z C.....
5. a) mean:
6.
7.
8.
9.
10.
e.g., The two middle intervals represent 60% of the data. The four middle intervals represent 86% of the data. Therefore, the data is normally distributed, since the intervals are dose to 68% and 95% of the data. Also, the graph of the data approximates a bell shape. 1) lntasval 20 24 25 29 30 34 35—39 40—44 45—49
standard deviation; cr = 18.3 points 73.6 points standard deviation: h) mean: = 79.3 points 12.7 points e.g., The mean is increased and the standard deviation is decreased. This signifies Jarome Iginla’s point production is more consistent when the two point totals list the lowest seasons are removed. BrandA =88.lkg,o’0.l6kgjBrandB;= 88.lkg,a= 0.11kg e.g., The means are approximately the same, but the standard deviation is slightly iower list B’s bags. So, the masses of B’s bags are more consistent. Mario should order Brand B. However, the standard deviations are low for each supplier, so Mario could order from either supplier. a) EverSmd>”.= 13.3h,o4.6h;EliteEducation:’I3.7h, cr”4.3h b) e.g., The standard deviation fist Elite Education is lower, so their clients are more consistent in their studying times. C. e.g., The mean test scores were approximately the same However, the standard deviation for Class A was lower. This means that the test scores for class A were more consistent and closer to the mean than those for Class B. So, Class A performed better, in this sense. a)_$37,ff’$12
b) kilps($)
Midpolntoflntenial,x
Frequencyf
10-20
3
15
f•j 45
20-30
4
25
100
30—40
6
35
210
40—50
6_
45
270
50—60
1
55
55
20
[
.!quey. oint
1
5
5
2
3
4
22
27
32
37
42
47
Data Distribution 10 ?
4) 0
—
6
-
.-
I-
I \
a)
I
Li..
-I
I
2
.-
I
n
———
—
20 30 40 50 Midpoint of Intervat
10
60
70
e.g., The two middle intervals represent 35% of the data. The four middle intervals represent 75% of the data. Therefore, the data is not normally distributed, since the intervals are not dose to 68% and 95% of the data. The graph of the data confirms this.
680
$34, u $32 c) e.g., The mean for the ungrouped data was higher than the mean for the grouped data. The standard deviation for the ungrouped data was much lower than the standard deviation list the grouped data. So, the mean and standard deviation fr the grouped data are not accurate representations of the actual mean and standard deviation for the data.
c)
Interval
1—6
7—12
13—18
19—24
25—30
1
8
5
8
4
4
3.5
9.5
15.5
21.5
27.5
33.5
Frásueflcy
MidPoint
31—36
C.. .
Lesson 5.4, page 122 1.
2.
b) 15.85% c) 0.15% a) 47.5% a)e.g.,Testlandtesr2haveequalmeans,buttest2’ssrandard deviation is almost twice that of test 1. So, the test 2 graph would be flatter than the test 1 graph. 6) e.g., Test 1 and test 3 have different means but the same standard deviation. So, the widths of the graphs are the same. C)
Test
Jaas*Ws Mask
3.
a)
90
2
3
+3.5cr
84
Inteival-
11
a) 0 Li-
JasmlnWs Mask (%)
M 2 s— 1 .5o
1
48 60 Midpoint of Interval
78
16- 20
15
Fuk
4
6
ijiit
13
18.0
21—25
26—30
15
18
23.0
28.0
e.g., The two middle intervals represent 43% of the data. The four middle intervals represent 83% of the data. Therefore, the data is not normally distributed, since the intervals are not dose to 68% and 95% of the data. The graph of the data confirms this.
36—40
31—35 9
4
33.0
38.0 4.
Data Distribution
20
::z::: ::::z::z 1 .
a) b)
tt
a)
l8.5goals,tr’ 11.3goals
Left Bosdaiy
—
16 C
t JL
j
—
—
—
C. 4).’
4o3o
=
Li
Interval (yean)
—25.2
—252—(— 14.4)
—14.4
—14.4—(—3.6)
Midpoint Frequency —19.5
0
—9.0
0
1.8
4
2o
=
—3.6
icr
=
7.2
7.2—18.0
12.6
7
18.0
18.0—28.8
23.4
7
jL =
—3.6—7.2
+
1cr
=
28.8
28.8—39.6
34.2
6
+
2cr
=
39.6
39.6—50.4
45.0
0
50.4
50.4—61.2
55.8
0
.A
..
÷ 3cr (5
—
—
—
10 278
Answers
—
20 30 40 50 Midpoint of Interval
60
(
70 Copyright © 2012 by Nelson Education Ltd.
c)
• Az-score indicates how many standard deviations a data value is above
Reggie’s Basketball Scores
or below the rncat. The a-score for a data value x is z
---fl”
E
C
EEtHEE
0 LI.
—
—————
—
—
,..
cb:b
;
6.
