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Ex. 6.1-1: NHL Goals
In 2010, there were 1319 games played in the National Hockey League’s regular season. Imagine selecting
one of these games at random and then randomly selecting one of the two teams that played in the game.
Define the random variable X = number of goals scored by a randomly selected team in a randomly selected
game. The table below gives the probability distribution of X:
Goals:
0
1
2
3
4
5
6
7
8
9
Probability: 0.061 0.154 0.228 0.229 0.173 0.094 0.041 0.015 0.004 0.001
Show that the probability distribution for X is legitimate.
Make a histogram of the probability distribution. Describe what you see.
What is the probability that the number of goals scored by a randomly selected team in a randomly selected
game is at least 6?
Ex. 6.1-2: More Roulette
Another wager players can make in Roulette is called a “corner bet.” To make this bet, a player places his
chips on the intersection of four numbered squares on the Roulette table. If one of these numbers comes up
on the wheel and the player bet $1, the player gets his $1 back plus $8 more. Otherwise, the casino keeps
the original $1 bet. If X = net gain from a single $1 corner bet, the possible outcomes are X = –1 or X = 8.
Here is the probability distribution of X:
Value:
–$1
$8
Probability: 34/38 4/38
What is the player’s average gain? In the long run, the player loses $1 in 34 of every 38 games and gains $8
in 4 of every 38 games. Imagine a hypothetical 38 bets. The player’s average gain is:
1  1   1  8  8  8  8
X 
38
34(1)  4(8)
X 
38
 34 
 4
 X  (1)    (8)  
 38 
 38 
 X  $0.05
If a player were to make $1 corner bets many, many times, the average gain would be about –$0.05 per bet.
In other words, in the long run, the casino keeps about 5 cents of every dollar bet in roulette.
Note: All of the bets in roulette have the same expected value! However, the bets have different standard
deviations, with the red/black bet having the smallest standard deviation and the “straight-up” bet on a
single value having the highest standard deviation.
Ex. 6.1-3, 4: NHL Goals
In an earlier alternate example, we defined the random variable X to be the number of goals scored by a
randomly selected team in a randomly selected game. The table below gives the probability distribution of
X:
Goals:
0
1
2
3
4
5
6
7
8
9
Probability: 0.061 0.154 0.228 0.229 0.173 0.094 0.041 0.015 0.004 0.001
Compute the mean of the random variable X and interpret this value in context.
Compute the standard deviation of the random variable X and interpret this value in context.
Ex. 6.1-6: Weights of Three-Year-Old Females
The weights of three-year-old females closely follow a Normal distribution with a mean of  = 30.7 pounds
and a standard deviation of 3.6 pounds. Randomly choose one three-year-old female and call her weight X.
Find the probability that the randomly selected three-year-old female weighs at least 30 pounds.
State:
Plan: The weight X of the three-year-old female we choose has the N(30.7, 3.6) distribution. We want to
find P(X  30). We will standardize this weight and use table A to find the shaded area.
Do:
Conclude:
Ex. 6.2-1, 2: El Dorado Community College
El Dorado Community College considers a student to be full-time if he or she is taking between 12 and 18
units. The number of units X that a randomly selected El Dorado Community College full-time student is
taking in the fall semester has the following distribution.
Number of Units (X) 12
13
14
15
16
17
18
Probability
0.25 0.10 0.05 0.30 0.10 0.05 0.15
Here is a histogram of the probability distribution along with the mean and standard deviation:
0.30
Probability
0.25
0.20
0.15
0.10
0.05
0.00
12
13
 X  12(0.25)  13(0.10)   18(0.15)  14.65
 X2  (12  14.65)2 (0.25)  (13  14.65)2 (0.10) 
14
15
16
Number of Units
17
18
 (18  14.65)2 (0.15)  4.23
 X  4.23  2.06
At El Dorado Community College, the tuition for full-time students is $50 per unit. That is, if T = tuition
charge for a randomly selected full-time student, T = 50X. Here are the details for T.
Tuition Charge (T) 600 650 700 750 800 850 900
Probability
0.25 0.10 0.05 0.30 0.10 0.05 0.15
30
Probability
25
20
15
10
5
0
600
650
700
750
800
Tuition Charge
850
900
T  600(0.25)  650(0.10)   (900)(0.15)  $732.50
 T2  (600  732.5) 2 (0.25)  (650  732.5) 2 (0.10)   (900  732.5)2 (0.15)  10568
 T  10568  $103
Compare the details of X with the details of T.
In addition to tuition charges, each full-time student at El Dorado Community College is assessed student
fees of $100 per semester. If C = overall cost for a randomly selected full-time student, C = 100 + T.
