Download Slide 1 - Algebra 2

Chapter 12: Test Your Proficiency
Directions:
•Select a section to work on.
•Work out each problem on a piece of paper.
•Click to check your answer.
•For detailed steps click on the provided link.
•Move on to the next problem or return to the menu.
Sections
12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations
12-3, 12-4, 12-5: Probability and Odds
12-6: Statistical Measures
12-7: The Normal Distribution
12-8: Binomial Experiments
12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations
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1. How many ways can you arrange 8 songs on a CD from a
list of 13 songs?
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12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations
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1. How many ways can you arrange 8 songs on a CD from a
list of 13 songs?
There are 51,891,840 ways to arrange the 8 songs.
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12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations
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1. How many ways can you arrange 8 songs on a CD from a
list of 13 songs?
Step 1: Determine whether the problem is a Permutation or Combination
The word “arrange” means to decide on an order, so this problem
is a Permutation.
Step 2: Recall the notation and formula for a Permutation
n!
P n , r  
where n is the total and r is how many are being taken
n  r  !
Step 3: Determine the values for n and r and plug them into the formula
Here n = 13 because there are a total of 13 songs and r = 8 because
that’s how many are being taken from the group and arranged
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13!
13!
P 13,8 

 51,891,840
13  8! 5!
There are 51,891,840 ways to arrange the 8 songs.
12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations
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2. How many 5 card hands can be made that include 4
diamonds and 1 club?
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12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations
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2. How many 5 card hands can be made that include 4
diamonds and 1 club?
There are 9295 different 5-card hands with 4 diamonds and 1 club.
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12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations
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2. How many 5 card hands can be made that include 4
diamonds and 1 club?
Step 1: Determine whether the problem is a Permutation or Combination
Since the order of the cards doesn’t matter, this is a combination problem.
However, because it asks for two different types of cards, it is the product
of two separate combinations (two events).
Step 2: Recall the notation and formula for a Combination
n!
C n , r  
where n is the total and r is how many are being taken
n

r
!

r
!


Step 3: Determine the values for n and r for each separate combination
and plug them in.
For the 4 diamonds: n = 13 (the total number of diamonds) and r = 4
For the 1 club: n = 13 (the total number of clubs) and r = 1
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C 13,4   C 13,1
4 diamonds
1 club
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12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations
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Step 4: Simplify and state the conclusion.
C 13,4   C 13,1
4 diamonds
1 club
13! 13!

 9295
9! 4! 12! 1!
There are 9295 different 5-card hands with 4 diamonds and 1 club.
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12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations
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3. An ice cream shop offers two types of cones, 15 flavors of
ice cream and 6 toppings. How many one-scoop cones with
two different toppings can be made?
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12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations
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3. An ice cream shop offers two types of cones, 15 flavors of
ice cream and 6 toppings. How many one-scoop cones with
two different toppings can be made?
There are 900 different cones that can be made.
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12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations
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3. An ice cream shop offers two types of cones, 15 flavors of
ice cream and 6 toppings. How many one-scoop cones with
two different toppings can be made?
Step 1: Determine the number of events.
There are four events because four decisions have to be made: type
of cone, flavor of ice cream, first topping, second topping
Step 2: Determine the number of choices for each event (decision)
and use the Fundamental Counting Principle.
2
 15

