Download Statistics “Pre-covery Packet”

Statistics “Pre-covery Packet”
Based on the given cumulative frequencies, find the median class and estimate the median .
Class
Frequency
CumRelFr
0.14
100-120
8
0.09
317
0.33
120-140
12
0.22
110-160
152
0.43
140-160
15
0.39
160-210
160
0.53
160-180
11
0.52
210-260
225
0.67
180-200
17
0.71
260-310
400
0.92
200-220
21
0.94
310-360
128
1.0
220-240
5
1.0
Class
Frequency
CumRelFr
20-50
217
0.14
50-80
317
0.33
80-110
152
0.43
110-140
160
0.53
140-170
225
0.67
170-200
400
0.92
200-230
128
1.0
Class
Frequency
CumRelFr
10-60
217
60-110
7.
a)
med  160 .5  .43

210  160 .53  .43
195  med
b)
Class
50-75
75-100
100-125
c)
Frequency
18
22
5
CumRelFr
0.12
0.27
0.30
125-150
31
0.51
150-175
27
0.69
175-200
200-225
31
15
med  160 .5  .39

180  160 .52  .39
177  med
med  125 .50  .30

150  125 .51  .30
149  med
0.90
1.0
med  110 .5  .43

140  110 .53  .43
131  med
Statistics “Pre-covery Packet”
Remember, you have to put in the “class marks” for x in L1.
Find the mean and standard deviation of each of the following population frequency distributions, to the nearest tenth.
Class
8.
Frequency
a)
Class
Frequency
50-75
18
510-520
30
520-530
25
75-100
22
530-540
50
100-125
5
540-550
40
125-150
31
550-560
15
150-175
27
560-570
25
175-200
31
570-580
10
200-225
15
Mean = 540.1
Standard Deviation = 17.2
b)
Class
Frequency
100-200
Mean = 142.7
Standard Deviation = 47.4
Class
Frequency
17
100-120
8
200-300
37
120-140
12
300-400
52
140-160
15
400-500
60
160-180
11
500-600
25
180-200
17
600-700
40
200-220
21
700-800
28
220-240
5
Mean = 454.6
Standard Deviation = 174.6
c)
Mean = 172.5
Standard Deviation = 35.5
Statistics “Pre-covery Packet”
0.15%
34
%
34
%
2.35%
2.35%
13.5%
32.6
27.9
Normally distributed
data with a mean of
112 and a standard
deviation of 19.3.
34
%
46.7
42
51.4
56.1
13.5%
73.4
54.1
34
%
34
%
2.35%
2.35%
13.5%
43.8
55.4
0.15%
0.15%
32.2
Find the interval of the
data that falls within
two standard
deviations of the
mean.
13.5%
67
78.6
43.8 < x < 90.2
90.2
101.8
c)
92.7
131.3
112
150.6
169.9
Sketch as a normal distribution, then use the Empirical Rule to
answer the question.
Normally distributed
data with a mean of
205 and a standard
deviation of 12.5.
0.15%
Sketch as a normal distribution, then use the Empirical Rule to
answer the question.
Normally distributed
data with a mean of 67
and a standard
deviation of 11.6.
2.35%
13.5%
92.7 < x < 131.3
37.3< x < 46.7
b)
34
%
2.35%
13.5%
37.3
Find the interval of the
data that falls within
one standard
deviation of the mean.
0.15%
Find the interval of the
data that falls within
one standard
deviation of the mean.
Sketch as a normal distribution, then use the Empirical Rule to
answer the question.
Find the interval of the
data that falls within
three standard
deviations of the
mean.
34
%
34
%
2.35%
167.5
13.5%
180
2.35%
0.15%
Normally distributed
data with a mean of
42 and a standard
deviation of 4.7.
a)
0.15%
Sketch as a normal distribution, then use the Empirical Rule to
answer the question.
0.15%
9.
13.5%
192.5
205
217.5
230
167.5 < x < 242.5
242.5
Statistics “Pre-covery Packet”
10. Sketch as a normal distribution, then use the Empirical Rule to
answer the question.
2.35%
13.5%
32.6
27.9
34
%
