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STAT 2300: Review Problems for Test #2
--
Solutions & Grading Guidelines
-- Spring 2017
Your Test 1 will cover Sections 5.2 – 8.1. This set of review problems includes all problems over these
sections from the released exams (Fall 2016 & Spring 2016) posted to the course website. Thus, this set
of review problems does not necessarily reflect the length and/or difficulty of Test 2. This review cannot
be guaranteed to be inclusive or exclusive of all that is covered on Test 2, it is merely meant to aid you
in the review process.
Part I: Multiple Choice. Circle the letter corresponding to the best answer of the choices given.
1. Find P(−.80 < Z < 1.25)
(A) .9798
(B) .8944
(C) .6825
(D) .2119
2. The normal probability plot for a random sample of 40 observations is shown below.
Which of the following statements is justified by this normal probability plot?
(A) Because the plot is approximately normal, the population data could be normally distributed.
(B) Because the plot is approximately normal, the sample data could be normally distributed.
(C) Because the plot is roughly linear, the sample data could have come from a population that is
normally distributed.
(D) Because the plot is not linear, the sample data could not have come from a population that is
normally distributed.
1
3. A homework problem in a probability and statistics textbook says that P(A) = .3 and P(B) = .4, and
nothing more, then asks students to find P(A or B). One student calculated P(A or B) assuming that
A and B are mutually exclusive. Another student calculated P(A or B) assuming that A and B are
independent. Calculate the difference in their answers.
(A) 0
(B) .10
(C) .12
(D) Cannot be determined from the given information
Use the following information to answer questions 4 – 5.
The men's basketball team at Clemson University holds its opponents to 64.1 points per game on
average. Answer the following questions assuming that opposing team scores follow a normal
distribution with a standard deviation of 16.3 points.
4. What is the probability that an opponent scores fewer than 50 points?
(A) −.1922
(B) .1922
(C) .8078
(D) .87
5. What is the minimum score an opposing team needs in order to be in the top 10% of points scored
against Clemson?
(A) 65 points
(B) 81 points
(C) 85 points
(D) 90 points
6. Find the value of 𝑧𝑧.5 .
(A) −.6915
(B) 0
(C) .50
(D) .6915
2
7. One of the questions on the STAT 2300 Student Survey asked students to report the number of hours
of sleep they got during the previous 24 hours. The results are shown in the histogram below.
Is it reasonable to use the normal distribution as a model for hours of sleep?
(A) Yes, because the histogram is roughly uniformly distributed.
(B) No, because the histogram is not uniformly distributed.
(C) Yes, because the histogram is approximately symmetric and bell-shaped.
(D) No, because the histogram is not bell-shaped.
8. Suppose you are conducting a clinical trial of a new blood pressure medication in an area where 29%
of the population has high blood pressure. Let X be the number of individuals selected from this area
in order to find the first person with high blood pressure. Is X a binomial random variable?
(A) Yes, X is a binomial random variable.
(B) No, because X is not discrete.
(C) No, because the number of trials in the experiment is not fixed.
(D) No, because there are more than two possible outcomes in each trial of the experiment.
9. As a variable, the number of Clemson students who voted in the 2016 South Carolina presidential
primaries is…
(A) Qualitative and Discrete
(B) Quantitative and Discrete
(C) Qualitative and Continuous
(D) Quantitative and Continuous
3
Use the following information to answer questions 10 – 11.
According to the National Center for Education Statistics, 11.5 million of the 20.2 million students
enrolled in American colleges and universities are female. Also, 59% of college students are expected to
graduate within 6 years of enrolling. Furthermore, 62% of females enrolled are expected to graduate
within 6 years, while 56% of males are expected to graduate within 6 years of enrolling. Suppose one
college student is randomly selected from all college students in the U.S.
10. Let M denote the event the selected student is male, F denote the event the selected student is
female, and G denote the event the selected student graduates within 6 years of enrolling. To two
decimal places, which of the following probability statements is/are correct?
I. P(M) = .43
II. P(F|G) = .62
(A) I only
(B) II only
(C) I and II
(D) Neither I nor II
11. What is the probability that the selected student will be a female who graduates within 6 years?
(A) .35
(B) .57
(C) .59
(D) .62
Use the following information to answer questions 12 – 13.
In the United States, 51.2% of newborn babies are boys. A sample of 5 newborn babies will be
randomly selected from the United States.
