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Final Exam
Elementary Statistics
May 15, 2014
PLEASE READ THE FOLLOWING INFORMATION.
• This is a 110-minute exam. Calculators are allowed. Books, notes, formula sheets, and
other aids are not allowed.
• You are required to show all your work and provide the necessary explanations everywhere
to get full credit.
1. The quality assurance engineer of a television manufacturer inspects TVs in lots of 100.
He selects 5 of the 100 TVs at random and inspects them thoroughly. Assuming that 6
of the 100 TVs in the current lot are defective, find the probability that exactly 2 of the
5 TVs selected by the engineer are defective.
Solution: We apply the Hypergeometric Distribution formula
b
a
·
n−x
x
f (x) =
a+b
n
with n = 5, a = 6, b = 94, and x = 2. We have
94
6
94
6
·
·
15 · 134, 044
2, 010, 660
5−2
2
3
2
= =
f (2) = =
≈ 0.027
6 + 94
100
75, 287, 520
75, 287, 520
5
5
So, there is a 2.7% chance that exactly 2 of the 5 TVs selected by the engineer will be
defective.
Info: The average in the class for this problem was 64.1%.
2. On New Year’s Eve, the probability of a person having a car accident is 0.09. The
probability of a person driving while intoxicated is 0.32 and probability of a person having
a car accident while intoxicated is 0.15. What is the probability of a person driving while
intoxicated or having a car accident?
Solution: Let I be the event that a person is driving while intoxicated and let A be the
event that a person is having a car accident. Then
P (I ∪ A) = P (I) + P (A) − P (I ∩ A) = 0.32 + 0.09 − 0.15 = 0.26
Info: The average in the class for this problem was 88.1%.
3. In the paper “Persistent Pulmonary Hypertension of the Neonate and Asymmetric Growth
Restriction” (Obstetrics & Gynecology, Vol. 91, No. 3, pp. 336-341), M. Williams et
al. reported on a study of characteristics of neonates. Infants treated for pulmonary
hypertension, called the PH group, were compared with those not so treated, called
the control group. One of the characteristics measured was head circumference. The
mean head circumference of the 10 infants in the PH group was 34.2 centimeters (cm).
Assuming that head circumferences for infants treated for pulmonary hypertension are
normally distributed with standard deviation 2.1 cm, determine the sample size required
to have a margin of error of 0.5 cm with a 95% confidence level.
z · σ 2
α/2
Solution: To find the sample size, we use the formula n =
. We know that
E
σ = 2.1 and E = 0.5. The confidence level is 0.95, which means that α = 0.05 and
zα/2 = z0.025 = 1.96. Thus
2
1.96 · 2.1
n=
≈ 67.7658
0.5
which, rounded up to the nearest whole number, is 68.
Info: The average in the class for this problem was 63.7%.
4. For a given population with σ = $15, we want to test the null hypothesis µ = $80 on the
basis of random sample of size n = 50. If the null hypothesis is rejected when x is greater
than or equal to $83 and otherwise it is accepted, find
(a) the probability of Type I error;
(b) the probability of Type II error when µ = $82;
(c) the probability of Type II error when µ = $84.
15
σ
Solution: First we note that σx = √ = √ .
n
50
(a) We have
83 − 80 √
√ = 2 ≈ 1.41
15/ 50
and the probability is 0.5 − 0.4207 = 0.0793.
z=
(b) We have
√
2
83 − 82
√ =
z=
≈ 0.47
3
15/ 50
and the probability is 0.5 + 0.1808 = 0.6808.
(c) We have
√
83 − 84
2
√ =−
z=
≈ −0.47
3
15/ 50
and the probability is 0.5 − 0.1808 = 0.3192.
Info: The average in the class for this problem was 50.6%.
5. The table below provides a sample of 19 exam grades randomly selected from a large
class.
76
76
86
92
74
78
84
82
82
72
62
74
96 66
52 68
76 78
88
8
4 6
4 6
6 6 8
8
Construct a stem-and-leaf display.
Solution:
5
6
7
8
9
2
2
2
2
2
6
4
2
6
8
Info: The average in the class for this problem was 96.7%.
6. Consider the following data for shipments of peanuts from a hypothetical U.S. exporter
to five Canadian cities.
City
Montreal
Ottawa
Toronto
Vancouver
Winnipeg
Peanuts (Thousands of Bags)
64.0
15.0
285.0
228.0
45.0
Let’s assume that shipments to the respective cities will be sold at the following profits
per thousand bags: $15.00, $13.50, $15.50, $12.00, and $14.00. Find the average profit
per thousand bags.