6

5
 900
cone flavor 1st topping 2nd topping
Step 3: State the conclusion
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There are 900 different cones that can be made.
12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations
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4. How many ways can the letters in the word INTERCEPT be
arranged?
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12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations
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4. How many ways can the letters in the word INTERCEPT be
arranged?
There are 90,720 ways to arrange the letters.
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12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations
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4. How many ways can the letters in the word INTERCEPT be
arranged?
Step 1: Identify the problem as a Permutation with Repetitions.
The word “arranged” means the letters are being put in an order, making this problem a
permutation. Since some of the letters look exactly the same the problem becomes a
permutation with repetitions.
Step 2: Recall the formula.
n!
where n is the total number of letters;
p ! q !
p and q represent the number of repetitions of letters that occur more than once
Note: If only one object (letter) has repetitions, there will only be a value for p. If
more than two objects have repetitions you must divide by additional factorials
beyond p and q.
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12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations
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4. How many ways can the letters in the word INTERCEPT be
arranged?
Step 3: Identify the values for n, p, q, etc. as needed.
n = 9 because there are 9 letters in the word
p = 2 because there are 2 t’s
q = 2 because there are 2 q’s
Step 4: Plug the values into the formula.
9!
 90,720
2 ! 2 !
Step 5: State the conclusion.
There are 90,720 ways to arrange the letters.
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12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations
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5. How many ways can you choose 9 books from a reading list
of 12 books?
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12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations
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5. How many ways can you choose 9 books from a reading list
of 12 books?
There are 220 ways to choose the 9 books.
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12-1, 12-2: Fundamental Counting Principle, Combinations, Permutations
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5. How many ways can you choose 9 books from a reading list
of 12 books?
Step 1: Identify the problem as a combination.
This problem is a combination because no order or position is being assigned to the chosen books.
Step 2: Recall the formula for a combination.
n!
C n , r  
where n is the total number of objects and
n  r  ! r !
r is how many are being chosen
Step 3: Identify the values for n and r and plug them into the formula.
n = 12 because there are a total of 12 books
r = 9 because that is how many books are being chosen
C 12, 9  
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12 !
12!

 220
12

9
!

9
!
3!

9!


There are 220 ways to choose the 9 books.
12-3, 12-4, 12-5: Probability and Odds
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1. A card is drawn from a standard deck. Find the
probability the card is a seven or a heart.
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12-3, 12-4, 12-5: Probability and Odds
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1. A card is drawn from a standard deck. Find the
probability the card is a seven or a heart.
4
The probability is
.
13
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12-3, 12-4, 12-5: Probability and Odds
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1. A card is drawn from a standard deck. Find the
probability the card is a seven or a heart.
Step 1: Set up the problem.
1) The use of the word “or” means we have to determine
whether the events are mutually exclusive or inclusive.
P(a seven or a heart)
2) Since the card drawn can be both a seven and a heart
at the same time, these events are inclusive.
Step 2: Recall the set up for the probability of inclusive events.
P  A or B   P  A   P B   P both A and B 
P  seven or heart   P  seven   P heart   P  both seven and heart 
Step 3: Determine each probability and simplify.
P  seven or heart  
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4 13 1 16 4




52 52 52 52 13
4
The probability is
.
13
12-3, 12-4, 12-5: Probability and Odds
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2. Two cards are drawn in succession from a standard
deck without replacement. Find the probability the cards
are both hearts.
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12-3, 12-4, 12-5: Probability and Odds
2. Two cards are drawn in succession from a
standard deck without replacement. Find the
probability the cards are both hearts.
1
The probability is
.
17
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12-3, 12-4, 12-5: Probability and Odds
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2. Two cards are drawn in succession from a standard deck without
replacement. Find the probability the cards are both hearts.
Step 1: Set up the problem.
P(a heart, then a heart)
1) The use of a comma or the word “and” means we have to
determine whether the events are independent or dependent.
2) Since the first card drawn is not being replaced the
deck has changed; therefore these events are dependent.
Step 2: Recall the set up for the probability of dependent events.
P  A and B   P  A   P B following A 
P  heart, then heart   P  heart   P heart following heart 
Step 3: Determine each probability and simplify.
P  heart, then heart  
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13 12
1