12. What is the probability that all 5 newborn babies selected are girls?
(A) .028
(B) .035
(C) .488
(D) .500
13. What is the probability that at least 1 of the 5 newborn babies selected is a girl?
(A) .028
(B) .035
(C) .965
(D) .972
4
Use the following information to answer questions 14 – 15.
Let X = the number of living grandparents that a randomly selected American has. Based on data from
recent General Social Surveys, the table below gives the probability distribution of X.
X
0
1
2
3
4
P(X)
.71
.15
.09
.03
.02
14. What is the probability that a randomly selected American has at least 2 living grandparents?
(A) .05
(B) .09
(C) .14
(D) .86
15. You can verify that the expected value of X is 0.5. Which of the following gives the best
interpretation of this value?
(A) In a sample of 1,000 randomly selected Americans, we would expect approximately 500 of
them to still have a living grandparent.
(B) In a large number of randomly selected Americans, we would expect the average number of
living grandparents the selected persons have to be approximately 0.5.
(C) In a sample of 10 randomly selected Americans, there will be a total of 5 living grandparents
among the persons in the sample.
(D) If you randomly select an American, the most likely number of living grandparents that person
has is 0.5.
16. The annual salary of a registered nurse working in the U.S. follows an approximately normal
distribution with a mean of $66,000 and a standard deviation of $11,000. Let 𝑥𝑥̅ denote the average
annual salary for a random sample of 10 registered nurses working in the U.S. Which of the
following statements is correct?
(A) The sampling distribution of 𝑥𝑥̅ will not necessarily be approximately normally distributed
because of the small sample size.
(B) The sampling distribution of 𝑥𝑥̅ will not necessarily be approximately normally distributed
because of the large population standard deviation.
(C) The sampling distribution of 𝑥𝑥̅ will be approximately normally distributed because the sample
size is large enough.
(D) The sampling distribution of 𝑥𝑥̅ will be approximately normally distributed because the sample is
selected from a population that is approximately normally distributed.
5
17. The three graphs below are approximate sampling distributions of the sample mean. Each graph was
created by selecting 1000 samples of the same size. All three graphs were created by selecting
samples from the same population, but each graph was based on samples of a different sample size n.
Graph I
Graph II
Graph III
Which graph was based on samples with the largest sample size?
(A) Graph I
(B) Graph II
(C) Graph III
(D) Cannot be determined from the given information
6
18. The table below gives the probability distribution of the random variable X.
X
0
1
2
3
4
5
P(X)
.42
.27
.18
.11
.05
−.03
Is this a legitimate probability distribution?
(A) Yes, because the probabilities sum to 1.
(B) Yes, because each outcome is positive.
(C) No, because not all outcomes are between 0 and 1.
(D) No, because not all probabilities are between 0 and 1.
19. Recall the sample space for the roll of one fair die: S = {1, 2, 3, 4, 5, 6}. Define the event E as
rolling an even number and the event G as rolling a number greater than 3. Identify the intersection
and the union of these two events.
(A) E and G = {4, 6}, E or G = {1, 2, 3, 4, 5, 6}
(B) E and G = {2, 4, 5, 6}, E or G = {4, 6}
(C) E and G = {4, 6}, E or G = {2, 4, 5, 6}
(D) E and G = {4, 6}, E or G = {2, 3, 4, 5, 6}
20. From past experience, a wheat farmer finds that his annual profit is $90,000 if the summer weather is
typical, $40,000 if the weather is unusually dry, and $20,000 if there is a severe storm that destroys
much of his crop. Weather bureau records indicate that the probability is .70 of typical weather, .20
of unusually dry weather, and .10 of a severe storm. Find the farmer's expected profit for next year.
(A) $90,000
(B) $73,000
(C) $50,000
(D) $31,000
21. According to the 2010 U.S. census, 49.2% of Americans are male and 50.8% are female. Also,
according to Wikipedia, 8% of males are color blind and .5% of females are color blind. If an
American is randomly selected, what is the probability that the person is a color blind male?
(A) .0025
(B) .0394
(C) .0406
(D) .5326
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22. Suppose P(A) = .5 and P(B) = .6. Which of the following must be true?
(A) A and B are mutually exclusive
(B) A and B are not mutually exclusive
(C) A and B are independent
(D) A and B are not independent
Use the following information to answer questions 23 – 24.