Solution: By the formula for the weighted mean, we get
(64)($15.0) + (15)($13.5) + (285)($15.5) + (228)($12.0) + (45)($14.0) $8946
=
≈ $14.04
64 + 15 + 285 + 228 + 45
637
Info: The average in the class for this problem was 87.3%.
7. An urn contains 6 red marbles and 4 black marbles. Two marbles are drawn without
replacement from the urn. What is the probability that both of the marbles are black?
Solution: Let B1 and B2 be the events that the first and second marbles are black,
respectively. Then we want P (B1B2 ). Note that we easily have that P (B1 ) = 4/10. Also,
P (B2 |B1 ) = 3/9 because the reduced sample space B1 consists of 3 black marbles and 6
red marbles, and we are asking for the probability of choosing a black marble. Therefore,
4 3
2 1
2
· = · =
10 9
5 3
15
10
4
.
/
Note, that we could also do it straightaway as
2
2
P (B1B2 ) = P (B1 )P (B2 |B1 ) =
Info: The average in the class for this problem was 79.4%.
8. In 1908, W. S. Gosset published the article “The Probable Error of a Mean” (Biometrika,
Vol. 6, pp. 125). In this pioneering paper, written under the pseudonym “Student,”
Gosset introduced what later became known as Students t-distribution. Gosset used the
following data set, which gives the additional sleep in hours obtained by a sample of 10
patients using laevohysocyamine hydrobromide.
1.9 0.8 1.1 0.1
4.4 5.5 1.6 4.6
-0.1
3.4
(a) Find the mean of the sample.
Solution: We have
x=
23.3
1.9 + 0.8 + 1.1 + 0.1 + (−0.1) + 4.4 + 5.5 + 1.6 + 4.6 + 3.4
=
= 2.33 hr
10
10
(b) Find the standard deviation of the sample.
Solution: We will use the formula
r
X
Sxx
, where Sxx =
x2 −
n−1
X
X
First we calculate
x and
x2 , getting
s=
X
and
X
X 2
x
n
x = 1.9 + 0.8 + 1.1 + 0.1 + (−0.1) + 4.4 + 5.5 + 1.6 + 4.6 + 3.4 = 23.3
x2 = 1.92 + 0.82 + 1.12 + 0.12 + (−0.1)2 + 4.42 + 5.52 + 1.62 + 4.62 + 3.42 = 90.37
Therefore
Sxx =
hence
X
x2 −
s=
r
X 2
x
n
Sxx
=
n−1
r
= 90.37 −
23.32
≈ 36.08
10
36.08
≈ 2.002 hr
9
(c) Obtain a 95% confidence interval for the additional sleep that would be obtained on
average for all people using laevohysocyamine hydrobromide.
Solution: We want a 95% confidence interval, so α = 1 − 0.95 = 0.05. For n = 10, we
have df= 10 − 1 = 9. From Table II,
tα/2 = t0.05/2 = t0.025 = 2.262
The confidence interval for µ is from
s
x − tα/2 · √
n
s
to x + tα/2 · √ .
n
We know that x = 2.33 and s = 2.002, therefore a 95% confidence interval for µ is from
2.002
2.33 − 2.262 · √
10
2.002
to 2.33 + 2.262 · √
10
or 0.90 to 3.76 hr.
(d) Interpret your answer from part (c).
Solution: It appears so because, based on the confidence interval, we can be 95% confident
that the mean additional sleep is somewhere between 0.90 hr and 3.76 hr and that, in
particular, the mean is positive.
Info: The average in the class for this problem was 73.1%.
9. Researchers G. Evans and E. Kantrowitz explored the health consequences for exposure
to many different environmental risks in the journal article “Socioeconomic Status and
Health: The Potential Role of Environmental Risk Exposure” (Annual Review of Public
Health, Vol. 23, No. 1, pp. 303-331). According to research, 65% of preschool children
living in poverty have been exposed to cigarette smoke at home. If 200 preschool children
living in poverty are selected at random, what is the probability that at least 125 have
been exposed to cigarette smoke at home?
Solution: We have n = 200 and p = 0.65, therefore
np = 200 · 0.65 = 130 and n(1 − p) = 200 · 0.35 = 70
Both np and n(1 − p) are greater than 5, so we can approximate the binomial distribution
by normal. We get
p
√
µ = np = 130 and σ = np(1 − p) = 200 · 0.65 · 0.35 ≈ 6.745
To make the correction for continuity, we subtract 0.5 from 125. Thus we need to find
the area under the normal curve with parameters µ = 130 and σ = 6.745 that lies to the
right of 124.5. We have
124.5 − 130
≈ −0.82
z=
6.745
therefore the probability is about 0.2939 + 0.5 = 0.7939 that least 125 have been exposed
to cigarette smoke at home.