,
11. 12.
Points
e.g., The two middle intervals represent 58% of the data. The four middle intervals represent 100% of the data. Therefore the data is normally distributed, since the intervals are dose to 68% and 95% of the data. The graph of the data confirms this, because it is approximately bell-shaped. Sycars
13. 14. 15. 16.
B. 2. B. 3. D. 4. A. 5. C. A. 7. C. 8. D. 9. B. 10. C. BrandA.jZkm BrandB:jZkm Whichrangeofukwidthsoc.cursmostftequenrly? 6.0—6.5 m Which ranges of trunk widths occur least frequently? 4.5—5.0 m and 7.0—7.5 m ZintcrvalsofwidthTh 16.35% The confidence interval is % ±.j4%,or from Z% to a) What is the mean price and standard deviation for each brand?
a) z 1.9685 c) z —0.4545 1) z 2.32 z —1.6667 Is) 98.98% c) 32.64% a) 97.64% d) 4.80% a) 7.48% Is) 9.02% a) z 1.04 c) z —0.175 Is) z —1.035
82% a) 94.4% Is) 1.3% D. a) accounting: z = 2.884...; chemistxy a = 2.295... 6) accounting: over 99.8%; themistry 98.9% c) accounting
DVDA($) (x-F
tore
Lesson 5.5, page 126 1. 2. 3. 4. 5. 6. 7. 8.
DVDB($)
(r-RF
1
32.94
5.06
34.59
930
2
31.97
1037
38.99
1.82
3
35.92
0.53
36.99
0.42
39.93
22.47
39.99
5.52
140.76
38.43
150.56
17.06
4
35.19
k
,/ 17.
(x—
3.10
—
1.0—2.0
a) 95% 6) 65% ± 2.9%, or 62.1% to 67.9% a)1.1L±0.1L.orl.OLtol.2L
6)
3. 4. 5. 6.
7. 8. 9.
Sample SIZe
MBrglnofError(L)
(cartons)
(mixed up)
c) 20630 to 22 556
MarglflofEivor(L) (correct)
75
±
0.15
±0.20
150
±
0.05
±
0.15
500
±
0.20
±
0.05
6
3.0—4.0
9
4.0—5.0
4
5.0—6.0
1
6.0—7.0
1
7.0—8.0
1
10 9 8
—
—
2.07
Hours Spent Practicing
c6 a, a, u-3
2 1
Chapter 5 Test Prep, page 135
Copyright C 2012 by Nelson Education Ltd.
37.64
2
2.0—3.0
Is)
e.g., As the sample size increases, the margin of error decreases. a) 80% 6) 80% ± 3.5%, or 76.5% to 83.5% c) 1913 to 2088 a) 90%;95% ± 3.7%,or9l.3%to98.7% 6) 7760500to8389500 carA a) 6.65 cm to 6.75 cm Is) .:0.05 cm B. 34% ± 3.2%, or 30.8% to 37.2% a) 95% b) 28% ± 3.2%, or 24.8% to 31.2% c) 8367 520 to 10526880
Qi: Al: Graph the frequency polygons for both data sets on the same graph and visually compare their distributions. A2: Calculate the standard deviation of each data set. A high (or low) standard deviation indicates a more (or j) dispersed set. Q2: • If the data set is well approximated, you can make predictions based on properties of nnnai curves. A normal distribution is symmetric about its mean, which also equals the median and the mode. About f% of the data is within one standard deviation of the mean, and about is within two standard deviations of the mean.
—
7)2
Is) e.g., Brand A is cheaper, but Brand B is more consistently priced. a) llme(h) Frequency:
Lesson 5.6 page 130 1. 2.
[
The sample size is is = 1ll2. The margin of error is i2. * The confidence level, based on the phrase “19 times out of 20”, is %. • The confidence interval for support for the Conservative party is 29.9% to •
Chapter 5 Test, page 136 1.
I 1)
5.
Q3:
=
T 1.5 2.5
i
3.5 4.5 5.5 6.5 7.5 Time(h)
c) 4.0 h
4) 18. 19.
20.
1.5b
e) e.g., The frequency polygon is not bell-shaped, so the data is not normally distributed. a)82% 6) ‘=0.13% e.g., A 2.5% repair rate means that 97.5% of the burners do not need repairs. This represents three standard deviations from the mean, which is 5.5 years. The warranty should be for 5 years. a) 90% 6) 75% ± 3.3% or 7 1.7% to 78.3% c) 717 to 783
Answers
279
6.