Based on exit interviews with graduating students, the university estimates that 92% of Clemson seniors
would choose Clemson again if they could start over as freshmen. Suppose that two Clemson seniors are
selected at random, such that their opinions are independent of one other.
23. What is the probability that both of these two seniors would choose Clemson again?
(A) .0064
(B) .8464
(C) .9936
(D) 1.840
24. What is the probability that at least one of these two seniors would choose Clemson again?
(A) .0064
(B) .0736
(C) .9200
(D) .9936
25. Greg is a kicker for the football team who makes 75% of his field goals from 40 yards. One day after
practice, he decided to kick as many field goals from 40 yards as it took for him to make 10 of them.
Assume that any one field goal attempt is independent of all other attempts. Is the number of field
goal attempts it takes to make 10 a binomial random variable?
(A) Yes, since there are only two possible outcomes for each field goal attempt
(B) Yes, since successive field goal attempts are independent
(C) No, since successive field goal attempts are independent
(D) No, since the number of field goal attempts is not fixed
8
26. Suppose that at the next South Carolina vs. Clemson football game, three fans are selected at random
from among those attending the game. The probability histogram below displays the probability
distribution for the number of Clemson fans in a random sample of three attendees. What is the
probability that there is more than one Clemson fan in the selected sample?
(A) 0.1
(B) 0.4
(C) 0.5
(D) 0.9
27. Find the z-score such that the area under the standard normal curve to its left is 0.67.
(A) −0.44
(B) 0.25
(C) 0.44
(D) 0.75
28. The lifetime of a 60-watt light bulb is approximately normally distributed with a mean of 580 hours
and a standard deviation of 20 hours. Let X represent the lifetime of a 60-watt light bulb. Which of
the following expressions is equivalent to the probability that a randomly selected 60-watt light bulb
lasts for at most 622 hours?
(A) P(X ≤ 622)
(B) P(Z ≤ 2.10)
(C) P(Z < 2.10)
(D) All of the above
9
Use the following information to answer questions 29 – 30.
On the driving range, Tiger Woods practices his swing with a particular club by hitting many, many
balls. Suppose that when Tiger hits his driver, the distance the ball travels follows a normal distribution
with mean 304 yards and standard deviation 8 yards.
29. Which of the following shaded regions corresponds to the proportion of Tiger's drives that travel less
than 290 yards?
(A)
(B)
(C)
(D)
30. What proportion of Tiger's drives travel less than 290 yards?
(A) −1.75
(B) .0401
(C) .9599
(D) .9981
31. What kind of samples more precisely estimate population parameters? That is, for what kind of
samples is the value of the sample statistic more likely to fall close to the population parameter?
(A) Samples selected randomly from normally distributed populations
(B) Samples selected randomly from uniform populations
(C) Larger samples
(D) Normally distributed samples
10
32. Let X be the profit from purchasing one ticket in the Pick 3 lottery. Use the given probability
distribution of X to determine the standard deviation of the profit from one ticket.
Win
Lose
X
$499
P(X)
0.001
−$1
0.999
(A) $249.75
(B) $16.00
(C) $15.80
(D) $15.84
33. A recent poll by Fox News claims that 40% of U.S. voters would vote for Donald Trump in the next
presidential election. Suppose that we have a random sample of 10 U.S. voters. What is the
probability that at least 8 of the selected voters will vote for Trump?
(A) 10C8(0.4)8(0.6)2 + 10C9(0.4)9(0.6)1 + 10C10(0.4)10(0.6)0
(B) 10C8(0.6)8(0.4)2 + 10C9(0.6)9(0.4)1 + 10C10(0.6)10(0.4)0
(C) 10C9(0.4)9(0.6)1 + 10C10(0.4)10(0.6)0
(D) 10C8(0.4)8(0.6)2
34. A waiter has kept track of how much he gets in tips and knows that his mean tip amount per hour is
$16.45 with a standard deviation of $8.14. He suspects that the distribution of his tips is skewed right
because he sometimes will get a really large tip. What is the probability that the average tip amount
per hour for his next 35 hours of work exceeds $20?
(A) .0049
(B) .3300
(C) .9951
(D) Cannot be determined because the distribution of tip amounts is skewed
35. Find P(Z > −2.74).
(A) .0031
(B) .0068
(C) .9932
(D) .9969
11
36. For which normal probability plot shown below is it most likely that the data are from a normally
distributed population?
