Download June 2006 Exam Review Questions

Name: _________________________
MDM 4UI
EXAM REVIEW
Friday, June 24, 2016
8:30 AM – 10:30 AM
TERMINOLOGY AND FORMULAS
Unit 1: Permutations
Factorial Notation:
Permutations:
Permutations with Like Objects:
Factorial
Indirect Reasoning
Permutation
Tree Diagrams
n! n  (n  1)  (n  2)  ...  2  1
n!
P(n, r ) 
(n  r )!
n!
a!b!c!...
Direct Reasoning
Unit 2: Combinations
 n
n!
C (n, r )    
 r  r!(n  r )!
Combinations:
n
n
(a  b) n    a n  r b r
r 0  r 
Binomial Theorem:
n( A  B)  n( A)  n( B)  n( A  B)
Additive Principle for Union of Two Sets:
Combination
Direct Reasoning
Pascal’s Triangle
Indirect Reasoning
Venn Diagrams
Pascal’s Method of Counting
Sample Space
Event Space
Binomial Theorem
Complementary Event
Venn Diagram
Intersection of Sets
Disjoint Sets
Union of Sets
Mutually Exclusive Events
Unit 3: Probability
Experimental Probability:
Theoretical Probability:
Probability of Complementary Event:
Additive Principle for Union of Two Sets:
Conditional Probability:
Probability of Independent Events:
Odds in Favour of A:
Odds Against A:
Experimental Probability
Random Variable
Theoretical Probability
Factorial
Indirect Reasoning
Pascal’s Triangle
Sample Space
Intersection of Sets
Mutually Exclusive Events
P( A) = # times event occurred  # of trials
n( A)
P( A) 
n( S )
P( A' )  1  P( A)
P( A  B)  P( A)  P( B)  P( A  B)
P( A  B)
P( B A) 
P( A)
P( A  B)  P( A)  P( B)
P(A) : P(A’)
P(A’) : P(A)
Simulation
Expected Value
Conditional Probability
Permutation
Tree Diagrams
Venn Diagrams
Event Space
Disjoint Sets
Trial
Simple Event
Independent Events
Direct Reasoning
Combination
Pascal’s Method of Counting
Complementary Event
Union of Sets
Unit 4: Probability Distributions with Discrete Variables
Expected Value of a Discrete Random Variable:
n
E ( X )   xi P ( X xi )
i 1
Expected Value of a Probability Distribution:
n
P( X  k )   ( p) k (1  p) nk
k 
E ( X )  np
Hypergeometric Probability Distribution:
𝑃(𝑥) =
Expected Value of a
Hypergeometric Probability Distribution:
𝐸(𝑥) =
Binomial Probability Distribution:
𝑎 𝐶𝑥
× 𝑛−𝑎 𝐶𝑟−𝑥
𝑛 𝐶𝑟
𝑟𝑎
𝑛
Discrete Random Variable
Probability Distribution
Uniform Distribution
Binomial Probability Distribution
Hypergeometric Probability Distribution
Unit 6: Organization of Data for Analysis
Population
Census
Inference
Cross-Sectional Study
Qualitative Variable
Quantitative Data
Discrete Data
Simple Random Sampling
Stratified Random Sampling
Strata
Multi-Stage Random Sampling
Destructive Sampling
Secondary Data
Open Questions
Information Questions
Checklist Questions
Rating Questions
Bias
Non-Response Bias
Household Bias
Unit 7: One Variable Data Analysis
Bin Width:
Sample
Longitudinal Study
Continuous Data
Systematic Random Sampling
Cluster Random Sampling
Primary Data
Closed Question
Ranking Questions
Sampling Bias
Response Bias
Range  Number of intervals
n
Mean:
x
x
i 1
i
n
n
Weighted Mean:
x
w x
i
i 1
n
i
w
i 1
Variance
i
s = σ2
n
Standard Deviation:
Z-Score:
Expected Value

z
 (x
i 1
i
 x) 2
n
xx

Bin Width
Mound-Shaped Distribution
Outlier
Weighted Mean
Left Skew
Quartile
