Pg 101
5a) Range = 94 – 41 = 53
b) Let’s say we want 10 intervals. 53/10 = 5.3
That give us an interval size of 5 and we would need 11 intervals
c)
Score
39.5 - 44.5
44.5 - 49.5
49.5 - 54.5
54.5 - 59.5
59.5 - 64.5
64.5 - 69.5
69.5 - 74.5
74.5 - 79.5
79.5 - 84.5
84.5 - 89.5
89.5 - 94.5
e)
d)
Tally
III
I
II
IIII
II
II
II
IIII
II
I
III
Freq.
3
1
2
4
2
2
2
5
2
1
3
f)
g) see textbook
#8
a)Grouped. So many different values. Grouped for an easier understanding
b)
Value
5 - 14.99
15 - 24.99
25 - 34.99
35 - 44.99
45 - 54.99
55 - 64.99
65 - 74.99
75 - 84.99
85 - 94.99
95 - 104.99
105 114.99
Tally
II
III
IIII
IIII
IIII
III
III
II
II
Freq
2
3
5
4
4
3
3
2
2
0
II
2
c)
d) See textbook
#9
a)
Speed
61 - 65
66 - 70
71 - 75
76 - 80
81 - 85
86 - 90
91 - 95
96 - 100
Tally
IIII
IIII III
IIIIIII
II
I
I
I
Freq.
0
4
8
7
2
1
1
1
b)
c)
d) 12 going over 60 but 75 and under
e) 5 going 80 or above
11a) The relative frequencies all add up to 1 (100%)
b) You are finding the percents of each category. If you add all the categories, they add up to 100%
12.
Score
35
45
55
65
75
85
95
Rel.
Freq
0.04
0.08
0.16
0.32
0.2
0.12
0.08
Frequence = Rel. Freq x 25
1
2
4
8
5
3
2
15.
a) 64.5 has the most scores. It has the largest increase.
b) looks like it goes from 8 – 17. Therefore 9 scores in the category.
Pg 109 # 1 – 5
1a January 2000 looks to have a value of about 111.5. Since Jan, 1992 has a base of 100, 11.5 would be the increase.
b) 11.5/100*100 = 11.5%
c) April 1998 has a value of about 108.5. Index in 1992 is 100. % increase = 8.5/100*100 = 8.5% increase.
Cost of item = 7.50 x 1.085 = 8.14
d) August, 1997 has a value of 108 May 2000 is 113.5. % increase = 5.5/108*100 = 5.1% increase.
Cost of item = 55 *1.051 = 57.80
2a) There is variety to get the best representation of the actual increase.
b). It’s an average price that’s also weighted. Different items change more than others. Also some are purchase more
often.
3) near the end of 1995. Not quite 1996. (when it reached 5000)
b) 1971 is 1000. 1997 is about 6000. Growth rate = (6000 – 1000)/1000 = 5 or 500%
c) 1999 – 2001 grew the most. It has the steepest line
d) Not a straight line. Most likely an exponential curve.
4) Inflation means things cost more because money is “worth” less. Items could be bought before for $1.00, but now
cost $3.00.
b) They show possible rate increases for items over a given time.
c) Both might show increases. Though houses might increase more.
d) All items tend to increase as everyone wants to be paid more for their items. Be it a house or a TV or even a wage at
a job.
5) See textbook
2.3 Pg 117 #1 – 5, 7 - 9
See textbook
2.4 Pg 123 # 1 – 4, 6 & article?
See textbook
2.5 pg 133 # 1b, 3, 5, 7, 9, 11
1b 𝑥̅ = (10+15+14+19+18+17+12+10+9+14+11+18)/16 = 14.6
Med = 9, 10, 10, 11, 12, 14, 14, 14, 15, 15, 17, 18, 18, 18, 19, 20
= (14 + 15) /2 = 14.5
Mode = 14 and 18 (both three times)
3. 70% of 87 = .7x87 = 60.9
30% of 71 = .3x71 = 21.3
Final mark = 60.9 +21.3 = 82.2%
5. Nadia [(4x2)+(4x2)+4]/5 = 4
Enzo [(5x2)+(4x2)+3]/5 = 4.2
Stephan = [(5x2)+(3x2)+4]/5 = 4
Enzo should get the job. Highest rating
7. a) .35x82+.25x71+.15x85+.25x75 = 28.7+17.75+12.75+18.75 = 77.95%
b) .7x77.95 + .3 x 65 = 54.565 + 19.5 = 74% (rounded)
9. 𝑥̅ = 224/28 = 8
Med = 7.5
Mode = 7 (there are nine of them)
b) The mode. Tells him what size he has the most of.
c) Size 10 is also important because there are 8 of them.
d) Use each shoe size as a category since there are not a lot
of different sizes. There are two tall columns because there
are a lot of size 7 and size 10 shoes.
11.
Salary Range
($000)
Cum. Freq.