7.
8.
9.
10.
e.g., The two middle intervals represent 60% of the data. The four middle intervals represent 86% of the data. Therefore, the data is normally distributed, since the intervals are dose to 68% and 95% of the data. Also, the graph of the data approximates a bell shape. 1) lntasval 20 24 25 29 30 34 35—39 40—44 45—49
standard deviation; cr = 18.3 points 73.6 points standard deviation: h) mean: = 79.3 points 12.7 points e.g., The mean is increased and the standard deviation is decreased. This signifies Jarome Iginla’s point production is more consistent when the two point totals list the lowest seasons are removed. BrandA =88.lkg,o’0.l6kgjBrandB;= 88.lkg,a= 0.11kg e.g., The means are approximately the same, but the standard deviation is slightly iower list B’s bags. So, the masses of B’s bags are more consistent. Mario should order Brand B. However, the standard deviations are low for each supplier, so Mario could order from either supplier. a) EverSmd>”.= 13.3h,o4.6h;EliteEducation:’I3.7h, cr”4.3h b) e.g., The standard deviation fist Elite Education is lower, so their clients are more consistent in their studying times. C. e.g., The mean test scores were approximately the same However, the standard deviation for Class A was lower. This means that the test scores for class A were more consistent and closer to the mean than those for Class B. So, Class A performed better, in this sense. a)_$37,ff’$12
b) kilps($)
Midpolntoflntenial,x
Frequencyf
10-20
3
15
f•j 45
20-30
4
25
100
30—40
6
35
210
40—50
6_
45
270
50—60
1
55
55
20
[
.!quey. oint
1
5
5
2
3
4
22
27
32
37
42
47
Data Distribution 10 ?
4) 0
—
6
-
.-
I-
I \
a)
I
Li..
-I
I
2
.-
I
n
———
—
20 30 40 50 Midpoint of Intervat
10
60
70
e.g., The two middle intervals represent 35% of the data. The four middle intervals represent 75% of the data. Therefore, the data is not normally distributed, since the intervals are not dose to 68% and 95% of the data. The graph of the data confirms this.
680
$34, u $32 c) e.g., The mean for the ungrouped data was higher than the mean for the grouped data. The standard deviation for the ungrouped data was much lower than the standard deviation list the grouped data. So, the mean and standard deviation fr the grouped data are not accurate representations of the actual mean and standard deviation for the data.
c)
Interval
1—6
7—12
13—18
19—24
25—30
1
8
5
8
4
4
3.5
9.5
15.5
21.5
27.5
33.5
Frásueflcy
MidPoint
31—36
C.. .
Lesson 5.4, page 122 1.
2.
b) 15.85% c) 0.15% a) 47.5% a)e.g.,Testlandtesr2haveequalmeans,buttest2’ssrandard deviation is almost twice that of test 1. So, the test 2 graph would be flatter than the test 1 graph. 6) e.g., Test 1 and test 3 have different means but the same standard deviation. So, the widths of the graphs are the same. C)
Test
Jaas*Ws Mask
3.
a)
90
2
3
+3.5cr
84
Inteival-
11
a) 0 Li-
JasmlnWs Mask (%)
M 2 s— 1 .5o
1
48 60 Midpoint of Interval
78
16- 20
15
Fuk
4
6
ijiit
13
18.0
21—25
26—30
15
18
23.0
28.0
e.g., The two middle intervals represent 43% of the data. The four middle intervals represent 83% of the data. Therefore, the data is not normally distributed, since the intervals are not dose to 68% and 95% of the data. The graph of the data confirms this.
36—40
31—35 9
4
33.0
38.0 4.
Data Distribution
20
::z::: ::::z::z 1 .
a) b)
tt
a)
l8.5goals,tr’ 11.3goals
Left Bosdaiy
—
16 C
t JL
j
—
—
—
C. 4).’
4o3o
=
Li
Interval (yean)
—25.2
—252—(— 14.4)
—14.4
—14.4—(—3.6)
Midpoint Frequency —19.5
0
—9.0
0
1.8
4
2o
=
—3.6
icr
=
7.2
7.2—18.0
12.6
7
18.0
18.0—28.8
23.4
7
jL =
—3.6—7.2
+
1cr
=
28.8
28.8—39.6
34.2
6
+
2cr
=
39.6
39.6—50.4
45.0
0
50.4
50.4—61.2
55.8
0
.A
..
÷ 3cr (5
—
—
—
10 278
Answers
—
20 30 40 50 Midpoint of Interval
60
(
70 Copyright © 2012 by Nelson Education Ltd.
c)
• Az-score indicates how many standard deviations a data value is above
Reggie’s Basketball Scores
or below the rncat. The a-score for a data value x is z
---fl”
E
C
EEtHEE
0 LI.
—
—————
—
—
,..
cb:b
;
6.