(A)
(B)
(C)
(D)
37. Which of the following is the best statement of the Central Limit Theorem for a sample mean?
(A) If a large sample is taken from a population that is somewhat symmetric, then the sampling
distribution of the sample mean will be uniformly distributed.
(B) If a large sample is taken from a population of any shape, then the sampling distribution of the
sample mean will be approximately normal.
(C) If a small sample is taken from a population of any shape, then the sampling distribution of the
sample mean will be approximately normal.
(D) If a small sample is taken from a normal population, then the sampling distribution of the sample
mean will not necessarily be approximately normal.
12
38. Which of the following is a discrete random variable?
(A) The amount of cola (in ounces) in a can of Coke
(B) The time it takes to fill a can of Coke with cola
(C) The number of cans of Coke on a delivery truck
(D) The height of a can of Coke
39. A University found the following probability distribution for the number of tattoos its students have.
What is the expected number of tattoos for a student at this school?
Number of tattoos
Probability
0
1
2
3
4
.85
.12
?
.01
.005
(A) 1.00
(B) 0.17
(C) 0.20
(D) Cannot be determined from the given information
Use the following information to answer questions 40 – 41.
The average of all 597 scores on Test 1 in STAT 2300 this semester was 69.6, the median was 71, and
the standard deviation was 14.6. The mean was lower than the median due to a few students that had
scores in the twenties and thirties. Suppose we repeatedly select 53 Test 1 scores and calculate the
averages for these samples.
40. What is true about the distribution of the population of all scores and the sampling distribution of the
sample mean?
(A) The population is symmetric and the sampling distribution is symmetric
(B) The population is skewed and the sampling distribution is symmetric
(C) The population is symmetric and the sampling distribution is skewed
(D) The population is skewed and the sampling distribution is skewed
41. Which of the following best describes the sampling distribution of the sample mean?
(A) Approximately normally distributed with mean 69.6 and standard error 14.6
(B) Approximately normally distributed with mean 69.6 and standard error 2
(C) Skewed left with mean 69.6 and standard error 14.6
(D) Skewed left with mean 69.6 and standard error 2
13
Part II: Free Response. Show all your work. Indicate clearly the methods you use, because you will be
graded on the correctness of your methods as well as on the accuracy and completeness of your results
and explanations. Answers with no justification will receive no credit.
1. The table below breaks down one section of STAT 2300 according to gender and class standing.
Freshman
Sophomore
Junior
Senior
Total
Male
11
13
4
4
32
Female
15
20
7
6
48
Total
26
33
11
10
80
One student will be randomly selected from this section of STAT 2300. For the questions below,
label your answers with appropriate probability notation, using the first letter of each category
name in the table to denote the event the randomly selected student is in that category.
(a) What is the probability that the student selected is a Male with Junior class standing? (2 pts)
P(M and J) =
4
= .05
80
1 pt for appropriate probability notation
1 pt for correct answer (fraction or decimal form)
(b) What is the probability that the student selected is a Male or a Junior? (2 pts)
P(M or J) = P(M) + P(J) − P(M and J)
32 11 4
39
=
+
−
=
= .4875
80 80 80 80
1 pt for appropriate probability notation
1 pt for correct answer (fraction or decimal form)
(c) What is the probability that a Senior is selected? (2 pts)
P(S) =
10
= .125
80
1 pt for appropriate probability notation
1 pt for correct answer (fraction or decimal form)
(d) If a Female is selected, what is the probability that she is a Senior? (2 pts)
P(S|F) =
6
= .125
48
1 pt for appropriate probability notation
1 pt for correct answer (fraction or decimal form)
(e) Are the events Female and Senior independent? Justify your answer. (2 pts)
Yes, they are independent because P(S) = P(S|F).
1 pt for concluding they are independent
1 pt for correct justification
Notes:
• Can also justify by showing P(F) = P(F|S) or P(F and S) = P(F)⋅P(S).
• No credit for incorrect reasoning or for not providing a justification.
14
2. According to the Centers for Disease Control and Prevention, 35% of U.S. adults are obese. Suppose
a sample of 45 U.S. adults is randomly selected. Let X represent the number of adults in the sample
that are obese.