Info: The average in the class for this problem was 42.9%.
10. A saleswoman has been offered a new job with a fixed salary of $290. Her records from
her present job show that her weekly commissions have the following probabilities:
Commission
Probability
0
$100 $200 $300 $400
0.05 0.15 0.25 0.45 0.1
Should she change her present job?
Solution: The expected value from her present job is
(0 · $0.05) + (0.15 · $100) + (0.25 · $200) + (0.45 · $300) + (0.1 · $400) = $240
Since $290 > $240, she should change her present job.
Info: The average in the class for this problem was 87.6%.
11. Data on investments in the high-tech industry by venture capitalists are compiled by
VentureOne Corporation and published in America’s Network Telecom Investor Supplement. A random sample of 18 venturecapital investments in the fiber optics business
sector yielded the following data, in millions of dollars.
5.60 6.27 5.96 10.51 2.04 5.48
5.74 5.58 4.13 8.63 5.95 6.67
4.21 7.71 9.21 4.98 8.64 6.66
Determine a 95% confidence interval for the mean amount, µ, of all venture-capital investments in the fiber optics business sector. Assume that the population standard deviation
is $2.04 million. (Note: The sum of the data is $113.97 million.)
Solution: From Table I, zα/2 = z0.05/2 = z0.025 = 1.96. The confidence interval for µ is
from
σ
σ
x − zα/2 · √
to x + zα/2 · √
n
n
We know σ = 2.04, n = 18, and zα/2 = 1.96. To compute x for the data in the Table
above, we apply the usual formula:
X
xi
113.97
=
≈ 6.332
x=
n
18
to one decimal place. Consequently, a 95% confidence interval for µ is from
2.04
2.04
6.332 − 1.96 · √
to 6.332 + 1.96 · √
18
18
or $5.389 million to $7.274 million.
Info: The average in the class for this problem was 76.7%.
12. The scores achieved by students in America make the news often, and all kinds of conclusions are drawn based on these scores. The ACT Assessment is designed to assess high
school students’ general educational development and their ability to complete collegelevel work. One of the categories tested is Science Reasoning. The mean ACT test score
for all high school graduates in 2008 in Science Reasoning was 20.8 with a standard deviation of 4.6. According to Chebyshev’s theorem, at least what percent of high school
graduates’ ACT scores in Science Reasoning were between 11.6 and 30.0?
Solution: Since
it follows that
30.0 − 20.8 = 20.8 − 11.6 = 9.2
k=
9.2
=2
4.6
therefore at least
3
1
= = 75%
2
2
4
of high school graduates’ ACT scores in Science Reasoning were between 11.6 and 30.0.
1−
Info: The average in the class for this problem was 78.7%.
13. A variable is normally distributed with mean 6 and standard deviation 2. Find the
percentage of all possible values of the variable that
(a) lie between 1 and 7.
Solution: We have
z1 =
1−6
7−6
= −2.5 and z2 =
= 0.5
2
2
therefore the probability is 0.4938 + 0.1915 = 0.6853, so the percentage is 68.53%.
(b) exceed 5.
Solution: We have
5−6
= −0.5
2
therefore the probability is 0.1915 + 0.5 = 0.6915, so the percentage is 69.15%.
z=
(c) are less than 4.
Solution: We have
4−6
= −1
2
therefore the probability is 0.5 − 0.3413 = 0.1587, so the percentage is 15.87%.
z=
Info: The average in the class for this problem was 75.6%.
14. If ice cream sundaes come in 5 flavors with 4 possible toppings, how many different
sundaes can be made with one flavor of ice cream and one topping?
Solution: By the Basic Principle of Counting, the number of different sundaes is 5·4 = 20.
Info: The average in the class for this problem was 96%.
15. Scores on the SAT test have a mean of 1518 and a standard deviation of 325. Scores on
the ACT test have a mean of 21.1 and a standard deviation of 4.8. Which is relatively
better: a score of 1840 on the SAT test or a score of 26.0 on the ACT test? Why?
Solution: We have
z1 =
1840 − 1518
26.0 − 21.1
≈ 0.99 and z2 =
≈ 1.02
325
4.8
The score of 26.0 is relatively better because it has the larger z score.
Info: The average in the class for this problem was 88.9%.
16. A math teacher gave her class two tests. 25% of the class passed both tests and 42% of
the class passed the first test. What percent of those who passed the first test also passed
the second test?
Solution: 0.25/0.42=0.6
Info: The average in the class for this problem was 79.8%.
17. In how many ways can the letters of the word “statistics” be arranged?
Solution: We have
10!
= 50400
3!3!2!
Info: The average in the class for this problem was 80.2%.