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MEDIAN

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by SwatiDenre » Thu Jun 16, 2011 8:40 am
The table shows the results of a poll which asked drivers how many accidents they had had over the previous 5 years. What is the median number of accidents per year?

A. 0.5
B. 1
C. 1.5
D. 2
E. 4

Correct Answer: C



Solution ?
Thanks in advance.
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by Frankenstein » Thu Jun 16, 2011 8:44 am
Hi,
Where is the table, dude...
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by baladon99 » Thu Jun 16, 2011 10:06 am
SwatiDenre wrote:The table shows the results of a poll which asked drivers how many accidents they had had over the previous 5 years. What is the median number of accidents per year?

A. 0.5
B. 1
C. 1.5
D. 2
E. 4

Correct Answer: C



Solution ?
Thanks in advance.
Can you please post the table ?
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by amit2k9 » Thu Jun 16, 2011 9:32 pm
since the number of years is 5, the median will be the number of accidents in the 3rd year.
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by edvhou812 » Thu Jun 16, 2011 10:03 pm
amit2k9 wrote:since the number of years is 5, the median will be the number of accidents in the 3rd year.
No. You will need to reorder the amounts from least to greatest, and then take the third largest number.
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by SwatiDenre » Sun Jun 19, 2011 3:51 pm
Hi,
very sorry for the mistake.
here goes the table -


No of accidents - No. of drivers

0 - 17
1 - 13
2 - 21
3 - 4
4 - 2
5 - 2
6 - 1
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by MBA.Aspirant » Sun Jun 19, 2011 4:10 pm
The middle of this set will = the number of drivers/2 = 43/2 = 21.5

This number is between two accidents values 1 and 2

so we take the average of them = 1+2/2 = 1.5
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by Frankenstein » Sun Jun 19, 2011 8:48 pm
SwatiDenre wrote:Hi,
very sorry for the mistake.
here goes the table -


No of accidents - No. of drivers

0 (1-17) - 17
1 (18-30) - 13
2 (31-51) - 21
3 (52-55) - 4
4 (56-57) - 2
5 (58-59) - 2
6 (60) - 1
Hi,
Total number of drivers is 17+13+21+... = 60
Median is the average of 30th and 31st elements
30th element is 1 and 31st element is 2
So, median is (1+2)/2 = 1.5

Hence, C
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