Standard Normal Distribution
Percentile
Sample
Class Interval
Histogram
Box-and-Whisker Plot
U-Shaped Distribution
Symmetric Distribution
Mean
Median
Right Skew
Interquartile Range
Z-Score
Statistics
Frequency
Bar Graph
Pictograph
Broken-Line Graph
Uniform Distribution
Skewed Distribution
Standard Deviation
Mode
Range
Normal Distribution
Z-Score Table
Data
Stem-and-Leaf Plot
Frequency Distribution
Circle Graph
Split Bar Graph
Unit 8: Two Variable Data Analysis
R1  y1  [m( x1 )  b]
Residual Value
Scatter Plot
Trend
Median-Median Line
Residual Value
Independent Variable
Line of Best Fit
Coefficient of Correlation
Dependent Variable
Correlation
Coefficient of Determination
Unit 9: Probability Distributions with Continuous Variables
Binomial Approx. (needs correction factor):
µ = np
and
Hypergeometric Approx. (needs correction factor):
𝜇 = 𝑛𝑝 and
Confidence Interval for a population:
𝑥̅ − 𝑧𝛼
2
𝑝̂ − 𝑧𝛼
Confidence Interval for a proportion:
𝜎
√𝑛
𝑛=(
√𝑝̂(1−𝑝̂)
2
𝑤
𝑁𝑃−𝑛
𝜎 = √𝑛𝑝𝑞 ( 𝑁𝑃−1)
< 𝜇 < 𝑥̅ + 𝑧𝛼
√𝑛
2
2𝑧𝛼 𝜎 2
Number in a Sample given standard deviation:
𝜎 = √𝑛𝑝𝑞.
2
𝜎
√𝑛
< 𝑝 < 𝑝̂ + 𝑧𝛼
2
√𝑝̂(1−𝑝̂)
√𝑛
)
𝑧𝛼 2
Number in a Sample given a proportion:
Binomial Approximation
Confidence Level
𝑛 = 4𝑝𝑞 ( 𝑤2 )
Hypergeometric Approximation
Confidence Interval
Correction Factor
Margin of Error
Note: All formulas along with their brief description will be provided on the exam. Definitions will
not be provided.
Statistical Analysis Review Questions
1.
Here are the sums of the two numbers from 50 rolls of a pair of standard dice.
11
4
4
10
8
7
6
6
7
9
8
8
4
7
9
11
3
7
6
9
5
8
6
8
7
5
11
2
5
5
6
6
11
9
6
5
5
5
3
9
a) Organize these data using a tally, frequency and relative frequency table..
b) Use a bar graph and a line graph to illustrate the information in the table.
5
12
2
5
8
2.
Given the following scatter plot, line of best fit and coefficient of correlation, describe the
relationship.
3.
Name each type of graph.
a)
b)
c)
d)
10
10
6
2
2
e)
f)
4.
Use the graphs in #3 to answer the following questions:
a) What intervals does the histogram show?
b) Approximately how much does it cost to deliver mail to a centralized point?
c) What trend seems to exist between number of officers and crime incidents?
d) What is the most popular game console?
e) During which month do you think students had their cell phones taken away?
f) Which movie made the most money on its opening weekend?
5.
What is the difference between a bar graph and a histogram?
6.
Explain the terms coefficient of correlation and coefficient of determination. How are these
values related, and what information is given by each of these coefficients?
7.
The following table shows the commuting distances to work for the employees at a company.
Commuting Distance (km)
17.8
4.2
12.6
10.1
5.0
6.3
7.2
15.0
27.4
11.4
25.4
9.1
13.0
23.9
2.3
13.3
13.2
3.4
20.0
9.4
19.6
19.4
0.4
2.3
15.9
9.9
12.9
22.3
15.8
20.5
23.5
11.3
19.5
13.6
7.5
12.8
8.2
4.9
21.0
Create a box-and-whisker plot to represent these data.
8.
7.6
Identify five different types of graphs that display data, and describe when each type would be
used.
9.