(a) Write out the formula (with the appropriate values substituted) for finding the probability that
exactly 15 of the adults in the sample are obese. You do not need to calculate the final
probability, instead you only need to write out the formula and label your answer with
appropriate probability notation. (4 pts)
X is a binomial random variable with n = 45 and p = .35.
Thus, P(𝑋𝑋 = 15) = 45 C15 (. 35)15 (. 65)30 .
1 pt for appropriate probability notation
1 pt for substituting the correct value of n in binomial formula
1 pt for substituting the correct value of p in binomial formula
1 pt for substituting the correct value of X in binomial formula
Subtract 2 pts for only giving the combination term of the binomial formula
Subtract 1 pt for using only the multiplication rule for independent events
(b) According to the rule of thumb we learned, would you expect the probability distribution of X to
be bell-shaped? Justify your answer. (2 pts)
𝑛𝑛𝑛𝑛(1 − 𝑝𝑝) = 45(. 35)(. 65) = 10.2375
Yes, because np(1 – p) ≥ 10.
1 pt for calculating np(1 – p)
1 pt for comparing to 10 AND concluding it is bell-shaped
(c) Find the mean and standard deviation of X. Round your answers to two decimal places. (2 pts)
𝜇𝜇 = 𝑛𝑛𝑛𝑛 = 45(. 35) = 15.75 people
𝜎𝜎 = �𝑛𝑛𝑛𝑛(1 − 𝑝𝑝) = �45(. 35)(.65) ≈ 3.20 people
1 pt for correctly finding the mean
1 pt for correctly finding the standard deviation
(d) In a sample of 45 U.S. adults, would it be unusual to observe 5 or fewer who are obese? Explain.
(2 pts)
1 pt for concluding it would be unusual
𝜇𝜇 − 2𝜎𝜎 = 15.75 − 2(3.20) = 9.35
1 pt for correct justification
Yes, because 5 is more than two standard
Notes:
deviations below the mean, which would
• It is also acceptable to calculate P(X ≤ 5) = .0003 and
be unusual in a bell-shaped distribution.
note that this is less than .05.
• It is not correct justification to note that P(X = 5) < .05.
• No credit for incorrect reasoning (e.g., noting that 5 is
far away from the expected value with no appeal to 𝜎𝜎).
15
3. Fish lengths in Lake Azul are approximately normally distributed with mean 10 inches and standard
deviation 3 inches. Fish lengths in Lake Blue are approximately normally distributed with mean 9
inches and standard deviation 2 inches. Fishing regulations require that any fish less than 6 inches be
thrown back into the lake.
(a) What is the probability that you will be able to keep a fish you catch in Lake Azul, assuming
you catch one fish at random? Show all your work and label your answer with appropriate
probability notation. (4 pts)
Let X = fish lengths in Lake Azul. Then X is approximately normal with 𝜇𝜇 = 10 and 𝜎𝜎 = 3.
𝑧𝑧 =
𝑥𝑥 − 𝜇𝜇 6 − 10
=
≈ −1.33
𝜎𝜎
3
P(X > 6) = P(Z > −1.33)
.0918
= 1 − .0918
= .9082
X
−1.33
0
Z
1 pt for finding correct z-score
1 pt for finding probability for z in table
1 pt for subtracting probability from 1
1 pt for correct probability notation
(b) What is the probability that you will be able to keep a fish you catch in Lake Blue, assuming
you catch one fish at random? Show all your work and label your answer with appropriate
probability notation. (4 pts)
Let X = fish lengths in Lake Blue. Then X is approximately normal with 𝜇𝜇 = 9 and 𝜎𝜎 = 2.
𝑧𝑧 =
𝑥𝑥 − 𝜇𝜇 6 − 9
=
= −1.50
𝜎𝜎
2
P(X > 6) = P(Z > −1.50)
.0668
= 1 − .0668
= .9332
X
−1.50
0
Z
1 pt for finding correct z-score
1 pt for finding probability for z in table
1 pt for subtracting probability from 1
1 pt for correct probability notation
(c) Based on your calculations in (a) and (b), at which lake are you more likely to be able to keep
your fish, assuming you catch one fish at random? (1 pt)
Lake Blue because it has the larger probability of catching a fish greater than 6 inches in length.
