=0.75k=3.75 minutes
75% of the students wait at most 3.75 minutes and 25% of the students wait at least 3.75 minutes.

Example 5

Problem 1

You can finish the uniform with a conditional. This reviews conditionals from Continuous Random Variables. What is the probability that a student waits more than 4 minutes when he/she has already waited more than 3 minutes?

Solution

Algebraically: P(X>4| X>3) = P(X>4ANDX>3) P(X>3) =P(X>4) P(X>3)P(X>4|X>3)=P(X>4ANDX>3) P(X>3)=P(X>4) P(X>3)

Note:
The students see it more clearly if you do the problem graphically. The lower value, a, changes from 0 to 3. The upper value stays the same (b=5(b=5). The function changes to: f ( x ) = 1 5 − 3 = 1 2 f(x)= 1 5 − 3 = 1 2
Figure 4: P ( X> 4 | X > 3 ) = ( base ) ( height ) = ( 5 - 4 ) ( 1 2 ) = 1 2 P(X>4|X>3)=(base)(height)=(5-4)( 1 2 )= 1 2
Figure 4 (graph5.PNG)

Introduce the Change Example

The exponential distribution is generally concerned with how a quantity declines or decays. Examples include the life of a car battery, the life of a light bulb, the length of time business long distance telephone calls last, and the amount of change a person is carrying. You can introduce the exponential by using the change example. Ask everyone in your classroom to count their change and record it. Then have them calculate the mean and standard deviation and graph the histogram. The histogram should appear to be declining. Let XX = the amount of change one person carries. Notation: XX ~ Exp(m)Exp(m) where mm is the parameter that controls the amount of decline or decay; m=1μm=1μ and μ=1mμ=1m. Also, μ=σμ=σ. (In the example, the calculated mean and standard deviation ought to be fairly close.)

Example 6

The function is where f(x)=me−mxf(x)=me−mx m≥0m≥0 AND x≥0x≥0. Find the probability that the amount of change one person has is less then $.50. Draw the graph.

Figure 5: The right tail extends indefinitely. There is no upper limit in x.
Figure 5 (graph6.PNG)
The formula is P(X<x)=1−emx P(X<.50)=P(X<x)=1−emxP(X<.50)= _________. The authors use technology to solve the probability problems. If you use the TI-83/84 calculator series, enter on the home-screen, 1−e−m⋅.501−e−m⋅.50. Fill in the mm with whatever the data produces ( m=1μm=1μ; replace μμ with the sample mean).

Ask the question, "Ninety percent of you have less than what amount?" and have them find the 90th percentile.

Draw the picture and let kk = the 90th percentile. P(X<k)=0.90P(X<k)=0.90. Solve the equation 1−emk=0.901−emk=0.90 for kk. On the home-screen of the TI-83/TI-84, enter ln(1−.90)(−m)ln(1−.90)(−m).

Note:

Have students fill in the blanks.
On average, a student would expect to have _________ . The word "expect" implies the mean. Ten students together would expect to have _________. (the mean multiplied by 10)

Assign Practice

Assign the Practice 1 and Practice 2 in class to be done in groups.

Assign Homework

Assign Homework . Suggested problems: 1 -13 odds, 15 - 20.

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