In 1995, the American estimate for the percentage of properly vaccinated American children was
raised from 67% to 75%. The new, higher estimate was based on 25 000 telephone interviews
which were followed up by immunization record checks. The earlier, lower estimate was based on
data collected from a more general national health survey. Describe what type of bias resulted in
the lower estimate and explain how it could have occurred.
10.
A high-speed internet service provider is interested in finding out how internet users feel about
pop-up advertising windows. The provider takes a simple random sample of 500 of its users and emails them the following question: "Many internet sites could not afford to function without being
sponsored by advertisers using pop-up windows. Do you agree that it is reasonable to have these
extra windows appear when you are using the internet so that you can view these sites?" For this
situation, identify the population and at least three sources of bias that are present and why they
occur.
11.
Explain how cluster sampling and stratified sampling both use simple random samples as part of
the sampling process.
12.
A study is to be conducted to compare how the mass of Canadian males varies with age. In
particular, a comparison will be made between those in their twenties and those in their thirties. A
sample of 1000 males will be taken. What variables need to be measured? Classify them as
qualitative or quantitative, and, if appropriate, discrete or continuous. What type of sampling
method would be appropriate? You may assume that census data is available on every Canadian.
13.
Organize these traffic light waiting times (in seconds) into five uniform intervals, than construct a
table with frequencies cumulative frequencies, and relative frequency percentages:
6, 11, 14, 12, 16, 25, 21, 24, 17, 19, 20, 2,
14, 11, 8
14.
Why is the mode, but not the mean or median an appropriate measure of central tendency for
qualitative data?
15.
Explain how you find the interquartile range for a set of ungrouped data.
16.
The masses of largemouth bass, in kilograms, are normally distributed with a mean of 1.1 and a
standard deviation of 0.3. Any bass caught with a mass between 0.8 kg and 1.7 kg must be
released. What percent of the bass does this represent?
17.
The amount of barbeque sauce on Spicee Wings is normally distributed with a mean of 20 mL.
What would you expect to find more of in a basket of these tasty morsels: wings with 19 mL of
sauce or wings with 21 mL of sauce? Explain.
18.
Use z-scores to determine who had a more disappointing quiz mark:
Alan had a mark of 8.5 on a quiz for which the mean mark was 10.1 and the standard deviation
was 2.5. Belle had a mark of 9.0 on a quiz for which the mean was 9.9 and the standard deviation
was 1.5.
19.
In a certain U.S. college, incoming students have a mean GPA (Grade Point Average) of 2.57 with
a standard deviation of 0.71. How many of the 850 new students have a GPA greater than 3.00 if
the GPAs are normally distributed?
20.
Suppose X~N(2, 32). Which set of data contains a greater percent of the data?
a) x
1 together with x 5 or b) 2 x 5?
21.
A jar contains 50 toonies and 150 loonies. Fifteen coins are removed. Confirm that you can use a
normal approximation and then use a normal approximation to determine the probability that
exactly 5 toonies are removed.
22.
A student randomly chooses each answer on a 100-question multiple choice test where there are
four choices for each question. Confirm that you can use a normal approximation and then use a
normal approximation to determine the probability that the student gets more than 30% correct?
23.
A politician conducts a poll to see how the public is supporting her latest initiatives. The poll
comes back with a 62% approval rating. This pleases the politician. However, the poll has a
margin of error of ±15%, 19 times out of 20. Based on this information, how many people did she
likely poll? How many people should she need to poll to ensure a margin of error of ±5%?
24.
A machine that produces control arms for a vehicle gas pedal generates pedals that have a length
with standard deviation of 0.08cm. Thirty pedals are tested to see if their lengths are acceptable.
The sample has a mean of 18.2cm. What would be the acceptable range of lengths for a 95%
confidence level for this length?
25.
An exit poll is done outside a voting location. People who have just voted are asked if they will
state who they voted for. In a close election an exit poll states that Larry Liberal has 48% of the
vote, while Constance Conservative has 46% of the vote, with the rest split up among other
candidates. The polling firm states that 500 people were polled. What is the margin of error and
confidence interval for each candidate if the poll is said to be correct 19 times out of 20? Is it
possible that Larry might actually be losing?
Counting Methods and Probability Review Questions
26.
