the main deal with 'at least one' problems - which come up more often on probability than on other problem types - is that they're very difficult to treat directly. instead, when you see 'at least one', you should treat the OPPOSITE situation - i.e., none.
so, because 'at least one' and 'none' are opposites, the following statements are exactly equivalent:
* there must be at least one
* it's IMPOSSIBLE to have NONE
the second is the easier way to think about it.
so:
in this problem, you should consider the case in which NONE of the homes was sold for whatever price is mentioned in the problem, and see whether it's IMPOSSIBLE.
NOTE: I AM NOT GOING TO WRITE THE THOUSANDS. so, '130' means $130,000. you'll thank me; this problem will be much easier to read.
(preface)
the median of 15 values is the value that comes 8th in the list. therefore, the first seven values are 130 or less, the 8th value is 130, and the 9th-15th values are 130 or more.
also, Sum = Average x Number of data points, so the sum of all the prices is 15 x 150 = 2250.
(i)
let's consider the case in which NONE of the homes was sold for more than 165.
the MAXIMUM sum of prices in this case would be 8(130) + 7(165), which is the case if all of the first 8 values are 130 each (the biggest they can be) and values 9-15 are 165 each.
that's a total of 1040 + 1155 = 2195.
not high enough.
therefore, it's IMPOSSIBLE to have NO prices over 165, so this statement must be true.
(ii)
let's try to create a list with NO such house prices.
let the first 7 prices be, say, 100 each.
the 8th is 130.
so the first 8 have a sum of 830, meaning that the highest 7 have a sum of 2250 - 830 = 1420.
there are all kinds of ways to do that with no values between 130 and 150, but the simplest is to make all seven of them equal to 1420 / 7, which is greater than 200.
so (ii) doesn't have to be true.
(iii)
let's try to create a list with NO such house prices.
this would mean that the first 8 prices are all 130.
so, the last 7 prices sum to 2250 - 8(130) = 1210.
that's an average of 1210/7, which is a shade over 170. you could let all 7 of the high prices equal that value, and it would work.
therefore, (iii) doesn't have to be true.
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this problem is a lot of work.
Ron has been teaching various standardized tests for 20 years.
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