Here is the probability distribution for C and a histogram of the probability distribution:
Overall Cost (C)
Probability
700 750 800 850 900 950 1000
0.25 0.10 0.05 0.30 0.10 0.05 0.15
30
Probability
25
20
15
10
5
0
700
750
800
850
Overall Cost
900
950
1000
The mean and standard deviation of this distribution are:
C  700(0.25)  750(0.10)   (1000)(0.15)  $832.50
 C2  (700  832.5) 2 (0.25)  (750  832.5) 2 (0.10) 
 C  10568  $103
Compare the details of T with the details of C.
Compare the details of X with the details of C.
 (1000  832.5)2 (0.15)  10568
Ex. 6.2-3: Scaling a Test
In a large introductory statistics class, the distribution of X = raw scores on a test was approximately
normally distributed with a mean of 17.2 and a standard deviation of 3.8.
The professor decides to scale the scores by multiplying the raw scores by 4 and adding 10.
(a) Define the variable Y to be the scaled score of a randomly selected student from this class. Find the
mean and standard deviation of Y.
(b) What is the probability that a randomly selected student has a scaled test score of at least 90?
Ex. 6.2-4: Speed Dating
To save time and money, many single people have decided to try speed dating. At a speed dating event,
women sit in a circle and men spend about 5 minutes getting to know a woman before moving on to the next
one.
What is the probability that the man is taller than the woman in a randomly selected speed dating couple?
To answer this question, we need to know about the distribution of men’s heights M, the distribution of
women’s heights W, and what happens when we combine these distributions.
Ex. 6.2-5, 6: El Dorado Community College
El Dorado Community College also has a campus downtown, specializing in just a few fields of study.
Full-time students at the downtown campus only take 3-unit classes. Let Y = number of units taken in the
fall semester by a randomly selected full-time student at the downtown campus. Here is the probability
distribution of Y:
Number of Units (Y) 12 15 18
Probability
0.3 0.4 0.3
The mean of this distribution is Y = 15 units, the variance is  Y2 = 5.40 units2 and the standard deviation is
 Y = 2.3 units.
If you were to randomly select 1 full-time student from the main campus and 1 full-time student from the
downtown campus and add their number of units, the expected value of the sum (S = X + Y) would be:
S   X  Y  14.65  15  29.65 .
Assume that X and Y are independent, which is reasonable since each student was selected at random. Here
are all the possible combinations of X and Y, followed by the probability distribution of S.
X
12
12
12
13
13
13
14
14
14
15
15
15
16
16
16
17
17
17
18
18
18
S
P(S)
24
0.075
25
0.03
26
0.015
27
0.19
P(X)
0.25
0.25
0.25
0.10
0.10
0.10
0.05
0.05
0.05
0.30
0.30
0.30
0.10
0.10
0.10
0.05
0.05
0.05
0.15
0.15
0.15
28
0.07
29
0.035
Y P(Y) S=X+Y P(S)=P(X)P(Y)
12 0.3
24
0.075
15 0.4
27
0.10
18 0.3
30
0.075
12 0.3
25
0.03
15 0.4
28
0.04
18 0.3
31
0.03
12 0.3
26
0.015
15 0.4
29
0.02
18 0.3
32
0.015
12 0.3
27
0.09
15 0.4
30
0.12
18 0.3
33
0.09
12 0.3
28
0.03
15 0.4
31
0.04
18 0.3
34
0.03
12 0.3
29
0.015
15 0.4
32
0.02
18 0.3
35
0.015
12 0.3
30
0.045
15 0.4
33
0.06
18 0.3
36
0.045
30
0.24
31
0.07
32
0.035
33
0.15
34
0.03
35
0.015
36
0.045
S  24(0.075)  25(0.03)   36(0.045)  29.65
 S2  (24  29.65) 2 (0.075)  (25  29.65) 2 (0.03)   (36  29.65) 2 (0.045)  9.63
Notice that S =  X  Y (29.65 = 14.65 + 15) and that  S2   X2   Y2 (9.63 = 4.23 + 5.40).
Ex. 6.2-7, 8, 9: Tuition, Fees, and Books
Let B = the amount spent on books in the fall semester for a randomly selected full-time student at El
Dorado Community College. Suppose that B  153 and  B  32 .
Recall from earlier that C = overall cost for tuition and fees for a randomly selected full-time student at El
Dorado Community College and C = 832.50 and  C = 103.
(a) Find the mean and standard deviation of the cost of tuition, fees, and books (C + B) for a randomly
selected full-time student at El Dorado Community College.
(b) At the downtown campus, full-time students pay $55 per unit. Let U = cost of tuition for a randomly
selected full-time student at the downtown campus. Find the mean and standard deviation of U.