The probability of it raining on a particular day is 0.3. The probability of the sun appearing is 0.6.
Determine the probability that the sun will appear and there will be no rain.
27.
Determine the probability that when a pair of dice is rolled that a sum of eight or doubles occurs.
Show all work and use the addition principle for probabilities.
28.
Statistics show that a team has a 30% chance of winning if it loses the previous game and a 60%
chance of winning if it wins the previous game. If the chance of winning its first game of the
season is 65%, determine the probability that it will win two out of its first three games.
29.
There are 10 students on the volleyball team and 15 on the basketball team. When planning a field
trip with both teams, the coach has to arrange transportation for a total of only 19 students.
a) Use a Venn diagram to illustrate this situation.
b) Determine how many students are on both teams.
c) Determine the number of students in the remaining regions of your diagram and explain what
these regions represent.
30.
Of the 140 grade 12 students at Vennville High School, 52 have signed up for biology, 71 for
chemistry, and 40 for physics. The science students include 15 who are taking both biology and
chemistry, 8 who are taking chemistry and physics, 11 who are taking biology and physics, and 2
who are taking all three science courses.
a) How many students are not taking any of these three science courses?
b) Illustrate the enrolments with a Venn diagram.
31.
A recent survey of a group of students found that many participate in baseball, football, and soccer.
The Venn diagram shows the results of the survey.
Baseball
Football
a) How many students participated in the survey?
b) How many of these students play both soccer
27
8
10
and baseball?
6
c) How many play only one sport?
3
4
d) How many play football and soccer?
5
e) How many play all three sports?
19
f) How many do not play soccer?
Soccer
S
32.
A bag contains 4 chocolate bars and 3 bags of chips. An item is drawn and then replaced.
Determine the probability that exactly one bag of chips is chosen.
33.
A media class of 25 students randomly selects 3 of its students to be the news anchor, sports
reporter, and weather person by pulling names out of a hat. Explain whether a permutation or
combination should be used to determine the number of ways that a selection can be made and
why.
34.
Kirk's Convenience store sells 15 different brands of chocolate bars. Of these, 6 brands contain
nuts, 5 brands contain caramel, and the rest are pure milk chocolate. If 8 chocolate bars are to be
purchased, determine the number of ways they can be purchased so that at least 2 of them contain
nuts.
35.
A focus group of three members is to be randomly selected from a medical team consisting of five
doctors and seven technicians.
a) What is the probability that the focus group will be comprised of doctors only?
b) What is the probability that the focus group will not be comprised of doctors only?
36.
A messy drawer contains three red socks, five white socks, and four black socks. What are the
odds in favour of randomly drawing a red sock?
37.
If the chance of snowstorm in Windsor, Ontario, in January is estimated at 0.4, what are the odds
against Windsor’s having a snowstorm next January?
38.
The odds of Rico’s hitting a home run are 2:7. What is the probability of Rico’s hitting a home
run?
39.
An athletic committee with three members is to be randomly selected from a group of six
gymnasts, four weightlifters, and eight long-distance runners. Determine the probability that
a) the committee is comprised entirely of runners?
b) the committee is represented by each of the three types of athletes.
40.
A hockey player has a success rate of 21% when taking a penalty shot. Determine the probability
of the hockey player scoring at least 1 of 3 penalty shots.
41.
Two four-sided dice are rolled and the sum observed is recorded.
a) List the possible outcomes for the two dice and the resulting sums.
b) Create a table showing the probability distribution of the random variable representing the
sum observed.
c) Calculate the expected value for the sum observed.
42.
How many different paths on the diagram below will spell PASCAL?
P
A A
S S S
C C
A A A
L L
43.
Apply the Binomial Theorem to expand (3a – 5b)6.
44.
Explain the similarities and differences between the hypergeometric probability distribution and
the binomial probability distribution.
45.
A baseball team has 25 players, 7 of whom bat left-handed. Suppose that the manager of this team
is frustrated with the way the team is playing, so he decides to randomly select 9 players to play in
the upcoming game. The random variable X will be the number of left-handed batters in the game.
a) What is the probability of creating a lineup with exactly 2 lefties?
b) What is the probability of creating a lineup with at least 2 lefties?
c) How many lefties would you expect to find in the lineup?
