Download Practice Exam 4 - Statistics

Lenarz
Math 102
Practice Exam # 4
Name:
Formulas:
P
x=
n
x
s
s=
Σ(x − x)2
n−1
z=
x−x
s
1. Discuss the following statement and tell what possible misuses or misinterpretations may
exist.
(a) The average depth of the pond is only 3 feet deep, so it is safe to go wading.
Solution: The pond may be very deep in some places, only the average depth
is 3 feet. It may not be safe to wade!
(b) More men than women are involved in automobile accidents. Therefore, women are
better drivers.
Solution: More men may drive than women, so they would understandably be
involved in more accidents. Also, men may drive in more dangerous conditions,
resulting in more accidents.
(c) At West High School, half the students are below average in mathematics. Therefore, the school should receive more federal aid to raise student scores.
Solution: Half of the student scores should be below the average if the data
are normally distributed, so the school should not receive more aid.
2. Use the histogram to answer the following questions:
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(a) How many homes were included in the survey?
Solution:
2 + 4 + 8 + 6 + 4 + 3 + 1 = 28
(b) In how many homes were four televisions observed?
Solution: 4
(c) What is the mode of the distribution?
Solution: 2
(d) How many total televisions were observed?
Solution:
2(0) + 4(1) + 8(2) + 6(3) + 4(4) + 3(5) + 1(6) = 75
3. The following data represents IQ scores from 50 sixth-graders:
80
81
87
88
89
89
89
90
91
92
92
93
94
94
94
95
95
97
97
97
97 100 102 106 110 120
98 100 103 108 113 120
99 100 103 108 114 122
100 100 103 108 114 128
100 101 104 109 119 135
(a) Construct a (grouped) frequency distribution with six intervals of length 10, starting
with the interval 80-89.
Solution:
IQ Score
80 − 89
90 − 99
100 − 109
110 − 119
120 − 129
130 − 139
Frequency
7
16
17
5
4
1
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(b) Construct a histogram using your frequency distribution in (a).
Solution:
4. Find (i) the mean, (ii) median, (iii) mode, and (iv) five number summary for the following
data
(a) 7, 8, 8, 10, 12, 12, 12, 23, 25
Solution:
(i)
117
7 + 8 + 8 + 10 + 12 + 12 + 12 + 23 + 25
=
= 13
9
9
(ii) 12
(iii) 12
(iv) The lower half is the set {7, 8, 8, 10} which has median 8+8
= 8, so Q1 = 8.
2
The upper half is the set {12, 12, 23, 25} which has median 12+23
= 17.5,
2
so Q3 = 17.5. Thus the five number summary is
7, 8, 12, 17.5, 25
(b) 1, 3, 5, 7, 9, 11, 13, 15
Solution:
(i)
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
64
=
=8
8
8
(ii)
7+9
16
=
=8
2
2
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(iii) no mode
(iv) The lower half is the set {1, 3, 5, 7} which has median 3+5
= 4, so Q1 = 4.
2
= 12, so
The upper half is the set {9, 11, 13, 15} which has median 11+13
2
Q3 = 12. Thus the five number summary is
1, 4, 8, 12, 15
(c) 1, 1, 1, 1, 4, 4, 4, 4, 6, 8, 10, 12, 15, 21
Solution:
(i)
1 + 1 + 1 + 1 + 4 + 4 + 4 + 4 + 6 + 8 + 10 + 12 + 15 + 21
92
=
≈ 6.6
14
14
(ii)
8
4+4
= =4
2
2
(iii) 1 & 4 (bimodal)
(iv) The lower half is the set {1, 1, 1, 1, 4, 4, 4} which has median 1, so Q1 = 1.
