1. 7.1 Discrete & Continuous Random Variables
AP Statistics
Spring 2010

2. Discrete Random Variables
When we roll two dice, we can define the sum to be a variable, X. X can take on any value from 2 through 12. Since we don’t know exactly what sum will appear on a given roll, we call X a random variable.

3. Terminology:
A random variable is a variable whose value is a numerical outcome of a random phenomenon where one and only one number is assigned to each outcome.
Discrete Random Variable—has a countable number of possible values.
Probability Distribution—list of values and their probabilities but can be in a table, graph, formula which specifies the probability associated with each possible value the RV can assume. probability mass function (p.m.f.)

4. x x1 x2 … xn P(x) P(x1) P(x2) Example: One form of a p.m.f. for x:
Important things to note:
Each of the events are mutually exclusive
The sum of all the probabilities is equal to 1
Note: If “x” is not in the range, then P(X = x) = 0 x x1 x2 … xn
P(x)
P(x1)
P(x2)

5. Example 1: Consider rolling two dice
Example 1: Consider rolling two dice. Define X to be the sum of the two dice. Construct the probability distribution of X and display it with a probability histogram. x 2 3 4 5 6 7 8 9 10 11 12
P(x)
1/36
2/36
3/36
4/36
5/36
6/36

6. P(X = 0) = P(X = 0 or 1) = P(X > 2) = P(X < 3) = P(X > 1) = x
Example 2: Consider flipping a coin 4 times and recording H or T. Define X to be the number of Heads flipped. Construct the probability distribution of X and use it to answer the following questions:
P(X = 0) =
P(X = 0 or 1) =
P(X > 2) =
P(X < 3) =
P(X > 1) = x 1 2 3 4
P(x)
1/16
4/16
6/16
1/16
5/16
5/16
15/16
15/16

7. Example 3: (7.16 p nd Ed. ) Weary of the low turnout in student elections, a college administration decides to choose a sample of three students to form an advisory board that represents student opinion. Suppose that 40% of all students oppose the use of student fees to fund student interest groups and that the opinions of the three students on the board are independent. Then the probability is 0.4 that each opposes the funding of interest groups.
Call the three students A, B, and C. What is the probability that A and B support funding and C opposes it?
(0.6)(0.6)(0.4) = 0.144

8. List all possible combinations of opinions that can be held by students A, B, and C. Then give the probability of each of these outcomes.
S = Support O = Oppose
S = {SSS, SSO, SOS, OSS, SOO, OSO, OOS, OOO}
P(SSS) =
P(SSO) = P(SOS) = P(OSS) =
P(SOO) = P(OSO) = P(OOS) =
P(OOO) =
(0.6)3
= 0.216
(0.6)2(0.4)
= 0.144
(0.6)(0.4)2
= 0.096
(0.4)3
= 0.064

9. P(x > 2) or P(x > 1) = 0.288 + 0.064 = 0.352
(c) Let the random variable X be the number of student representatives who oppose the funding of interest groups. Give the probability distribution of X.
Express the event “a majority of the advisory board opposes funding” in terms of X and find its probability.
P(x > 2) or P(x > 1)
= = 0.352 x 1 2 3
P(x)
0.216
0.432
0.288
0.064

10. Continuous Random Variables
Rolling dice and flipping coins result in random variables whose outcomes are countable. Some situations result in outcomes that can take on any value over a given interval.
Terminology:
Continuous Random Variable—represent numerical values that go on forever; ex: time, distance. The values of these variables are intervals, not discrete numbers.
Probability Distribution of a Continuous R. V.—described by a density curve since we cannot list all possible outcomes. We view the areas under the curve as the probability values.

11. Special Notes: The total area under a density curve is 1.
The probability of an individual outcome is 0.

12. P(x > .25) = P(x < .5) = P(.25 < x < .5) = 1.75 x .5
Example 4: Draw a rectangular density curve whose height is .5. What is the length?
Find:
P(x > .25) =
P(x < .5) =
P(.25 < x < .5) =
0.5 2
1.75 x .5
= 0.875
.5 x .5
= 0.25
0.25 x 0.5
= 0.125

13. Example 5: Suppose that an opinion poll asks a random sample of 1500 adults, “Do you happen to jog?” Suppose that the population proportion who jog is 0.15 and that this distribution is approximately normally distributed with mean µ = 0.15 and standard deviation
N~(µ = 0.15, σ = .0092)
Find P( X ≥ .16)
Normalcdf(.16, 1E99, .15, .0092)  
Find P( .14 ≤ X ≤ .16)
Normalcdf(.14, .16, .15, .0092) 
= .1385
= .7229

14. Example 6: Let the random variable X represent the profit made on a randomly selected day by a certain store. Assume X is normal with a mean of $360 and standard deviation $50. The probability is approximately 0.6 that on a randomly selected day the store will make less than x0 amount of profit. Find x0.
X ~ N(µ = 360, σ = 50) P(z < x0) = 0.6 invNorm(0.6, 360, 50) x0 = $372.67

15. Example 7: According to a recent AP poll, approximately 40% of American adults indicated they used the internet to get news and information about political candidates. Suppose 40% of all American adults use this method to get their political information. What would happen if you randomly sampled a group of 1500 American adults and asked them if they used the internet to get this information? Define X to be the % of your sample that would respond that the internet was their primary source.  
Suppose we are told that the distribution of X is approximately N(0.4, ). Use this information to sketch the probability distribution of X and answer the following questions:

16. What is P(X > 0. 42). Normalcdf(. 42, 1E99, 0. 4, 0. 01265) = 0
What is P(X > 0.42)? Normalcdf(.42, 1E99, 0.4, ) = What is P(X < 0.35)? Normalcdf(-1E99, 0.35, 0.4, ) ≈ 0 What is P(your result is within 5% of the actual % who use the internet as a primary source)? P(0.35 < x < 0.45) =