(c) Calculate the mean and standard deviation of the total amount of tuition for a randomly selected fulltime student at the main campus and for a randomly selected full-time student at the downtown campus.
(d) Suppose we randomly select one full-time student from each of the two campuses. What are the mean
and standard deviation of the difference in tuition charges, D = T – U? Interpret each of these values.
Ex. 6.2-10: A Graphing Calculator Simulation
You can imitate the Computer Simulation example on a graphing calculator. For example, in the heading of
L1, enter RandNorm(3, 0.9, 500) to randomly generate 500 values from a Normal distribution with a mean
of 3 and a standard deviation of 0.9. Likewise, entering RandNorm(1, 1.2) in the heading of L2 will
randomly generate 500 values from a Normal distribution with a mean of 1 and a standard deviation of 1.2.
In the heading of L3, enter L1+L2 and in the heading of L4, enter L1–L2. Before looking at graphs, review
the values in the first row of each list to make sure students understand where each value came from. Now,
look at histograms of L1, L2, L3 and L4 and you should see the same things as the Computer Simulation
example.
Ex. 6.2-11: Apples
Suppose that the weights of a certain variety of apples have weights that are approximately Normally
distributed with a mean of 9 ounces and a standard deviation of 1.5 ounces.
If bags of apples are filled by randomly selecting 12 apples, what is the probability that the sum of the
weights of the 12 apples is less than 100 ounces?
State: What is the probability that a random sample of 12 apples has a total weight less than 100 ounces?
Plan: Let X = weight of a randomly selected apple. Then X1 = weight of first randomly selected apple, etc.
We are interested in the total weight T = X1 + X2 + + X12. Our goal is to find P(T < 100).
Do:
Conclude:
Note: to get full credit on the AP exam when using the calculator command normalcdf, students must clearly
identify the shape (Normal), center (mean = 108) and spread (standard deviation = 5.2) somewhere in their
work.
Ex. 6.2-12: Speed Dating
Suppose that the height M of male speed daters follows a Normal distribution with a mean of 70 inches and
a standard deviation of 3.5 inches and the height F of female speed daters follows a Normal distribution
with a mean of 65 inches and a standard deviation of 3 inches.
What is the probability that a randomly selected male speed dater is taller than the randomly selected female
speed dater with whom he is paired?
State: What is the probability that a randomly selected male speed dater is taller than the randomly selected
female speed dater with whom he is paired?
Plan: We’ll define the random variable D = M – F to represent the difference between the male’s height and
the female’s height. Our goal is to find P(M > F) or P(D > 0).
Do:
Conclude: There is about an 86% chance that a randomly selected male speed dater will be taller than the
female he is randomly paired with. Or, in about 86% of speed dating couples, the male will be taller than
the female.
Ex. 6.3-1: Dice, cars, and hoops
Determine whether the random variables below have a binomial distribution. Justify your answer.
(a) Roll a fair die 10 times and let X=the number of sixes.
(b) Shoot a basketball 20 times from various distances on the court. Let Y=number of shots made.
(c) Observe the next 100 cars that go by and let C=color.
Ex. 6.3-2: Rolling Doubles
In many games involving dice, rolling doubles is desirable. Rolling doubles mean the outcomes of two dice
are the same, such as 1&1 or 5&5. The probability of rolling doubles when rolling two dice is 6/36 = 1/6.
If X = the number of doubles in 4 rolls of two dice, then X is binomial with n = 4 and p = 1/6.
What is P(X = 0)?
That is, what is the probability that all 4 rolls are not doubles?
Since the probability of not getting doubles is 1 – 1/6 = 5/6, P(X = 0) = P(FFFF) = (5/6)(5/6)(5/6)(5/6) =
(5/6)4 = 0.482.
Note: F represents a failure and S represents a Success.
What is P(X = 1)?
There are four different ways to roll doubles once in four tries.
For example, the doubles could occur on the first try (SFFF), the second try (FSFF), the third try (FFSF) or
the fourth try (FFFS).
P(SFFF) = (1/6)(5/6)(5/6)(5/6) = (1/6)(5/6)3
P(FSFF) = (5/6)(1/6)(5/6)(5/6) = (1/6)(5/6)3
P(FFSF) = (5/6)(5/6)(1/6)(5/6) = (1/6)(5/6)3
P(FFFS) = (5/6)(5/6)(5/6)(1/6) = (1/6)(5/6)3
Thus, the probability of rolling doubles once in 4 attempts is P(X = 1) = 4(1/6)(5/6)3 = 0.386.
Ex. 6.3-3, 4: Rolling Doubles
When rolling two dice, the probability of rolling doubles is 1/6. Suppose that a game player rolls the dice 4
times, hoping to roll doubles.
(a) Find the probability that the player gets doubles twice in four attempts.