The upper half is the set {4, 6, 8, 10, 12, 15, 21} which has median 10, so
Q3 = 10. Thus the five number summary is
1, 1, 4, 10, 21
(d) 9, 9, 9, 9, 9
Solution:
(i)
9+9+9+9+9
45
=
=9
5
5
(ii) 9
(iii) no mode or 9 (either is acceptable)
(iv) The lower half is the set {9, 9} which has median 9+9
= 9, so Q1 = 9. The
2
9+9
upper half is the set {9, 9} which has median 2 = 9, so Q3 = 9. Thus
the five number summary is
9, 9, 9, 9, 9
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5. A mean of 60 on seven exams is needed to pass a course. On her first six exams, Sheryl
received grades of 51, 72, 80, 62, 57, and 69.
(a) What grade must she receive on her last exam to pass the course (earn a mean
grade of 60)?
Solution: Let x be the score she needs on the seventh exam.
51 + 72 + 80 + 62 + 57 + 69 + x
7
391 + x
7
391 + x
x
= 60
= 60
= 420
= 29
(b) A mean of 70 is needed to get a C in the course. Is it possible for Sheryl to get a
C? If so, what score must she receive on the seventh exam?
Solution: Let x be the score she needs on the seventh exam.
51 + 72 + 80 + 62 + 57 + 69 + x
7
391 + x
7
391 + x
x
= 70
= 70
= 490
= 99
So Sheryl will get a C in the course if she scores a 99 on the seventh exam.
(c) If her lowest grade of the exams already taken is to be dropped, what grade must
she receive on her last exam to get a C in the course?
Solution: Sheryl’s lowest score on the first six exams is 51, so we will eliminate
it from the data set. Let x be the score she needs on the seventh exam.
72 + 80 + 62 + 57 + 69 + x
6
340 + x
6
340 + x
x
= 70
= 70
= 420
= 80
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6. Find (i) the range and (ii) the standard deviation of the following data sets
(a) 7, 8, 8, 10, 12, 12, 12, 23, 25
Solution:
(i)
Range = 25 − 7 = 18
(ii) Mean (from question 4 part (a) ) is x = 13
x
f
7
1
8
2
10 1
12 3
23 1
25 1
n= 9
s
s=
x−x
7 − 13 = −6
8 − 13 = −5
10 − 13 = −3
12 − 13 = −1
23 − 13 = 10
25 − 13 = 12
Σ(x − x)2
=
n−1
r
(x − x)2
f (x − x)2
36
36
25
50
9
9
1
3
100
100
144
144
2
Σ(x − x) =
342
342
=
9−1
r
342 √
= 42.75 ≈ 6.54
8
(b) 1, 3, 5, 7, 9, 11, 13, 15
Solution:
(i)
Range = 15 − 1 = 14
(ii) Mean (from question 4 part (b) ) is x = 8
x
1
3
5
7
9
11
13
15
n=8
x−x
1 − 8 = −7
3 − 8 = −5
5 − 8 = −3
7 − 8 = −1
9−8=1
11 − 8 = 3
13 − 8 = 5
15 − 8 = 7
(x − x)2
49
25
9
1
1
9
25
49
Σ(x − x)2 = 168
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s
s=
Σ(x − x)2
=
n−1
r
168
=
8−1
r
168 √
= 24 ≈ 4.90
7
(c) 1, 1, 1, 1, 4, 4, 4, 4, 6, 8, 10, 12, 15, 21
Solution:
(i)
Range = 21 − 1 = 20
(ii) Mean (from question 4 part (c) ) is x ≈ 6.6
x
f
x−x
(x − x)2
f (x − x)2
1
4 1 − 6.6 = −5.6
31.36
125.44
4
4 4 − 6.6 = −2.6
6.76
27.04
6
1 6 − 6.6 = −0.6
0.36
0.36
8
1
8 − 6.6 = 1.4
1.96
1.96
10
1 10 − 6.6 = 3.4
11.56
11.56
12
1 12 − 6.6 = 5.4
29.16
29.16
15
1 15 − 6.6 = 8.4
70.56
70.56
21
1 21 − 6.6 = 14.4
207.36
207.36
2
n = 14
Σ(x − x) =
473.44
s
s=
Σ(x − x)2
=
n−1
r
473.44
=
14 − 1
r
473.44
≈ 6.03
13
(d) 9, 9, 9, 9, 9
Solution:
(i)
Range = 9 − 9 = 0
(ii) Mean (from question 4 part (d) ) is x = 9
x
f
9
5
n= 5
s
s=
x−x
9−9=0
Σ(x − x)2
=
n−1
(x − x)2
f (x − x)2
0
0
2
Σ(x − x) =
0
r
0
=
5−1
r
0 √
= 0=0
4
Math 102
Page 8
7. Jiffy Lube has franchises in two different parts of the city. The number of oil changes
made daily, for 25 days, is given below.