1 pt for identifying the correct lake based on calculated probabilities
16
4. The number of hours that college students spend studying in a typical week is skewed right with a
mean of 𝜇𝜇 = 12.6 hours and a standard deviation of 𝜎𝜎 = 9.5 hours. Let 𝑥𝑥̅ represent the mean number
of hours spent studying in a typical week for a random sample of 100 college students.
(a) Will 𝑥𝑥̅ follow an approximately normal distribution? Why or why not? (3 pts)
Yes, because n = 100 ≥ 30.
1 pt for comparing the sample size to 30
1 pt for noting the sample size is large enough
1 pt for concluding 𝑥𝑥̅ is approximately normal
Note: No credit awarded for incorrect reasoning.
(b) Find the mean and standard deviation of 𝑥𝑥̅ . Label each value with the appropriate symbol. (3 pts)
𝜇𝜇𝑥𝑥̅ = 𝜇𝜇 = 12.6 hours
1 pt for correctly finding the mean
1 pt for correctly finding the standard deviation
1 pt for correct symbols
𝜎𝜎𝑥𝑥̅ = 𝜎𝜎� = 9.5�
= .95 hours
√𝑛𝑛
√100
(c) Find the probability that for a random sample of 100 college students the average amount of time
spent studying in a typical week is less than 15 hours. Show all your calculations and label your
answer with appropriate probability notation. (4 pts)
𝑧𝑧 =
.9943
𝑥𝑥̅ − 𝜇𝜇 15 − 12.6
=
≈ 2.53
𝜎𝜎𝑥𝑥̅
. 95
P(𝑥𝑥̅ < 15) = P(Z < 2.53) = .9943
𝑥𝑥̅
0
2.53
Z
1 pt for substituting correct values in z-score formula
1 pt for finding the correct value of the z-score
1 pt for finding the area below z from table
1 pt for appropriate probability notation
Note: If a response does not recognize that the
probability must be calculated from a sampling
distribution, no credit will be awarded.
Only deduct 1 pt total for using x instead of 𝑥𝑥̅ in both the symbols in (b) and the probability notation in (c)
17
5. The table below breaks down one section of STAT 2300 according to gender and political affiliation.
Democratic
Party
Republican
Party
Libertarian
Party
Green
Party
Total
Male
7
15
7
1
30
Female
22
25
5
2
54
Total
29
40
12
3
84
One student will be randomly selected from this section of STAT 2300. For the questions below,
label your answers with appropriate probability notation, using the first letter of each category
name in the table to denote the event the randomly selected person is in that category. Answer with
a fraction or a decimal rounded to three decimal places.
(a) What is the probability that the selected student is a Male or a Republican? (2 pts)
30
40
15
55
P(M or R) = P(M) + P(R) – P(M and R) = 84 + 84 − 84 = 84 ≈ .655
1 pt for appropriate probability notation
1 pt for correct answer (fraction or decimal form)
(b) Are the categories Male and Republican mutually exclusive? Explain. (2 pts)
No, because P(M and R) ≠ 0. That is, a person can be both Male and Republican.
1 pt for concluding they are NOT mutually exclusive
1 pt for correct justification
No credit for incorrect reasoning or for not providing a justification.
(c) What is the probability that the selected student is a Democrat? (2 pts)
29
P(D) = 84 ≈ .345
1 pt for appropriate probability notation
1 pt for correct answer (fraction or decimal form)
(d) If a Female is selected, what is the probability that the selected student is a Democrat? (2 pts)
22
P(D|F) = 54 ≈ .407
1 pt for appropriate probability notation
1 pt for correct answer (fraction or decimal form)
(e) Are the categories Female and Democrat independent? Explain mathematically. (2 pts)
No, because P(D|F) ≠ P(D). Knowing that the selected student is female makes it more likely that
the selected student is a Democrat.
1 pt for concluding they are not independent
1 pt for correct justification
Notes:
Can also justify by showing P(F|D) ≠ P(F) or P(F and D) ≠ P(F)⋅P(D).
No credit for incorrect reasoning or for not providing a justification.
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6. A box contains 5 small bags of snacks: one bag of Doritos, two bags of Cheetos, and two bags of
Lays. A bag is selected without looking in the box and given to a child, then a second bag is selected
and given to a second child.