(b) Should the player be surprised if he gets doubles more than twice in four attempts? Justify your answer.
If X = number of doubles in 4 attempts, then X follows a binomial distribution with n = 4 and p = 1/6. Here
is the probability distribution and a histogram of the probability distribution.
Value (x)
0
1
2
3
4
Probability 0.482 0.386 0.116 0.015 0.001
0.5
Probability
0.4
0.3
0.2
0.1
0.0
0
1
2
Number of Doubles
The mean of this distribution is  X = 0(0.482) + 1(0.386) +
3
4
+ 4(0.001) = 4(1/6) = 0.667
The variance of this distribution is  X2 = (0 – 0.667)2(0.482) + (1 – 0.667)2(0.386) +
0.667)2(0.001) = 4(1/6)(5/6) = 0.556 so the standard deviation is  X  0.556 = 0.746.
Verify the mean and standard deviation using your calculator.
+ (4 –
Ex 6.3-5: Tastes as Good as the Real Thing?
The makers of a diet cola claim that its taste is indistinguishable from the full calorie version of the same
cola. To investigate, an AP Statistics student named Emily prepared small samples of each type of soda in
identical cups. Then, she had volunteers taste each cola in a random order and try to identify which was the
diet cola and which was the regular cola. Overall, 23 of the 30 subjects made the correct identification.
If we assume that the volunteers really couldn’t tell the difference, then each one was guessing with a 1/2
chance of being correct.
Let X = the number of volunteers who correctly identify the colas.
(a) Explain why X is a binomial random variable.
(b) Find the mean and the standard deviation of X. Interpret each value in context.
(c) Of the 30 volunteers, 23 made correct identifications. Does this give convincing evidence that the
volunteers can taste the difference between the diet and regular colas?
Ex. 6.3-6: NASCAR Cards and Cereal Boxes
In the NASCAR Cards and Cereal Boxes example from section 5.1, we read about a cereal company that
put one of 5 different cards into each box of cereal. Each card featured a different driver: Jeff Gordon, Dale
Earnhardt, Jr., Tony Stewart, Danica Patrick, or Jimmie Johnson. Suppose that the company printed 20,000
of each card, so there were 100,000 total boxes of cereal with a card inside.
If a person bought 6 boxes at random, what is the probability of getting no Danica Patrick cards?
Let X be the number of Danica Patrick cards obtained from 6 different boxes of cereal. Since we are
sampling without replacement, the trials are not independent, so the distribution of X is not quite binomial—
but it is close.
6
If we assume X is binomial with n = 6 and p = 0.2, then P(X = 0) =   (0.2)0 (0.8) 6 = 0.262144
0
The actual probability, using the general multiplication rule is:
 80,000  79,999  79,998  79,997  79,996  79,995 
P(no Danica Patrick cards) = 





 = 0.262134.
 100,000  99,999  99,998  99,997  99,996  99,995 
These two probabilities are quite close!
Ex. 6.3-7: Dead Batteries
Almost everyone has one—a drawer that holds miscellaneous batteries of all sizes. Suppose that your
drawer contains 8 AAA batteries but only 6 of them are good. You need to choose 4 for your graphing
calculator.
If you randomly select 4 batteries, what is the probability that all 4 of the batteries you choose will work?
 4
Explain why the answer isn’t P(X = 4) =   (0.75) 4 (0.25)0 = 0.3164. The actual probability is 0.2143.
 4
Ex. 6.3-8: Teens and Debit Cards
In a survey of 506 teenagers ages 14-18, subjects were asked a variety of questions about personal finance.
(http://www.nclnet.org/personal-finance/66-teens-and-money/120-ncl-survey-teens-and-financial-education)
One question asked teens if they had a debit card.
Suppose that exactly 10% of teens ages 14-18 have debit cards.
Let X = the number of teens in a random sample of size 506 that have a debit card.
(a) Show that the distribution of X is approximately binomial.
(b) Check the conditions for using a Normal approximation in this setting.
(c) Use a Normal distribution to estimate the probability that 40 or fewer teens in the sample have debit
cards.
Note: to get full credit on the AP exam when using the calculator command normalcdf, students must clearly
identify the shape (Normal), center (mean = 50.6) and spread (standard deviation = 6.75) somewhere in
their work.
Ex. 6.3-9, 10: Monopoly
The random variable of interest in this example is Y = number of attempts it takes to roll doubles one time.
Each attempt is one trial of the chance process.
Let’s check the conditions for a geometric setting:
Binary?
Independent?
Trials?
Success?
(a) Find the probability that it takes 3 turns to roll doubles.
(b) Find the probability that it takes more than 3 turns to roll doubles and interpret this value in context.