Filling in Blanks

Here’s an interesting mean/median question. Assume your lab assistant took five readings, but only recorded three of them: -3, 5, and 7. The assistant also recorded that the mean was 6, and (for reasons lost on you) that the median was half of one of the missing readings. What were the readings?

Actually, the problem was posed without the background story, but I like it better this way. At any rate, here’s what we know. Let’s call the missing readings x and y. So:

  • The mean of (-3, 5, 7, x, y) is 6.
  • The median of (-3, 5, 7, x, y) is x/2.

Finding the mean is straightforward algebra: \(\frac{-3 + 5 + 7 + x + y}{5} = 6 \rightarrow x + y + 9 = 30\) \(\rightarrow x + y = 21\).

So we know that x + y = 21. But to find the median, we need to know how these values relate to the known values. If x and y are both at least 7, for instance, then the median will be 7, but if x or y are less than 7, then the median could be 5, x, or y. We need to consider each possible range of x, y, and x/2 with regards to 5 and 7.

In the number lines that follow, red indicates the possible values for x, from which we can determine the possible values for y (blue) and for x/2 (green). First, consider the case where \(x < 5\):

medianpuzzle1

In this case, there are two numbers less than 5 (-3 and x) and two numbers greater than 5, so 5 is the median. However, x/2 must be less than 5, so it is not possible that \(\frac{x}{2} = 5\). Therefore this is not a possible scenario.

In the next case, consider \(5 \le x < 7\):

medianpuzzle2

Since there are two values less than or equal to x (-3 and 5) and two values more than x (7 and y), x is the median. However, x/2 must be less than x, so it can’t be equal to the median. Therefore this is not a possible scenario, and x is at least 7.

What if y is less than 7? We again have two cases to consider: y is less than 5, and y is at least 5 but less than 7. Let’s look at the first case:

medianpuzzle3

Once again, the median is 5. This time, though, x/2 is more than 5, so we can rule this case out as well. Here’s the other case:

medianpuzzle4

In this case, since there are two values more than y (7 and x) and two values that can’t be more than y (-3 and 5), y is the median. y is less than 7, but x/2 is more than 7, so this case doesn’t work either.

At this point, we’ve considered all cases where either x or y is less than 7. Since we didn’t get any solutions, we know that x and y are both greater than or equal to 7, and so 7 is the median: \((-3, 5) < 7 \le (x, y)\).

medianpuzzle5

Once we know this, we can finish the problem. If the median is 7, then x is two times this, or 14. \(x = 14 \rightarrow y = 21 – 14 = 7\).

Now we know everything about the data: The set is (-3, 5, 7, 7, 14), the mean is 6, the median is 7, the mode is 7, and the range is 17.

1 Comment

  1. Faiza

    Hi thank you so much for the detailed description… I am preparing for an exam.. I have another problem for which I need assistance. Can you please help me with that too.

    Reply

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