46.
Explain the difference between a discrete and a continuous variable.
47.
The hypergeometric probability distribution is used in acceptance sampling. Suppose that a
machine shop orders 500 bolts from a supplier. To determine whether to accept the shipment of
bolts, the manager of the facility randomly selects 12 bolts. If none of the 12 randomly selected
bolts is found to be defective, he concludes that the shipment is acceptable.
a) If 10% of the bolts in the population are defective, what is the probability that none of the
selected bolts are defective?
b) If 20% of the bolts in the population are defective, what is the probability that none of the
selected bolts are defective?
48.
A weighted die is tossed 10 times. The probability of rolling a 6 with this unfair die is 0.25. What
is the probability of observing a 6 more than 3 times for the 10 tosses?
49.
A student calculates the probability that exactly 8 parts out of a sample of 100 will be defective
given that 11% of the total production is known to be defective. The next question he encounters
asks what the probability is for 4 parts out of a sample of 50 being defective. Having a sound grasp
of the concept of equivalent fractions, the student copies her answer to the first question and moves
on. Calculate both probabilities.
ANSWERS
1.
a)
SUM
TALLY
FREQUENCY
2
3
4
5
6
7
8
9
10
11
12
||||
||
|||
||||| ||||
||||| |||
|||||
||||| |
|||||
|||
||||
|
4
2
3
9
8
5
6
5
3
4
1
RELATIVE
FREQUENCY
8
4
6
18
16
10
12
10
6
8
2
b)
2.
r2 = 0.75
r = – 0.87
This is a moderate negative correlation. As speed increases, time decreases.
75% of the variation in time is due to the variation in speed.
3.
a)
b)
c)
d)
e)
f)
Broken Line Graph
Split Bar Graph
Scatter Plot
Pie Graph
Bar Graph
Histogram
4.
a)
b)
c)
d)
e)
f)
5 mark intervals (10-14, 15-19, 20-24)
$110
Positive (as number of officers increases, crime increases)
Xbox 360 (always ‘start’ at 12:00 and read through the legend clockwise)
October (shows no increase in text messages)
Breaking Dawn (top bar is budget, bottom bar is revenue – this is the only one where revenue
is larger than budget)
5.
Bar Graph: graphs the frequency of qualitative or discrete quantitative data. There may be gaps
between the bars.
Histogram: graphs the frequency of continuous quantitative data. There are no gaps between the
bars.
6.
The coefficient of correlation, denoted r, gives the strength and direction of the relationship
between two variables. An r value close to –1 indicates a strong negative correlation. A value close
to +1 indicates a strong positive correlation. An r value close to 0 indicates no correlation between
the variables.
The coefficient of determination is calculated as . It gives the relative strength of the
relationship between two variables. Specifically, it determines the amount of variation in the
dependent variable that is due to variation in the independent variable.
7.
Minimum = 0.4 km
Maximum = 27.4 km
Median (average of 20th and 21st data values) = 12.85
Median of lower half (average of 10th and 11th data values) = 7.55
Median of upper half (average of 30th and 31st data values) = 19.45
8.
Possible Answers:
Bar Graph: Used to display the frequency of data in discrete categories
Circle Graph: Used to display the relative frequency of various categories in a sample
Box-and-Whisker Plot: Used to display the spread of data, including minimum and maximum
values, median, and medians of the lower and upper halves of the data
Histogram: Used to display the frequency of data in continuous class intervals
Broken-Line Graph: Used to show the trend of data over time
9.
The earlier estimate occurred because of response bias. The interviews that resulted in the new
estimate were followed up with record checks. Thus, we can assume that they are reasonably
accurate estimates of the number of properly vaccinated children. Because the previous estimate
relied on data from a more general survey (and was presumably not verified), respondents likely
did not accurately remember when or if their children had been vaccinated.
10.
The population is all internet users. The wording of the question leads to response bias because it
encourages the respondent to say that the advertisements are reasonable by first justifying the popup windows’ existence. Sampling bias also exists. Only this internet service provider’s users are
surveyed, not a sample of the entire population of internet users. In particular, dial-up users are
missed, and they may find pop-up windows more annoying as users have to wait for them to load.