2.35%
46.7
42
51.4
56.1
73.4
54.1
13.5%
92.7
32.2
55.4
2.35%
13.5%
67
169.9
78.6
81.5%
90.2
Sketch as a normal distribution, then use the Empirical Rule to
answer the question.
Normally distributed
data with a mean of
205 and a standard
deviation of 12.5.
101.8
0.15%
34
%
13.5%
43.8
c)
Find the percent of
data that falls
between 55.4 and
90.2.
0.15%
0.15%
Normally distributed
data with a mean of 67
and a standard
deviation of 11.6.
2.35%
150.6
97.5%
Sketch as a normal distribution, then use the Empirical Rule to
answer the question.
34
%
131.3
112
0.3%
b)
2.35%
13.5%
13.5%
37.3
34
%
Find the percent of
data that falls
between 180 and 230.
34
%
34
%
2.35%
167.5
13.5%
180
192.5
2.35%
0.15%
2.35%
Find the percent of
data that falls above
73.4 .
0.15%
34
%
0.15%
0.15%
34
%
Sketch as a normal distribution, then use the Empirical Rule to
answer the question.
Normally distributed
data with a mean of
112 and a standard
deviation of 19.3.
Find the percent of
data that falls below
27.9 and above 56.1.
0.15%
Normally distributed
data with a mean of
42 and a standard
deviation of 4.7.
a)
13.5%
205
95%
217.5
230
242.5
Statistics “Pre-covery Packet”
34
%
2.35%
13.5%
51.4
60.6
79
88.2
97.4
106.6
25.5
39
52.5
137
13.5%
161
185
24
93
106.5
209
233
257
Find the standard
deviation.
34
%
0.15%
2.35%
13.5%
113
79.5
Normally distributed
data with a mean of
85 and that 2.5% of
the data falls above
109.
0.15%
0.15%
34
%
2.35%
66
c) Sketch as a normal distribution, then use the Empirical Rule to
answer the question.
Find the standard
deviation.
34
%
2.35%
13.5%
13.5
Sketch as a normal distribution, then use the Empirical Rule to
answer the question.
Normally distributed
data with a mean of
185 and that 68% of
the data falls between
161 and 209.
34
%
13.5%
9.2
b)
34
%
2.35%
13.5%
69.8
Find the standard
deviation.
34
%
2.35%
2.35%
13.5%
49
61
73
13.5%
85
12
97
109
121
0.15%
2.35%
0.15%
34
%
Sketch as a normal distribution, then use the Empirical Rule to
answer the question.
Normally distributed
data with a mean of 66
and that 99.7% of the
data falls between 25.5
and 106.5.
Find the standard
deviation.
0.15%
0.15%
Normally distributed
data with a mean of
79 and that 95% of
the data falls between
60.6 and 97.4.
a)
0.15%
11. Sketch as a normal distribution, then use the Empirical Rule to
answer the question.
Statistics “Pre-covery Packet”
Input the data into your calculator (Stat, Edit, Calc, 1-Var Stats) and find the 5-number summary.
1.
a)
63
65
69
64
66
62
25
20
18
26
42
34
66
64
67
70
75
73
38
56
29
30
30
40
72
74
77
79
78
71
50
28
21
38
19
20
76
81
89
87
86
84
18
18
83
85
88
82
88
87
87
95
93
99
18, 20, 28.5, 38, 56
62, 69, 77.5, 87, 99
b)
7
8
9
9
9
10
5
7
5
6
10
7
8
2
8
7
5
3
6
c)
60
61
62
66
59
68
9
60
66
72
72
63
68
9
9
70
71
60
75
66
65
8
10
68
66
10
2, 6, 8, 9, 10
59, 61.5, 66, 69, 75
Statistics “Pre-covery Packet”
All data goes in L1.
Find the requested statistics for the given sample data. Round answers to nearest tenth. All
parts of a problem MUST be correct in order to get credit for a problem.
2.
a)
201
208
120
143
Mean
Mean==145.4
Mean==174.2
Mean
158
144
133
117
Std.
Std.Dev.
Dev.==25.7
Std.Dev.
Dev.==27.2
Std.
192
195
169
171
Range
==208 – 115 = 93
167
199
Range
=
Range
Range