(a) List all of the possible outcomes for the types of snacks that the two children get using the letters
D, C, and L. (3 pts)
1st child
2nd child
Outcome
C
DC
L
DL
D
CD
C
CC
L
CL
D
LD
C
LC
L
LL
D
C
L
S = {DC, DL, CD, CC, CL, LD, LC, LL}
Subtract 1 pt for not listing final outcomes
Subtract 1 pt for each missing or incorrect outcome
(b) What is the probability that the second child gets Doritos if we know that the first child got
Cheetos? Label your answer with appropriate probability notation. (2 pts)
1
P(D second | C first) = 4 = .25
1 pt for appropriate probability notation
1 pt for correct answer (fraction or decimal form)
(c) What is the probability that one child gets Doritos and the other gets Lays? Label your answer
with appropriate probability notation. (4 pts)
P(DL or LD) = P(DL) + P(LD)
1
2
2
1
=5×4+5×4
2
2
= 20 + 20
4
1
= 20 = 5 = .20
1 pt for appropriate probability notation
1 pt for finding P(DL) correctly
1 pt for finding P(LD) correctly
1 pt for adding P(DL) and P(LD) to get final answer
Note: If the outcomes are assumed to be equally likely, then no
credit will be awarded for finding P(DL) or P(LD) correctly.
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7. A coin is flipped 4 times. Let X be the number of heads observed in the 4 coin flips.
(a) What are the possible values of X? (2 pts)
0, 1, 2, 3, and 4
Subtract 1 pt for each missing or incorrect outcome (max of 2 pts)
Now suppose that the coin is biased such that the probability of observing heads is 0.6.
(b) Find the expected number of heads observed in 4 flips of the coin. Label your answer with the
appropriate symbol. (2 pts)
X is binomial with n = 4 & p = 0.6
𝜇𝜇 = 𝑛𝑛𝑛𝑛 = 4(0.6) = 2.4 heads
1 pt for labeling answer with appropriate symbol
1 pt for correctly finding the mean
(c) Find the standard deviation of the number of heads observed in 4 flips of the coin. Label your
answer with the appropriate symbol and round your answer to two decimal places. (2 pts)
𝜎𝜎 = �𝑛𝑛𝑛𝑛(1 − 𝑝𝑝) = �4(0.6)(0.4) = √0.96 ≈ 0.98 heads
1 pt for labeling answer with appropriate symbol
1 pt for correctly finding the standard deviation
(d) What is the probability of observing exactly 3 heads? Show all your calculations and label
your answer with appropriate probability notation. (3 pts)
𝑃𝑃(𝑋𝑋 = 3) = 4 C3 (0.6)3 (0.4)1 = 4(0.216)(0.4) = 0.3456
1 pt for appropriate probability notation
1 pt for correctly substituting values into binomial formula
1 pt for the correct final answer
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8. Cherry trees in a certain orchard have heights that follow a normal distribution with a mean of 112
inches and a standard deviation of 14 inches.
(a) What proportion of trees are between 90 and 100 inches tall? Show all your work and label
your answer with appropriate probability notation. (5 pts)
Let X = height of cherry trees in this orchard.
Then X is approximately normal with 𝜇𝜇 = 112 and 𝜎𝜎 = 14.
.1949
.0582
X
-1.57 -0.86
0
Z
𝑧𝑧 =
𝑧𝑧 =
𝑥𝑥 − 𝜇𝜇 90 − 112
=
≈ −1.57
𝜎𝜎
14
𝑥𝑥 − 𝜇𝜇 100 − 112
=
≈ −0.86
𝜎𝜎
14
P(90 < X < 100) = P(−1.57 < Z < −0.86)
= .1949 − .0582
= .1367
1 pt for finding correct z-score for 90
1 pt for finding correct z-score for 100
1 pt for finding probability for calculated z-scores in table
1 pt for subtracting the two probabilities
1 pt for appropriate probability notation
(b) An agricultural scientist wants to study the tallest 10% of trees to determine whether they have a
certain gene that allows them to grow taller. To do this, she needs to study all the trees above a
certain height. What height is this? (5 pts)
𝑥𝑥 − 𝜇𝜇
𝑧𝑧 =
𝜎𝜎
.1000
1.28 =
.9000
X
0
1.28
Z
𝑥𝑥 − 112
14
𝑥𝑥 = 1.28(14) + 112 = 129.92
129.92 inches
2 pts for finding correct z value from table
2 pts for correctly substituting values into z-score formula
1 pt for the correct final answer
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