,
11. 12.
Points
e.g., The two middle intervals represent 58% of the data. The four middle intervals represent 100% of the data. Therefore the data is normally distributed, since the intervals are dose to 68% and 95% of the data. The graph of the data confirms this, because it is approximately bell-shaped. Sycars
13. 14. 15. 16.
B. 2. B. 3. D. 4. A. 5. C. A. 7. C. 8. D. 9. B. 10. C. BrandA.jZkm BrandB:jZkm Whichrangeofukwidthsoc.cursmostftequenrly? 6.0—6.5 m Which ranges of trunk widths occur least frequently? 4.5—5.0 m and 7.0—7.5 m ZintcrvalsofwidthTh 16.35% The confidence interval is % ±.j4%,or from Z% to a) What is the mean price and standard deviation for each brand?
a) z 1.9685 c) z —0.4545 1) z 2.32 z —1.6667 Is) 98.98% c) 32.64% a) 97.64% d) 4.80% a) 7.48% Is) 9.02% a) z 1.04 c) z —0.175 Is) z —1.035
82% a) 94.4% Is) 1.3% D. a) accounting: z = 2.884...; chemistxy a = 2.295... 6) accounting: over 99.8%; themistry 98.9% c) accounting
DVDA($) (x-F
tore
Lesson 5.5, page 126 1. 2. 3. 4. 5. 6. 7. 8.
DVDB($)
(r-RF
1
32.94
5.06
34.59
930
2
31.97
1037
38.99
1.82
3
35.92
0.53
36.99
0.42
39.93
22.47
39.99
5.52
140.76
38.43
150.56
17.06
4
35.19
k
,/ 17.
(x—
3.10
—
1.0—2.0
a) 95% 6) 65% ± 2.9%, or 62.1% to 67.9% a)1.1L±0.1L.orl.OLtol.2L
6)
3. 4. 5. 6.
7. 8. 9.
Sample SIZe
MBrglnofError(L)
(cartons)
(mixed up)
c) 20630 to 22 556
MarglflofEivor(L) (correct)
75
±
0.15
±0.20
150
±
0.05
±
0.15
500
±
0.20
±
0.05
6
3.0—4.0
9
4.0—5.0
4
5.0—6.0
1
6.0—7.0
1
7.0—8.0
1
10 9 8
—
—
2.07
Hours Spent Practicing
c6 a, a, u-3
2 1
Chapter 5 Test Prep, page 135
Copyright C 2012 by Nelson Education Ltd.
37.64
2
2.0—3.0
Is)
e.g., As the sample size increases, the margin of error decreases. a) 80% 6) 80% ± 3.5%, or 76.5% to 83.5% c) 1913 to 2088 a) 90%;95% ± 3.7%,or9l.3%to98.7% 6) 7760500to8389500 carA a) 6.65 cm to 6.75 cm Is) .:0.05 cm B. 34% ± 3.2%, or 30.8% to 37.2% a) 95% b) 28% ± 3.2%, or 24.8% to 31.2% c) 8367 520 to 10526880
Qi: Al: Graph the frequency polygons for both data sets on the same graph and visually compare their distributions. A2: Calculate the standard deviation of each data set. A high (or low) standard deviation indicates a more (or j) dispersed set. Q2: • If the data set is well approximated, you can make predictions based on properties of nnnai curves. A normal distribution is symmetric about its mean, which also equals the median and the mode. About f% of the data is within one standard deviation of the mean, and about is within two standard deviations of the mean.
—
7)2
Is) e.g., Brand A is cheaper, but Brand B is more consistently priced. a) llme(h) Frequency:
Lesson 5.6 page 130 1. 2.
[
The sample size is is = 1ll2. The margin of error is i2. * The confidence level, based on the phrase “19 times out of 20”, is %. • The confidence interval for support for the Conservative party is 29.9% to •
Chapter 5 Test, page 136 1.
I 1)
5.
Q3:
=
T 1.5 2.5
i
3.5 4.5 5.5 6.5 7.5 Time(h)
c) 4.0 h
4) 18. 19.
20.
1.5b
e) e.g., The frequency polygon is not bell-shaped, so the data is not normally distributed. a)82% 6) ‘=0.13% e.g., A 2.5% repair rate means that 97.5% of the burners do not need repairs. This represents three standard deviations from the mean, which is 5.5 years. The warranty should be for 5 years. a) 90% 6) 75% ± 3.3% or 7 1.7% to 78.3% c) 717 to 783
Answers
279