52 51 17
1
The probability is
.
17
12-3, 12-4, 12-5: Probability and Odds
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3. A card is drawn from a standard deck.
Find P(red card or face card)
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12-3, 12-4, 12-5: Probability and Odds
3. A card is drawn from a standard deck.
Find P(red card or face card)
8
The probability is
.
13
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12-3, 12-4, 12-5: Probability and Odds
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3. A card is drawn from a standard deck.
Find P(red card or face card)
Step 1: Set up the problem.
P(red card or face card)
1) The use of the word “or” means we have to determine
whether the events are mutually exclusive or inclusive.
2) Since the card drawn can be both red and a face card
at the same time, these events are inclusive.
Step 2: Recall the set up for the probability of inclusive events.
P  A or B   P  A   P B   P  both A and B 
P  red or face card  P  red  P  face card  P  both red and face card 
Step 3: Determine each probability and simplify.
P  red or face card 
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26 12 6 32 8




52 52 52 52 13
8
The probability is
.
13
12-3, 12-4, 12-5: Probability and Odds
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4. Michelle has a collection of 15 CDs: 6 contemporary
and 9 classical. If she selects 6 of the CDs at random,
find: P(4 contemporary and 2 classical).
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12-3, 12-4, 12-5: Probability and Odds
4. Michelle has a collection of 15 CDs: 6 contemporary
and 9 classical. If she selects 6 of the CDs at random,
find: P(4 contemporary and 2 classical).
108
The probability is
.
1001
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12-3, 12-4, 12-5: Probability and Odds
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4. Michelle has a collection of 15 CDs: 6 contemporary
and 9 classical. If she selects 6 of the CDs at random,
find: P(4 contemporary and 2 classical).
Step 1: Set up the problem.
1) The absence of the word “then” means the CDs are being taken at the same time, therefore
order is not important. Since more than one CD is being selected at the same time,
combinations must be used.
2) The use of the word “and” means we need to multiply the number of contemporary
combinations by the number of classical combinations, and then put this over the total
number of ways to select 6 CDs.
P  4 contemporary and 2 classical

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 # ways to select 4 contemporary    # ways to select 2 classical
# ways to select any 6 CDs
C  6, 4  C  9,2  15  36 108



C  15,6 
5005 1001
108
The probability is
.
1001
12-3, 12-4, 12-5: Probability and Odds
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5. Michelle has a collection of 15 CDs: 6 contemporary
and 9 classical. If she selects 6 of the CDs at random,
find: P(5 contemporary or 3 classical).
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12-3, 12-4, 12-5: Probability and Odds
5. Michelle has a collection of 15 CDs: 6 contemporary
and 9 classical. If she selects 6 of the CDs at random,
find: P(5 contemporary or 3 classical).
1734
The probability is
.
5005
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12-3, 12-4, 12-5: Probability and Odds
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5. Michelle has a collection of 15 CDs: 6 contemporary
and 9 classical. If she selects 6 of the CDs at random,
find: P(5 contemporary or 3 classical).
Step 1: Identify the characteristics of the problem.
1)
2)
3)
In this problem, the use of the word “or” separates the groups of CDs being chosen making
each group less than the total number of 6 being chosen. It also means we will add.
We need to make up the total of 6 for each group:
5 contemporary becomes 5 contemporary and 1 classical
3 classical becomes 3 classical and 3 contemporary
Since more than one CD is being selected at once we will need to use combinations.
Step 2: Set up the problem.
P  5 contemporary or 3 classical
 P  5 contemporay and 1 classical or 3 classical and 3 contemporary 
C  6, 5  C  9, 1  C  9,3  C  6, 3  6  9  84  20 1734



