Zamala wrote:
On a test, I have encountered the following question. I am not entirely sure if I remember 100% of the details of the question, however, I know what was asked. Please point out if there is missing information, which is required for solving it.
Question:
An aquarium has 7 fishes inside. The four heaviest fishes have a total weight of 117 pounds. The four lightest fishes have an average weight of 20.5 pounds.
What is (what could be?) the median weight of the 7 fishes.
A) 19.0
B) 19.5
C) 20.5
D) 21.5
E) 22.0
I forgot the precise numbers of the answer choices, they were in the given range. The numbers in the questions are correct.
Hi...
Although it is explained quite nicely by Brent, let me pitch in my bit.
Median is the center value and we are given four lightest and then the four heaviest, so the common fish in the TWO groups is the median as 4th fish out of 7 is the median.So CAN we be sure of a specific value .... NO
But we can surely think of a range..
Let the fishes in ascending order be A, B, C, D, E, F, G and we are looking for D.
1) If we take the lightest four-D, E, F, G-, the lightest out of this four, D is the median.
the largest value of D will be when all these fishes weigh the same.
So the largest value of median is \(\frac{117}{4}=28.75\)
2) If we take the heaviest four-A, B, C, D-, the heaviest out of this four, D is the median.
the smallest value of D will be when all these fishes weigh the same.
So the smallest value of median is 20.5 itself
The range is \(20.5\leq{median}\leq{28.75}\)..
Thus C, D and E , all three choices fit in..
C. 20.5 ------- 20.5, 20.5, 20.5, 20.5, 28, 30, 38.5
D. 21.5 ------- 20, 20, 20.5, 21.5, 28, 30, 37.5
E. 22 ------- 19.5, 20, 20.5, 22, 28, 30, 37
Note - If the question was asked in a multiple choice question and could have more than 1 answer, all three choices C, D and E fit in. However if we had to pick only one, the choices must be different or there may be some restrictions on the weight in initial statements.