Non-response bias will also exist in this situation as many users will not respond to the e-mail.
11.
In both cluster sampling and stratified sampling, the population is divided into groups—clusters or
strata. In cluster sampling, a simple random sample of the clusters is taken. Then every member of
the population in the chosen clusters forms the sample. In stratified sampling, every stratum is
used. Then from each stratum, a simple random sample of members of that stratum is chosen to
participate.
12.
Mass and age are the variables that need to be measured. Mass is quantitative and continuous. For
purposes of this study, age can be measured as a qualitative variable: 20-29 and 30-39. Assume
that a list of all Canadians and their ages is available. Then, a stratified sample of 1000 people, 500
randomly chosen from males in their twenties and 500 randomly chosen from males in their
thirties, would be appropriate.
13.
Waiting
Time (s)
Frequency
Cumulative
Frequency
Relative
Frequency
0.5–5.5
1
5.5–10.5
2
10.5–15.5
5
15.5–20.5
4
20.5–25.5
3
1
7%
3
13%
8
33%
12
27%
15
20%
14.
The mean and median need the data to have properties of numbers to be evaluated, for example,
the capacity to be added and divided, and order. The mode can be evaluated with any type of data
since only frequency is required.
15.
Arrange the data from smallest to largest, then find the median, Q2. Find the median of the data
below Q2. This gives Q1. Find the median of the data above Q2. This gives Q3. The interquartile
range is Q3 Q1.
16.
0.8 is one standard deviation below the mean and 1.7 is two standard deviations above the mean.
Therefore, the percent released is 68% + 13.5% = 81.5%.
17.
You would expect to find neither. Normal curves are symmetrical about the mean and since 19 and
21 are equidistant from the mean of 20, there should be the same number of each sort of wing in
the basket.
18.
Computing z-scores:
Alan:
Belle:
Alan’s z-score is smaller. Therefore, he had the more disappointing quiz mark.
19.
Therefore, 230 new students will have a GPA over 3.00.
20.
The first set consists of data not within one standard deviation of the mean, so it contains 32% of
the data. The second set consists of the data within one standard deviation above the mean, and it
contains 34% of the data. Therefore, the set in b) contains a greater percent of the data.
21.
This is a discrete hypergeometric distribution.
50
150
𝑛 = 15
𝑝 = 200 = 0.25
𝑞 = 200 = 0.75
𝑁𝑃 = 200
0.1𝑁𝑃 = 0.1(200) = 20
𝑛 < 0.1𝑁𝑃 ∴ We can approximate with a normal distribution
𝜇 = 𝑛𝑝
= 15(. 25)
𝑁𝑃−𝑛
𝜎 = √𝑛𝑝𝑞 ( 𝑁𝑃−1)
200−15
= √15(0.25)(0.75) ( 200−1 )
= 3.75
= 1.62
𝑃(𝑋 = 5) = 𝑃(4.5 < 𝑋 < 5.5)
4.5−3.75
5.5−3.75
= 𝑃(
<𝑧<
)
1.62
1.62
= 𝑃(0.46 < 𝑧 < 1.08)
= 𝑃(𝑧 < 1.08) − 𝑃(𝑧 < 0.46)
= 0.8599 − 0.6775
= 0.1827
∴ There is about an 18.3% chance that exactly 5 are toonies.
22.
This is a discrete binomial distribution.
1
3
𝑛 = 100
𝑝 = 4 = 0.25
𝑞 = 4 = 0.75
𝑛𝑝 = 100(0.25) = 25 > 5
𝑛𝑞 = 100(. 75) = 75 > 5
∴ We can approximate with a normal distribution
𝜇 = 𝑛𝑝
𝜎 = √𝑛𝑝𝑞
= 100(. 25)
= √100(0.25)(0.75)
= 25
= 4.33
𝑃(𝑋 > 30) = 𝑃(𝑥 > 30.5)
30.5−25
= 1 − 𝑃 (𝑧 < 4.33 )
= 1 − 𝑃(𝑧 < 1.27)
= 1 − 0.8980
= 0.102
∴ There is about a 10.2% chance of getting more than 30% by guessing.