C  15,6 
5005
5005
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The probability is
1734
.
5005
12-3, 12-4, 12-5: Probability and Odds
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6. The probability an event will happen is
Find the odds the event will happen.
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8
.
15
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12-3, 12-4, 12-5: Probability and Odds
6. The probability an event will happen is
8
.
15
Find the odds the event will happen.
The odds are 8:7.
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12-3, 12-4, 12-5: Probability and Odds
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8
6. The probability an event will happen is .
15
Find the odds the event will happen.
Step 1: Recall the definitions for probability and odds.
Given s = success and f = failure
P s  
s
s f
and Odds = s : f
Step 2: Using the given information to find the values for s and f.
8
s

therefore, s  8 and s  f  15
15 s  f
So, 8  f  15  f  7
P s  
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Step 3: Use the definition of odds to answer the problem.
Odds = s : f  8 : 7
The odds are 8:7.
12-6: Statistical Measures
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1. In his last eight basketball games Jeremy scored the following
number of points: 13, 8, 17, 10, 24, 6, 17, 20. Find the mean,
median, and mode for this set of data.
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12-6: Statistical Measures
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1. In his last eight basketball games Jeremy scored the following
number of points: 13, 8, 17, 10, 24, 6, 17, 20. Find the mean,
median, and mode for this set of data.
The mean is 14.375, the median is 15 and the mode is 17.
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12-6: Statistical Measures
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1. In his last eight basketball games Jeremy scored the following
number of points: 13, 8, 17, 10, 24, 6, 17, 20. Find the mean,
median, and mode for this set of data.
Step 1: Recall the definitions for mean, median, and mode.
Mean = the sum of the data values divided by the number of data values
Median = the middle value when the data are in numerical order
Mode = the data value(s) with the greatest frequency
Step 2: Apply the definitions to find each value.
Put the data in order: 6, 8, 10, 13, 17, 17, 20, 24
115
Mean =
 14.375
8
13+17
Median =
 15; here you must average the two middle values
2
Mode = 17 because there are two 17s and only one of everything else
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The mean is 14.375, the median is 15 and the mode is 17.
12-6: Statistical Measures
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2. Find the variance and standard deviation for the set of data {28,
31, 15, 44, 22}. Round to the nearest tenth if necessary.
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12-6: Statistical Measures
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2. Find the variance and standard deviation for the set of data {28,
31, 15, 44, 22}. Round to the nearest tenth if necessary.
The variance is 94 and the standard deviation is about 9.7.
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12-6: Statistical Measures
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2. Find the variance and standard deviation for the set of data {28,
31, 15, 44, 22}. Round to the nearest tenth if necessary.
Step 1: Recall the steps for finding the variance.
1)
2)
3)
4)
Find the mean
Find the difference between the mean and each data value
Square the differences
Find the mean of the squares
Step 2: Follow the steps to find the variance.
step 1) Find the mean:
steps 2  4) variance
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28  31 15  44  22
 28
5
2
2
2
2
2
28  28    31 28    28  15    44  28    28  22 


