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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
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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 \(median=\frac{1+2}{2}\). Answer: C. 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



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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



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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 01 accidents and 30 drivers with 26 accidents so you know that your median will fall exactly between 1 and 2. Hence the answer of 1.5.



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09 Apr 2017, 08:50
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Re: The table shows the results of a poll which asked drivers [#permalink]
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18 Apr 2017, 16:12
anaik100 wrote: Attachment: p0069.gif 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 To determine where the median falls in a data set, we can use the following expression in which n = the total number of data points in the set (here, n = 60 since 60 drivers were surveyed): median position = (n + 1)/2 (60 + 1)/2 = 61/2 = 30.5 Thus, the median is halfway between the 30th and 31st data points, which is the same as the average of the 30th data point and the 31st data point when the data are sorted from least to greatest. Looking at the chart, we see that the 30th data point is 1 accident and the 31st data point is 2 accidents. Thus, the median is (1+2)/2 = 1.5. Answer: C
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