23.
p = 0.62
q = 0.38
w = 2 (0.15) = .30
19 out of 20 is a 95% confidence level. This corresponds to a 𝑧𝛼 = 1.96
2
𝑧𝛼 2
1.96 2
𝑛 = 4𝑝𝑞 ( 𝑤2 ) = 4(0.62)(0.38) ( 0.3 ) = 40.22
∴ They surveyed about 41 people.
For a margin of error of ±5%, change: w = 2 (0.05) = 0.10
𝑧𝛼 2
1.96 2
𝑛 = 4𝑝𝑞 ( 𝑤2 ) = 4(0.62)(0.38) ( 0.1 ) = 362.03
∴ They will need to survey about 363 people.
24.
𝜎 = 0.08
𝑛 = 30
𝑥̅ = 18.2
95% confidence level corresponds to a 𝑧𝛼 = 1.96
2
𝑥̅ − 𝑧𝛼
𝜎
< 𝜇 < 𝑥̅ + 𝑧𝛼
𝜎
2 √𝑛
2 √𝑛
0.08
0.08
18.2 − 1.96
< 𝜇 < 182 + 1.96
√30
√30
18.2 − 0.0286 < 𝜇 < 18.2 + 0.0286
18.17 < 𝜇 < 18.22
∴ The average length of a pedal will be between 18.17 cm and 18.22 cm, 95% of the time, if a
sample of 30 pedals is taken.
25.
Larry: 𝑝̂ = 0.48, 1 − 𝑝̂ = 0.52, 𝑛 = 500
95% confidence level corresponds to a 𝑧𝛼 = 1.96
2
𝑝̂ − 𝑧𝛼
√𝑝̂ (1 − 𝑝̂ )
√𝑛
√0.48(0.52)
< 𝑝 < 𝑝̂ + 𝑧𝛼
√𝑝̂ (1 − 𝑝̂ )
√𝑛
√0.48(0.52)
0.48 − 1.96
< 𝑝 < 0.48 + 1.96
√500
√500
0.48 − 0.044 < 𝑝 < 0.48 + 0.044
0.436 < 𝑝 < 0.524
Constance: 𝑝̂ = 0.46, 1 − 𝑝̂ = 0.54, 𝑛 = 500
2
2
95% confidence level corresponds to a 𝑧𝛼 = 1.96
2
𝑝̂ − 𝑧𝛼
√𝑝̂ (1 − 𝑝̂ )
2
√𝑛
√0.46(0.54)
< 𝑝 < 𝑝̂ + 𝑧𝛼
√𝑝̂ (1 − 𝑝̂ )
√𝑛
√0.46(0.54)
0.46 − 1.96
< 𝑝 < 0.48 + 1.96
√500
√500
0.46 − 0.044 < 𝑝 < 0.46 + 0.044
0.416 < 𝑝 < 0.504
2
Since the ranges for Larry and Constance overlap, it could be that Larry is not winning.
26.
The probability that the sun will appear and there will be no rain is 0.42.
27.
P(sum of eight or doubles) = P(sum of eight) + P(doubles)
28.
P(win two out of its first three games) = (0.65)(0.6)(0.6) + (0.65)(0.4)(0.3) + (0.65)(0.6)(0.4) +
(0.35)(0.3)(0.6) = 0.531.
29.
a)
Volleyball
4
6
P(sum of eight and doubles)
Basketball
9
S
30.
b)
Since there are 10+15=25 players on both teams and they only need spots for 19 players, there
must be 6 players on both teams (25-19=6)
c)
There are 10 players on the Volleyball team and 6 are on both teams so there must be 4 who
are only on the volleyball team. There are 15 players on the Basketball team and 6 on both, so
there must be 9 who are only on the basketball team.
a)
Chemistry
Biology
28
13
9
9
S
31.
2
50
6
23
Physics
b)
There are 9 students not taking any science courses.
a)
b)
c)
d)
e)
f)
27+8+10+3+6+4+19+5=82
6+4=10
27+10+19=56
6+3=9
6
5+27+8+10=50
3
.