5
02  32  132  162  62 470


 94
5
5
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12-6: Statistical Measures
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2. Find the variance and standard deviation for the set of data {28,
31, 15, 44, 22}. Round to the nearest tenth if necessary.
Step 3: Recall the definition of the standard deviation.
The standard deviation is the square root of the variance.
Step 4: Find the standard deviation.
standard deviation = 94  9.7
Step 5: State the conclusion.
The variance is 94 and the standard deviation is about 9.7.
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12-7: The Normal Distribution
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1. The scores on 800 tests are normally distributed with a
mean score of 71 and a standard deviation of 8. What
percent of those tested scored between 71 and 95?
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12-7: The Normal Distribution
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1. The scores on 800 tests are normally distributed with a
mean score of 71 and a standard deviation of 8. What
percent of those tested scored between 71 and 95?
About 49.5% scored between 71 and 95.
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12-7: The Normal Distribution
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1. The scores on 800 tests are normally distributed with a
mean score of 71 and a standard deviation of 8. What
percent of those tested scored between 71 and 95?
Step 1: Make a number line using the given mean and standard
deviation and the corresponding percentages for a normal distribution.
0.5%
2%
47
13.5%
55
34%
63
13.5%
34%
71
mean
79
2%
87
0.5%
95
Step 2: Identify whether the answer should be a percentage or whole
number.
For this problem the phrase “what percent” means the answer is a percentage
number.
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12-7: The Normal Distribution
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1. The scores on 800 tests are normally distributed with a
mean score of 71 and a standard deviation of 8. What
percent of those tested?
Step 3: Identify the correct intervals. “scored between 71 and 95”
0.5%
2%
47
13.5%
55
34%
63
13.5%
34%
71
mean
79
2%
87
0.5%
95
Step 4: Add the percentages and state the conclusion.
34% + 13.5% + 2% = 49.5%
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About 49.5% scored between 71 and 95.
12-7: The Normal Distribution
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2. The scores on 800 tests are normally distributed with a
mean score of 71 and a standard deviation of 8. What
percent of those tested scored less than 79?
Check Your Answer
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12-7: The Normal Distribution
2. The scores on 800 tests are normally distributed with a
mean score of 71 and a standard deviation of 8. What
percent of those tested scored less than 79?
About 84% scored less than 79.
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12-7: The Normal Distribution
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2. The scores on 800 tests are normally distributed with a
mean score of 71 and a standard deviation of 8. What
percent of those tested scored less than 79?
Step 1: Make a number line using the given mean and standard
deviation and the corresponding percentages for a normal distribution.
0.5%
2%
47
13.5%
55
34%
63
13.5%
34%
71
mean
79
2%
87
0.5%
95
Step 2: Identify whether the answer should be a percentage or whole
number.
For this problem the phrase “what percent” means the answer is a percentage
number.
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See the rest of the explanation
12-7: The Normal Distribution
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2. The scores on 800 tests are normally distributed with a
mean score of 71 and a standard deviation of 8. What
percent of those tested scored less than 79?
Step 3: Identify the correct intervals. “scored less than 79”
0.5%
2%
47
13.5%
55
34%
63
13.5%
34%
71
mean
79
2%
87
0.5%
95
Step 4: Add the percentages and state the conclusion.
0.5% + 2% + 13.5% + 34% + 34% = 84%
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About 84% scored less than 79.
12-7: The Normal Distribution
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3. The scores on 800 tests are normally distributed with a
mean score of 71 and a standard deviation of 8. How
many of those tested scored between 55 and 95?
Check Your Answer
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12-7: The Normal Distribution
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3. The scores on 800 tests are normally distributed with a
mean score of 71 and a standard deviation of 8. About
how many of those tested scored between 55 and 95?
About 776 of those tested scored between 55 and 95.
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12-7: The Normal Distribution
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3. The scores on 800 tests are normally distributed with a
mean score of 71 and a standard deviation of 8. About
how many of those tested scored between 55 and 95?
Step 1: Make a number line using the given mean and standard
deviation and the corresponding percentages for a normal distribution.
0.5%
2%
47
13.5%
55
34%
63
13.5%
34%
71
mean
79
2%
87
0.5%
95
Step 2: Identify whether the answer should be a percentage or whole
number.
For this problem the phrase “about how many” means the answer is a whole number.
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12-7: The Normal Distribution
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3. The scores on 800 tests are normally distributed with a
mean score of 71 and a standard deviation of 8. About
how many of those tested scored between 55 and 95?
Step 3: Identify the correct intervals. “scored between 55 and 95”
0.5%
2%
47
13.5%
55
34%
63
13.5%
34%
71
mean
79
2%
87
0.5%
95
Step 4: 1) Add the percentages. 2) Find this percentage of the total number
of scores on the test. 3) State the conclusion.
Note: if the
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1) 13.5% + 34% + 34% + 13.5% + 2% = 97%
2) 97% of 800 scores = 0.97(800) = 776
answer in part 2
is a decimal,
round to the
nearest whole
number
About 776 of those tested scored between 55 and 95.
12-7: The Normal Distribution
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4. The scores on 800 tests are normally distributed with a
mean score of 71 and a standard deviation of 8. About
how many of those tested scored more than 79?
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12-7: The Normal Distribution
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4. The scores on 800 tests are normally distributed with a
mean score of 71 and a standard deviation of 8. How
many of those tested scored more than 79?
About 128 of those tested scored more than 79.
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12-7: The Normal Distribution
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4. The scores on 800 tests are normally distributed with a
mean score of 71 and a standard deviation of 8. How
many of those tested scored more than 79?
Step 1: Make a number line using the given mean and standard
deviation and the corresponding percentages for a normal distribution.
0.5%
2%
47
13.5%
55
34%
63
13.5%
34%
71
mean
79
2%
87
0.5%
95
Step 2: Identify whether the answer should be a percentage or whole
number.
For this problem the phrase “about how many” means the answer is a whole number.
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12-7: The Normal Distribution
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4. The scores on 800 tests are normally distributed with a
mean score of 71 and a standard deviation of 8. How
many of those tested scored more than 79?
Step 3: Identify the correct intervals. “scored more than 79”
0.5%
2%
47
13.5%
55
34%
63
13.5%
34%
71
mean
79
2%
87
0.5%
95
Step 4: 1) Add the percentages. 2) Find this percentage of the total number
of scores on the test. 3) State the conclusion.
Note: if the
1) 13.5% + 2% + 0.5% = 16%
2) 16% of 800 scores = 0.16(800) = 128
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About 128 of those tested scored more than 79.
answer in part 2
is a decimal,
round to the
nearest whole
number
12-8: Binomial Experiments
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1. The probability that Julio makes a goal is 1/5. If he takes
7 shots at the goal, what is the probability that he makes
exactly 4 goals?
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12-8: Binomial Experiments
1. The probability that Julio makes a goal is 1/5. If he takes
7 shots at the goal, what is the probability that he makes
exactly 4 goals?
The probability is
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448
.
15,625
See a detailed explanation
12-8: Binomial Experiments
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1. The probability that Julio makes a goal is 1/5. If he takes 7 shots at
the goal, what is the probability that he makes exactly 4 goals?
Step 1: Define the two possible outcomes and identify the power of the
corresponding binomial.
Let x = makes a goal
Let y = misses a goal
The power of the binomial is 7 because Julio is taking 7 shots at the goal.
Step 2: Set up the binomial and expand using the binomial theorem.
x  y 
7
x7
7x 6 y 7  6x 5 y 2 7  6  5  x 4 y 3 7  6  5  4  x 3 y 4





