Problem 1.7
(a)
Number
Nonconforming
Frequency
RelativeFrequency(Freq/60)
0
7
0.117
1
12
0.200
2
13
0.217
3
14
0.233
4
6
0.100
5
3
0.050
6
3
0.050
7
1
0.017
8
1
0.017
doesn't add exactly to 1 because relative frequencies have been rounded
→
1.001
Problem 1.11
(b) A histogram of this data, using classes of width 1000 separated at 0, 1000, 2000, and
6000 is shown below.
The proportion of subdivisions with total length less than 2000
is (12+11)/47 = .489, or 48.9%.
Between 2000 and 4000, the proportion is (10 +
7)/47 = .362, or 36.21%.
6000
5000
4000
3000
2000
1000
0
10
5
0
length
Frequency
Problem 1.13
(a)
The frequency distribution is:
Relative
Relative
Class
Frequency
Class
Frequency
0< 150
.193
900<1050
.019
150< 300
.183
1050<1200
.029
300< 450
.251
1200<1350
.005
450< 600
.148
1350<1500
.004
600< 750
.097
1500<1650
.001
750< 900
.066
1650<1800
.002
1800<1950
.002
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The relative frequency distribution is almost unimodal and exhibits a large positive
skew.
The typical middle value is somewhere between 400 and 450, although the
skewness makes it difficult to pinpoint more exactly than this.
(b) The proportion of the fire loads less than 600 is .193+.183+.251+.148 = .775.
The
proportion of loads that are at least 1200 is .005+.004+.001+.002+.002 = .014.
Problem 1.19
(a)
The density curve forms a rectangle over the interval [4, 6].
For this reason,
uniform densities are also called
rectangular densities
by some authors.
Areas under
uniform densities are easy to find (i.e., no calculus is needed) since they are just areas
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 Spring '08
 marzban
 Frequency, Frequency distribution, 1%, Histogram, 1 1 k

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