A car insurance company conducted a survey to find out how many car accidents people had been involved in. They selected a sample of 32 adults between the ages of 30 and 70 and asked each person how many accidents they had been involved in in the past ten years. The following data were obtained. Construct a frequency and relative frequency distribution using classes based on a single value. 01 03 2102 1 1 1 0 2041 2001 0 2 1 3 1 3001054 OA. O C. Number Frequency Relative Frequency X f f/n 0 1 2 3 4 5 Number X 0 1 2 3 4 5 11 10 6 3 1 1 Frequency f 11 10 5 3 2 1 11/32 34% 10/32 31% 6/32 19% 3/32 9% 1/32 3% 1/32 3% Relative Frequency f/n 11/100 = 11% 10/100 = 10% 5/100 = 5% 3/100 = 3% 2/100 = 2% 1/100=1% *** OB. O D. Number Frequency Relative Frequency f f/n X 0 1 2 3 4 5 Number X 0 1 2 3 4 5 11 10 5 3 2 1 Frequency f 10 11 5 3 2 1 11/32 34% 10/32 31% 5/32 16% 3/32 9% 2/32 6% 1/32 3% Relative Frequency f/n 10/32 31% 11/32 34% 5/32 16% 3/32 9% 2/32 6% 1/32 3%
A car insurance company conducted a survey to find out how many car accidents people had been involved in. They selected a sample of 32 adults between the ages of 30 and 70 and asked each person how many accidents they had been involved in in the past ten years. The following data were obtained. Construct a frequency and relative frequency distribution using classes based on a single value. 01 03 2102 1 1 1 0 2041 2001 0 2 1 3 1 3001054 OA. O C. Number Frequency Relative Frequency X f f/n 0 1 2 3 4 5 Number X 0 1 2 3 4 5 11 10 6 3 1 1 Frequency f 11 10 5 3 2 1 11/32 34% 10/32 31% 6/32 19% 3/32 9% 1/32 3% 1/32 3% Relative Frequency f/n 11/100 = 11% 10/100 = 10% 5/100 = 5% 3/100 = 3% 2/100 = 2% 1/100=1% *** OB. O D. Number Frequency Relative Frequency f f/n X 0 1 2 3 4 5 Number X 0 1 2 3 4 5 11 10 5 3 2 1 Frequency f 10 11 5 3 2 1 11/32 34% 10/32 31% 5/32 16% 3/32 9% 2/32 6% 1/32 3% Relative Frequency f/n 10/32 31% 11/32 34% 5/32 16% 3/32 9% 2/32 6% 1/32 3%
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
Related questions
Question
![A car insurance company conducted survey to find out how many car accidents people had been involved in. They selected a sample of 32 adults between the ages of 30 and 70 and asked each person how
many accidents they had been involved in in the past ten years. The following data were obtained. Construct a frequency and relative frequency distribution using classes based on a single value.
01 0 3
2 1 0 2
1 1 1 0 2 0 4 1
20 0 1 0 2
1 3
1 3 0 0
1 0
5 4
O A.
C.
Number
X
1
2
3
X
5
Frequency
f
Number Frequency
f
1
2
3
4
5
11
10
6
3
1
1
11
10
5
3
2
1
Relative Frequency
f/n
11/32 34%
10/32 ~ 31%
6/32≈ 19%
3/32~9%
1/32 ~ 3%
1/32≈ 3%
Relative Frequency
f/n
11/100 = 11%
10/100 = 10%
5/100 = 5%
3/100 = 3%
2/100 = 2%
1/100 = 1%
B.
D.
Number
X
0
1
2
3
4
5
Number
X
1
2
3
4
5
Frequency
f
11
10
5
3
2
1
Frequency
f
10
11
LO
3
2
1
Relative Frequency
f/n
11/32 34%
10/32 31%
5/32≈ 16%
3/32~ 9%
2/32~6%
1/32≈ 3%
Relative Frequency
f/n
10/32 31%
11/32≈ 34%
5/32≈ 16%
3/32 ~ 9%
2/32 ≈ 6%
1/32 ~ 3%](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0a9267cf-8e34-41e0-92ec-06b114eb620d%2F0ebeb5c6-0b4d-4646-8728-7dea5f3c52ba%2Fz9a4kbs_processed.png&w=3840&q=75)
Transcribed Image Text:A car insurance company conducted survey to find out how many car accidents people had been involved in. They selected a sample of 32 adults between the ages of 30 and 70 and asked each person how
many accidents they had been involved in in the past ten years. The following data were obtained. Construct a frequency and relative frequency distribution using classes based on a single value.
01 0 3
2 1 0 2
1 1 1 0 2 0 4 1
20 0 1 0 2
1 3
1 3 0 0
1 0
5 4
O A.
C.
Number
X
1
2
3
X
5
Frequency
f
Number Frequency
f
1
2
3
4
5
11
10
6
3
1
1
11
10
5
3
2
1
Relative Frequency
f/n
11/32 34%
10/32 ~ 31%
6/32≈ 19%
3/32~9%
1/32 ~ 3%
1/32≈ 3%
Relative Frequency
f/n
11/100 = 11%
10/100 = 10%
5/100 = 5%
3/100 = 3%
2/100 = 2%
1/100 = 1%
B.
D.
Number
X
0
1
2
3
4
5
Number
X
1
2
3
4
5
Frequency
f
11
10
5
3
2
1
Frequency
f
10
11
LO
3
2
1
Relative Frequency
f/n
11/32 34%
10/32 31%
5/32≈ 16%
3/32~ 9%
2/32~6%
1/32≈ 3%
Relative Frequency
f/n
10/32 31%
11/32≈ 34%
5/32≈ 16%
3/32 ~ 9%
2/32 ≈ 6%
1/32 ~ 3%
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