0!
1!
2!
3!
4!
7  6  5  4  3  x 2 y 5 7  6  5  4  3  2  xy 6 7! y 7



5!
6!
7!
 x  y 7  x 7  7x 6 y  21x 5 y 2  35x 4 y 3  35x 3 y 4  21x 2 y 5  7xy 6  x 7
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12-8: Binomial Experiments
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1. The probability that Julio makes a goal is 1/5. If he takes 7 shots at
the goal, what is the probability that he makes exactly 4 goals?
Step 3: Determine which term(s) should be used. “makes exactly 4
goals”
4 3
Use 35x y
The term with x4 should be used because x stands for making a goal and we are finding the
probability of making 4 goals.
Step 4: Find the probabilities for making a goal and missing a goal.
1
(this is given information)
5
1 4
P  missing a goal  P  y   1 
5 5
P  making a goal  P  x  
Step 5: Plug in the probabilities for x and y and simplify.
4
3
35 1 64
448
 1  4 
Use 35x 4 y 3  35      



1 625 125 15,625
5 5
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The probability is
448
.
15,625
12-8: Binomial Experiments
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2. The probability that Julio makes a goal is 1/5. If he takes
7 shots at the goal, what is the probability that he makes 2
or 3 goals?
Check Your Answer
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12-8: Binomial Experiments
2. The probability that Julio makes a goal is 1/5. If he takes
7 shots at the goal, what is the probability that he makes 2
or 3 goals?
The probability is
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30,464
.
78,125
See a detailed explanation
12-8: Binomial Experiments
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2. The probability that Julio makes a goal is 1/5. If he takes
7 shots at the goal, what is the probability that he makes 2
or 3 goals?
Step 1: Define the two possible outcomes and identify the power of the
corresponding binomial.
Let x = makes a goal
Let y = misses a goal
The power of the binomial is 7 because Julio is taking 7 shots at the goal.
Step 2: Set up the binomial and expand using the binomial theorem.
x  y 
7
x7
7x 6 y 7  6x 5 y 2 7  6  5  x 4 y 3 7  6  5  4  x 3 y 4