7
32.
The probability that exactly one bag of chips is chosen is
33.
A permutation should be used because order matters in this case. The event of 3 people being
selected for these positions is different than the same 3 people being selected but having these
positions in a different way. Therefore, using a combination where you would divide by 3! once
the 3 people are selected is not appropriate. A permutation in which 25 24 23 are multiplied
and the result is not divided by anything is appropriate.
34.
Number of ways = (total number of ways)
of ways with 1 bar with nuts)
35.
a)
n(S) = 12C3 = 220
P(A) = 10/220 = 0.045
b)
P(at least one technician) = 1 – P(no technicians) = 1 – P(all doctors) = 1 – 0.045 = 0.955
36.
P(A) = 3/12
P(A’) = 9/12
P(A) : P(A’) = 3/12 : 9/12 = 3 : 9 = 1:3
37.
P(A) = 0.4
P(A’) = 0.6
P(A’) : P(A) = 0.6 : 0.4 = 6 : 4 = 3 : 2
38.
P(A) = 2/x
P(A) + P(A’) = 1
2/x + 7/x = 1
9/x = 1
9=x
(number of ways with zero bars with nuts)
n(A) = 5C3 = 10
P(A’) = 7/x
So, P(A) = 2/9
39.
(number
a)
n(S) = 18C3 = 816
P(A) = 56/816 = 0.069
n(A) = 8C3 = 56
b)
n(S) = 18C3 = 816
P(A) = 192/816 = 0.235
n(A) = 6C1 × 4C1 × 8C1 = 192
40.
P(scoring at least once)
or P(scoring at least once)
41.
a)
There are sixteen possible outcomes.
1+1=2
2+1=3
3+1=4
1+2=3
2+2=4
3+2=5
1+3=4
2+3=5
3+3=6
1+4=5
2+4=6
3+4=7
b)
X
P(X)
c)
E(X) = 2
2
3
+3
4
+4
5
+5
4+1=5
4+2=6
4+3=7
4+4=8
6
+6
7
+7
8
+8
=5
42.
1
1 1
1 2 1
3 3
3 6 3
9 9
The numbers show how many paths lead to each individual letter. The two final letter Ls can each
be reached by 9 possible paths for a total of 18.
43.
(3a – 5b)6 =
44.
Binomial: Independent trials, probability of success is the same for each trial
Hypergeometric: Dependent trials, probability of success changes for each trials
45.
a)
𝑃(2) =
b)
𝑃(𝑋 ≥ 2) = 1 − 𝑃(𝑋 < 2) = 1 − 𝑃(0) − 𝑃(1)
7 𝐶2
× 18 𝐶7
25 𝐶9
= 0.3271
=1−
c)
𝐸(𝑋) =
𝑟𝑎
𝑛
=
9×7
25
7 𝐶0
× 18 𝐶9
25 𝐶9
−
A discrete variable takes on natural number values.
A continuous variable takes on real (decimal) values.
47.
a)
50 𝐶0
× 450 𝐶12
500 𝐶12
= 0.2783
a = 20% of 500 = 100 defective bolts
r = 12
n = 500
𝑃(0) =
49.
= 0.693
a = 10% of 500 = 50 defective bolts
r = 12
n = 500
𝑃(0) =
48.
× 18 𝐶8
25 𝐶9
= 2.52
46.
b)
7 𝐶1
100 𝐶0
× 400 𝐶12
500 𝐶12
= 0.0664
Binomial Distribution – trials are independent; probability of success stays the same for each trial.
p = 0.25
q = 0.75
P(X > 3) = 1 – P(X ≤ 3) = 1 – P(0) – P(1) – P(2) – P(3)
10
10
10
10
= 1 − ( ) (0.25)0 (0.75)10 − ( ) (0.25)1 (0.75)9 − ( ) (0.25)2 (0.75)8 − ( ) (0.25)3 (0.75)7
0
3
1
2
= 1 – 0.0563 – 0.1877 – 0.2816 – 0.2503
= 0.2241