0
$\begingroup$

This is a homework question but it is worded in a way that says "Find the answer" and not "Is there an answer". Heuristic knowledge on the course and how it's written indicates an answer as well so my friends and I are extremely curious if we're missing some angle.

The question:

We have a sample (1.0, 1.2, 1.4, 1.6, 1.6, 1.8, 2.0, 2.6, 2.8, 2.8, 3.0) and, changing only one value, make the average smaller than the median.

So the two approaches we started off with is either increasing the median (1.8) or decreasing the average (1.98).

Following the increase of the median, the limits we have on the median are 1.6 to 2.0. Increasing the median to 2 with the smallest increase in number (to keep the average low) would be 1.8 to 2.0. This however makes the average = the median (actually the answer to the next question in the exercise, making us further believe there must be an answer for the question).

This means we can't change the median to anything to solve this problem. The only other option we have is drop the average.

I'm on a phone so if someone could write this better, that would be great.

(x1+x2+x3+x4+x5+x6+x7+x8+x9+x10+x11)/11 < x6

x1+x2+x3+x4+x5+x7+x8+x9+x10+x11 < 10x6

We know x6 has to be 1.8 since we can't change the median. So that means;

20 < 18

We need to drop the average by more than 2 to get it lower than the median. We can only drop numbers on the right side of the median down to 1.8. This is at most a change of 3 - 1.2.

Additionally the left side of the median can only drop by 1.6.

Through this exhaustion of deduction, we figure the only way to get the average lower is to drop a number into the negative. But that seems like a bit of a cheat. Is it possible to do this without going into negative?

$\endgroup$
  • 3
    $\begingroup$ Why would dropping a number into negatives be a cheat? Also, note that the median is currently 1.7, not 1.8 $\endgroup$ – David Robinson Jan 17 '17 at 14:27
  • $\begingroup$ Sorry, forgot a 2.8 in the sample. And it feels like a cheat because up until this point we've been using realistic values. If I now have a valid answer of - 1000000, it doesn't really fit the pattern of questions we've been getting. I suppose it's a knowing my course kind of thing. $\endgroup$ – Pejman Poh Jan 17 '17 at 14:36
  • 2
    $\begingroup$ Part of this exercise seems to be showing how the sample mean is very much affected by an outlier while the sample median is affected much less. You need to add the self study tag. $\endgroup$ – Michael R. Chernick Jan 17 '17 at 14:41
  • $\begingroup$ Alright, added the tag. $\endgroup$ – Pejman Poh Jan 17 '17 at 14:43
  • 2
    $\begingroup$ I'm unclear on something-- when you say you can only change one value, I assume that that means you can only change one of the numbers in the sample (as opposed to only being allowed to change the median or the average). Is that correct? Overall, it does sound like you are missing an angle, but that angle is related to your meta-assessment of the problem. Your assumptions about information not contained in the problem as stated leaves the math unable to produce a valid answer. Which are you more confident in: your assumptions about the problem, or your knowledge of arithmetic? $\endgroup$ – Upper_Case Jan 17 '17 at 15:36
4
$\begingroup$

There are no solutions without dropping one of the numbers below 0. Besides your arithmetic solution to show this, you can run a simulation in R.

The following simulation tries adding various numbers between -3 and 3 (at .05 intervals) to each item in your list, then compares the mean to the median.

x <- c(1.0, 1.2, 1.4, 1.6, 1.6, 1.8, 2.0, 2.6, 2.8, 2.8, 3.0)

# change each of the numbers by some value between -3 and 3
# in this case, -3, -2.95, -2.98, -2.97, etc.
changes <- seq(-3, 3, .05)

num_tries <- length(changes) * length(x)
i <- seq_len(num_tries)

# create a matrix of 11 by all the possible tries
m <- replicate(num_tries, x)
m[cbind(i %% 11 + 1, i)] <- m[cbind(i %% 11 + 1, i)] + rep(changes, each = 11)

# remove any containing negatives
m <- m[, colSums(m < 0) == 0]

# are there any cases where the mean equals the median?
sum(apply(m, 2, mean) == apply(m, 2, median))
# Result: yes, 1, the one you already found 

# are there any cases where the mean is less than the median?
sum(apply(m, 2, mean) < apply(m, 2, median))
# Result: no, the answer is 0.

It's OK if you don't know R and can't follow this example, but it's just more evidence that you're not missing an arithmetic trick. With the numbers as you give them, you can't change one value to make the mean < median.

The only answers to the problem as you state them involve sending a number negative: for example, changing the 1.0 to -2.0.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.