33
19
43
40
15
East Side
59 27 30
42 25 22
27 57 37
67 38 44
31 49 41
42
32
52
43
35
38
38
39
30
31
West Side
46 38 38
38 37 39
36 40 37
34 42 45
46 28 45
30
31
47
29
48
(a) Construct a (grouped) frequency distribution for each store with 11 intervals of
length 5 starting with 15.
Solution:
East Side
Oil Changes Frequency
15 − 19
2
20 − 24
1
25 − 29
3
30 − 34
4
35 − 39
3
40 − 44
7
45 − 49
1
50 − 54
1
55 − 59
2
60 − 64
0
65 − 69
1
(b) Draw a histogram for each store.
Solution:
West Side
Oil Changes Frequency
15 − 19
0
20 − 24
0
25 − 29
2
30 − 34
5
35 − 39
10
40 − 44
2
45 − 49
6
50 − 54
0
55 − 59
0
60 − 64
0
65 − 69
0
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(c) Using the histogram, which store appears to have a greater mean or do the means
appear about the same?
Solution: The mean of the distributions appears to be about the same.
(d) Using the histogram, determine which store appears to have the standard deviation.
Solution: The standard deviation for the East side appears greater – the data
are more widely spread apart from the mean.
(e) Calculate the mean for each store and determine if your answer in part (c) was
correct.
Solution: For the East side:
1
x =
33 + 59 + 27 + 30 + 42 + 19 + 42 + 25 + 22 + 32 + 43 + 27 + 57
25
+37 + 52 + 40 + 67 + 38 + 44 + 43 + 15 + 31 + 49 + 41 + 35
=
950
25
= 38
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For the West side:
1
x =
38 + 46 + 38 + 38 + 30 + 38 + 38 + 37 + 39 + 31 + 39 + 36 + 40
25
+37 + 47 + 30 + 34 + 42 + 45 + 29 + 31 + 46 + 28 + 45 + 48
=
950
25
= 38
(f) Calculate the standard deviation for each store at determine if your answer in part
(d) was correct.
Solution: For the East side: Mean is x = 38
x
f
15
1
19
1
22
1
25
1
27
2
30
1
31
1
32
1
33
1
35
1
37
1
38
1
40
1
41
1
42
2
43
2
44
1
49
1
52
1
57
1
59
1
67
1
n = 25
x−x
15 − 38 = −23
19 − 38 = −19
22 − 38 = −16
25 − 38 = −13
27 − 38 = −11
30 − 38 = −8
31 − 38 = −7
32 − 38 = −6
33 − 38 = −5
35 − 38 = −3
37 − 38 = −1
38 − 38 = 0
40 − 38 = 2
41 − 38 = 3
42 − 38 = 4
43 − 38 = 5
44 − 38 = 6
49 − 38 = 11
52 − 38 = 14
57 − 38 = 19
59 − 38 = 21
67 − 38 = 29
(x − x)2
f (x − x)2
529
529
361
361
256
256
169
169
121
242
64
64
49
49
36
36
25
25
9
9
1
1
0
0
4
4
9
9
16
32
25
50
36
36
121
121
196
196
361
361
441
441
841
841
2
Σ(x − x) =
3832
Math 102
Page 11
s
s=
Σ(x − x)2
=
n−1
r
3832
=
25 − 1
r
3832
≈ 12.64
24
For the West side: Mean is x = 38
x
f
28
1
29
1
30
2
31
2
34
1
36
1
37
2
38
5
39
2
40
1
42
1
45
2
46
2
47
1
48
1
n = 25
s
s=
(x − x)2
f (x − x)2
100
100
81
81
64
128
49
98
16
16
4
4
1
2
0
0
1
2
4
4
16
16
49
98
64
128
81
81
100
100
2
Σ(x − x) =
858
r
r
Σ(x − x)2
858
858
=
=
≈ 5.98
n−1
25 − 1
24
x−x
28 − 38 = −10
29 − 38 = −9
30 − 38 = −8