=
186
–
117
=
69
123
126
IQR
IQR==170 - 123.5 = 46.5
IQR== 197 – 160.5 = 36.5
IQR
163
180
183
186
168
115
150
124
b)
c)
167
182
122
156
188
199
135
173
174
177
133
137
Mean =
Mean = 161.9
Std. Dev. =
Std. Dev. = 24.8
Range =
Range = 199 – 122 = 77
IQR =
IQR = 179.5 – 136 = 43.5
169
171
188
199
192
195
123
126
135
173
167
199
Mean =
Std. Dev. =
Mean = 169.8
Std. Dev. = 27.8
Range =
IQR =
Range = 199 – 123 = 76
IQR = 193.5 – 151 = 42.5
Statistics “Pre-covery Packet”
Identify the data set as “unimodal” (one mode), “bimodal” (two modes), or “multi-modal”
(more than two modes). Then identify the mode or modes. Both parts must be correct.
a)
3.
230
241
14
17
There are 3 numbers that
257
280
Unimodal
occur twice. Therefore,
15
21
this data set is
280
256
25
18
“multi-modal”.
271
219
Mode = 280
17
19
233
267
13
23
The modes are:
211
247
14
15
14, 15, & 17.
b)
c)
174
182
69
71
18
19
19
19
122
156
23
12
13
13
16
19
188
199
133
173
174
177
133
137
Bimodal
The modes are:
133 & 174
Unimodal
Mode = 19
Statistics “Pre-covery Packet”
All data goes in L1.
Complete the frequency table for the given data. Be careful… all frequencies MUST be correct in
order to get credit for the problem.
a)
4. 120
Class
Frequency
Class
Frequency
143
201
208
Class
Frequency
Class
Frequency
110-125
4
100-130
1
133
117
100-130
158
144
110-125
125-140
2
169
171
125-140
140-155
2
192
195
123
126
140-155
155-170
1
155-170
170-185
2
167
199
170-185
185-200
1
163
180
168
115
183
186
150
124
185-200
b)
130-160
130-160
160-190
160-190
190-220
190-220
2
4
5
c)
167
182
122
156
188
199
135
173
174
177
133
137
Class
Class
120-140
120-140
140-160
140-160
160-180
160-180
180-200
180-200
Frequency
Frequency
4
169
171
18!
199
192
195
1
123
126
135
173
167
199
4
3
Class
Class
Frequency
Frequency
110-130
110-130
2
130-150
130-150
1
150-170
150-170
2
170-190
170-190
3
190-210
190-210
4
Statistics “Pre-covery Packet”
Based on the given frequency table, find the relative frequency of the requested class. Round
final answers to nearest hundredth.
a)
5.
Class
Frequency
0-50
32
50-100
57
100-150
68
150-200
14
200-250
81
250-300
17
Total = 269
Relative
Frequency
for the
150-200
class is:
Class
Frequency
90-140
417
140-190
231
190-240
117
240-290
432
14
 0.05
269
b)
Relative
Frequency for the
140-190 class is:
231
 0.19
1197
c)
Class
Frequency
25-50
112
50-75
142
75-100
157
100-125
93
Relative
Frequency for
the 100-125
class is:
93
 0.18
504
Class
Frequency
0-500
1250
500-1000
2325
1000-1500
1875
1500-2000
1435
2000-2500
2145
Relative
Frequency for
the 500-1000
class is:
2325
 0.26
9030
Statistics “Pre-covery Packet”
Find the cumulative relative frequency for the specified class., to the nearest hundredth.
Class
6.
b)
Frequency
510-520
30
520-530
25
530-540
50
540-550
40
550-560
15
560-570
25
570-580
10
550-560 class
30
30+25=55
55+50=105
105+40=145
145+15=160
a)
160
 .82
195
Class
Frequency
50-75
18
75-100
22
100-125
5
125-150
31
150-175
27
175-200
31
200-225
15
Class
Frequency
Class
Frequency
100-200
17
100-120
8
200-300
37
120-140
12
300-400
52
140-160
15
400-500
60
160-180
11
500-600
25
180-200
17
600-700
40
200-220
21
700-800
28
220-240
5
300-400 class
52
 0.20
259
c)
150-175 class
103
 0.69
149
180-200 class
63
 0.71
89