0!
1!
2!
3!
4!
7  6  5  4  3  x 2 y 5 7  6  5  4  3  2  xy 6 7! y 7



5!
6!
7!
 x  y 7  x 7  7x 6 y  21x 5 y 2  35x 4 y 3  35x 3 y 4  21x 2 y 5  7xy 6  x 7
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12-8: Binomial Experiments
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2. The probability that Julio makes a goal is 1/5. If he takes
7 shots at the goal, what is the probability that he makes 2
or 3 goals?
Step 3: Determine which term(s) should be used. “makes 2 or 3 goals”
Use 35x 3 y 4  21x 2 y 5
The terms with x2 and x3 should be used because x stands for making a goal and we are
finding the probability of making 2 or 3 goals.
Step 4: Find the probabilities for making a goal and missing a goal.
1
(this is given information)
5
1 4
P  missing a goal  P  y   1 
5 5
P  making a goal  P  x  
Step 5: Plug in the probabilities for x and y and simplify.
3
4
 1  4 
 1
Use 35x y  21x y  35       21 
5 5
5
35 1 256 21 1 1024 30, 464



 


1 125 625 1 25 3125 78,125
3
4
2
5
2
4
 
5
5
The probability is
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30,464
.
78,125
12-8: Binomial Experiments
3. The probability that Juan makes a goal is 3/5. If he takes
6 shots at the goal, what is the probability that he makes at
least 4 goals?
Check Your Answer
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12-8: Binomial Experiments
3. The probability that Juan makes a goal is 3/5. If he takes
6 shots at the goal, what is the probability that he makes at
least 4 goals?
The probability is
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1701
.
3125
See a detailed explanation
12-8: Binomial Experiments
3. The probability that Juan makes a goal is 3/5. If he takes
6 shots at the goal, what is the probability that he makes at
least 4 goals?
Step 1: Define the two possible outcomes and identify the power of the
corresponding binomial.
Let x = makes a goal
Let y = misses a goal
The power of the binomial is 6 because Juan is taking 6 shots at the goal.
Step 2: Set up the binomial and expand using the binomial theorem.
x  y 
6
x6
6x 5 y 6  5x 4 y 2 6  5  4  x 3 y 3 6  5  4  3  x 2 y 4





0!
1!
2!
3!
4!
6  5  4  3  2  x 1y 5 6! y 6


5!
6!
 x  y 6  x 6  6x 5 y  15x 4 y 2  20x 3 y 3  15x 2 y 4  6xy 5  x 6
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12-8: Binomial Experiments
3. The probability that Juan makes a goal is 3/5. If he takes
6 shots at the goal, what is the probability that he makes at
least 4 goals?
Step 3: Determine which term(s) should be used. “at least 4 goals”
Use x 6  6x 5 y  15x 4 y 2
The terms with x6 and x5 and x4 should be used because x stands for making a goal and we
are finding the probability of making at least 4 goals (4 or 5 or 6 goals).
Step 4: Find the probabilities for making a goal and missing a goal.
3
(this is given information)
5
3 2
P  missing a goal  P  y   1 
5 5
P  making a goal  P  x  
Step 5: Plug in the probabilities for x and y and simplify.
6
5
4
3
3 2
3 2
Use x  6x y  15x y     6       15     
5
5 5
5 5
729
6 243 2 15 81 4
8505
1701

 
 




15,625 1 3125 5 1 625 25 15,625 3125
6
5
4
2
2
The probability is
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1701
.
3125