31 − 38 = −7
34 − 38 = −4
36 − 38 = −2
37 − 38 = −1
38 − 38 = 0
39 − 38 = 1
40 − 38 = 2
42 − 38 = 4
45 − 38 = 7
46 − 38 = 8
47 − 38 = 9
48 − 38 = 10
8. In 2005, 1,475,623 college-bound seniors took the SAT exam. The distribution of scores
in the math section of the SAT was approximately normal with mean x = 520 and
standard deviation s = 115.
(a) Find the percentile corresponding to a test score of 750.
Solution:
750 − 520
230
=
=2
115
115
Using Table 15.16 on page 754, this z-score corresponds to 0.477 = 47.7%, so
47.7% of the scores are between z = 0 and z = 2. The percentage of scores
below z = 2 is then given by 50% + 47.7% = 97.7%.
z750 =
Math 102
Page 12
(b) Estimate the 75th percentile score on the exam.
Solution: The 75th percentile would correspond to a positive z-score with
75% − 50% = 25% of the data between it and z = 0. From Table 15.16 on
page 754, this would correspond to a z-score of around z = 0.67. So we have
x − 520
115
77.05 = x − 520
597.05 = x
0.67 =
So a score of around 597 is at approximately the 75th percentile.
9. Assume the amount of cornflakes in a box is normally distributed with a mean of 16
ounces with a standard deviation of 0.1 ounce.
(a) Determine the percent of boxes that will contain between 15.83 ounces and 16.32
ounces of corn flakes.
Solution:
15.83 − 16
= −1.7
0.1
which corresponds to 0.455 = 45.5% in Table 15.16. Thus 45.5% of the data lie
between z = −1.7 and z = 0.
z15.83 =
z16.32 =
16.32 − 16
= 3.2
0.1
which corresponds to 0.499 = 49.9% in Table 15.16. Thus 49.9% of the data
lie between z = 0 and z = 3.2. So the percent of boxes that will have between
15.83 and 16.32 ounces is
45.5% + 49.9% = 95.4%
(b) Determine the percent of boxes that will contain more than 16.16 ounces of corn
flakes.
Solution:
16.16 − 16
= 1.6
0.1
which corresponds to 0.445 = 44.5% in Table 15.16. Thus 44.5% of the data lie
between z = 0 and z = 1.6. Also note that 50% of the data lie to the right of
z = 0. Thus
50% − 44.5% = 5.5%
z16.16 =
Math 102
Page 13
will have more than 16.16 ounces.
(c) If the manufacturer produces 300,000 boxes, how many of them will contain less
than 15.83 ounces of corn flakes?
Solution: z15.83 = −1.7 which corresponds to 0.455 = 45.5% in Table 15.16.
Thus 45.5% of the data lie between z = −1.7 and z = 0. Since 50% of the
data lies to the left of z = 0, we have 50% − 45.5% = 4.5% of the data below
z = −1.7. Thus, the number of boxes with less than 15.83 ounces is
300, 000 × 0.045 = 13, 500
(d) If the manufacturer produces 300,000 boxes, how many of them will contain more
than 16.16 ounces of corn flakes?
Solution: From part (b), 5.5% have more than 16.16 ounces, so we have
300, 000 × 0.055 = 16, 500