Midpoint
Frequency
fixi
12
20 - 30
25
12
30 - 40
35
24
300
36
840
68
40 - 50
45
32
50 - 60
55
19
1440
87
1045
96
60 - 70
65
9
70 - 80
75
3
585
99
225
99
80 - 90
85
0
0
90 - 100
95
1
95
∑𝑖 𝑓𝑖 = 100
∑𝑖 𝑓𝑖 𝑥𝑖 = 4530
100
a) 𝑥̅ = 4530/100 = 45.3 The average salary is $45300
b) the 50th employee would be the median. This falls into the 40 – 50 range. Therefore the median salary is $45000.
c) If you count the last range as an outlier the table changes as follows.
Salary Range
($000)
20 - 30
30 - 40
40 - 50
50 - 60
60 - 70
70 - 80
Midpoint
25
35
45
55
65
75
Frequency
12
24
32
19
9
3
∑𝑖 𝑓𝑖 = 99
Cum. Freq.
12
36
68
87
96
99
fx
300
840
1440
1045
585
225
∑𝑖 𝑓𝑖 𝑥𝑖 = 4435
𝑥̅ = 4435/99 = 44.8 The average salary is now $44800. About $500 less. The median salary does not change.
Ch 2.6 pg 148 #1 – 7, 9, 10, 15
1a
Data
x-u
5
7
9
6
5
10
8
2
11
8
7
7
6
9
5
8
u= 7.1
-2.1
-0.1
1.9
-1.1
-2.1
2.9
0.9
-5.1
3.9
0.9
-0.1
-0.1
-1.1
1.9
-2.1
0.9
(x - u)2
4.41
0.01
3.61
1.21
4.41
8.41
0.81
26.01
15.21
0.81
0.01
0.01
1.21
3.61
4.41
0.81
74.96
Var. 4.997333
S.D. 2.235472
1b.
data
12.55
15.31
21.98
45.35
19.81
33.89
29.53
30.19
38.2
u = 27.42
x-u
-14.87
-12.11
-5.44
17.93
-7.61
6.47
2.11
2.77
10.78
(x - u)2
221.1169
146.6521
29.5936
321.4849
57.9121
41.8609
4.4521
7.6729
116.2084
946.9539
Var.
S.D.
118.3692
10.87976
2a
3
4
5
6
6
8
9
11
15
Q1 = 4.5
Median has been used, therefore ignore
median
Q3 = 10
Median has been used, therefore ignore
Interquartile range = 10 - 4.5
= 5.5
semi-interquartie range = 5.5/2
=2.25
2b.
43
48
56
59
62
64
67
71
72
75
75
78
81
84
88
90
Q1 = (59+62)/2 = 60.5
Median wasn't used (we took the average)
Therefore use the first 8 numbers
median = (71+72)/2 = 71.5
Q3 = (78 + 81)/2 = 79.5
Median wasn't used (we took the average)
Therefore use the last 8 numbers
Interquartile range = 79.5 - 60.5
= 19
semi-quartile range = 19.5/2
= 9.5
3. a) in the first quartile
b) 81 is in the second quartile
c) 105 is in the third quartile
4. a) Q1 = 25th percentile
b) Median = 50th percentile
c) Q3 = 75th percentile
5.
Data
x u18
15
26
20
21
(x - u)2
-2
-5
6
0
1
4
25
36
0
1
66
Data
var.
SD.
16.50
4.06
x-u
18
15
26
20
21
-2
-5
6
0
1
Z-Score
-0.49
-1.23
1.48
0.00
0.25
u = 20
6.
Data
18
35
42
44
51
52
63
68
72
75
80
96
110
125
260
x-u
(x - u)2
-61.4 3769.96
-44.4 1971.36
-37.4 1398.76
-35.4 1253.16
-28.4
806.56
-27.4
750.76
-16.4
268.96
-11.4
129.96
-7.4
54.76
-4.4
19.36
0.6
0.36
16.6
275.56
30.6
936.36
45.6 2079.36
180.6 32616.36
u = 79.4
46331.6
Data
Var.
S.D.
3309.4
57.52738
18
35
42
44 Q1
51
52
63
68 Median
72
75
80
96 Q3
110
125
260
Interquartile range = 96 - 44
= 52
Semi-interquartile range = 52/2
= 26
7. u = 35.8, var = 35, S.D. = 5.9, Median = 36.5, Inter = 3, semi = 1.5
9.
S.D = 38
Var = 1445.56
Inter = 1.5, Semi = 0.75
10.
Chi-Yan seems to have better control over her drives (a lower standard deviation in their distances). If their putting abilities are essentially equal, then
Chi-Yan is more likely to have a better score in a round of golf.
15.
a) One class may have a tight grouping of marks, while the other class has a larger distribution of marks.
b) The larger standard deviation means the data is more spread out. This should make a larger interquartile range as
the quartiles will also me be more spread out.