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# The table shows the results of a poll which asked drivers

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The table shows the results of a poll which asked drivers [#permalink]

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23 Jul 2010, 05:52
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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 driver?

A. 0.5
B. 1
C. 1.5
D. 2
E. 4
[Reveal] Spoiler: OA

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Re: median number of accidents per driver? [#permalink]

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23 Jul 2010, 08:07
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i actually messed up the question the first time cause i didn't add correctly. in either case, i got C also.

in order to find the median we must put the values in order from lowest to highest values.

6 5 4 3 2 1 0 = number of accidents
-----------------------------
1 2 2 4 21 13 17 = values representing people

we can now find the median by finding the middle number. there is a total of
1 + 2 + 2 + 4 + 21 + 13 + 17 = 60 people.
middle value will be represented as average in 30th and 31st person.

so 30th person had 2 accidents while 31st person had one. median will thusly be:

2 + 1 / 2 = 1.5
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Re: median number of accidents per driver? [#permalink]

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27 Sep 2010, 08:10
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anaik100 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 driver?

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

If a set has odd number of terms the median of a set is the middle number when arranged in ascending or descending order;
If a set has even number of terms the median of a set is the average of the two middle terms when arranged in ascending or descending order.

There are total of $$17+13+21+4+2+2+1=60$$ drivers (60 terms) so the median number of accidents per driver would be the average of accidents of 30th and 31st drivers (as we have even # of terms).

30th term equals to 1 and 31st term equals to 2 so $$median=\frac{1+2}{2}$$.

To elaborate more, you can imagine these data points as:
0, ..., 0, 1, ..., 1, 2, ..., 2, 3, ..., 3, 4, ..., 4, 5, ..., 5, 6, ..., 6 --> 17 zeros, 13 ones, 21 twos and so on, total of 60 data points. Median would be the average of 30th and 31st terms: $$median=\frac{1+2}{2}$$.

Hope it's clear.
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Re: median number of accidents per driver? [#permalink]

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27 Sep 2010, 08:54
can we just estimate and don't do the calculation, the frequencies at 1 accident is 13 and 2 accidents is 21 (most concentration of all) , then the average will be arround 1 to 2 then i chose 1.5
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Re: median number of accidents per driver? [#permalink]

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27 Sep 2010, 09:28
Mikko wrote:
can we just estimate and don't do the calculation, the frequencies at 1 accident is 13 and 2 accidents is 21 (most concentration of all) , then the average will be arround 1 to 2 then i chose 1.5

You should be careful with such approximations. For example if it were: 1-14 and 2-20 (instead of 1-13 and 2-21) the answer would be B(1) and not C(1.5).
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Re: The table shows the results of a poll which asked drivers [#permalink]

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18 Jun 2013, 01:41
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: The table shows the results of a poll which asked drivers [#permalink]

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20 Jul 2014, 09:46
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Re: The table shows the results of a poll which asked drivers [#permalink]

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06 Aug 2015, 12:31
A quicker way to do this is to note that there are 30 drivers with 0-1 accidents and 30 drivers with 2-6 accidents so you know that your median will fall exactly between 1 and 2. Hence the answer of 1.5.
Re: The table shows the results of a poll which asked drivers   [#permalink] 06 Aug 2015, 12:31
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