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coldweather999
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bkk145
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coldweather999
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I'm getting 3/10, which is none of the answers.

(number driver, accident per year)
(17, 0)
(13, 1/5)
(21, 2/5)
(4, 3/5)
(2, 4/5)
(2, 1)
(1, 6/5)
The mid point of between 13 and 21 because 17+13 = 30 and 21+4+2+2+1 = 30
Therefore, the median is [(1/5)+(2/5)]/2 = 3/10
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GMATBLACKBELT
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coldweather999
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Answer 1.5.


We have 17 0's, 13 1's, 21 2's, 4 3's, 2 4's, 2 5's, and 1 6.

We can add some of these up to make this more clear:

17 13 21 9 . This adds up to 60 so we know the median is not going to be the average of the two middle numbers.

If we add 17 adn 13 we get 30. So since 1 is at the very end of this 30 we have 1 as one of the middle numbers.

2 is the next number above 1 so the two middle numbers are 1 and 2

1+2/2 = 3/2 = 1.5

Took me bout 4 min, but I think this is correct.

Ans C.
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GMATBLACKBELT
coldweather999
See the question below :

Answer 1.5.


We have 17 0's, 13 1's, 21 2's, 4 3's, 2 4's, 2 5's, and 1 6.

We can add some of these up to make this more clear:

17 13 21 9 . This adds up to 60 so we know the median is not going to be the average of the two middle numbers.

If we add 17 adn 13 we get 30. So since 1 is at the very end of this 30 we have 1 as one of the middle numbers.

2 is the next number above 1 so the two middle numbers are 1 and 2

1+2/2 = 3/2 = 1.5

Took me bout 4 min, but I think this is correct.

Ans C.
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Two things...
(1) The part in red, if you have even number of integers, wouldn't the median is in between two numbers?
1,2 => median is 1.5
(2) The chart list the accidents in 5 years and the question ask for accidents per year
How do you justify that to your answer?

Am I missing something?
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GMATBLACKBELT
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Honestly, I ignored this, b/c the question is wierd to begin with.

Whyd it give us 7 years of data, etc...
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ywilfred
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# of elements = 60
So the median would be the mean of the 30th and 31st term.

30th term = 1, 21st tem = 2, so median = 1.5



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