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## 6th grade (Eureka Math/EngageNY)

### Unit 6: Lesson 2

Topic B: Summarizing a distribution that is approximately symmetric using the mean and mean absolute deviation- Statistics intro: Mean, median, & mode
- Mean, median, & mode example
- Calculating the mean
- Calculating the mean
- Calculating the mean: data displays
- Missing value given the mean
- Mean as the balancing point
- Missing value given the mean
- Median & range puzzlers
- Median & range puzzlers
- Impact on median & mean: removing an outlier
- Impact on median & mean: increasing an outlier
- Effects of shifting, adding, & removing a data point
- Mean absolute deviation (MAD)
- Mean absolute deviation example
- Mean absolute deviation (MAD)

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# Median & range puzzlers

Sal solves an interesting median and range challenge problem.

## Want to join the conversation?

- That was the most confusing video on here!(25 votes)
- What were you confused about? Maybe I can help :)(9 votes)

- I think that when you have an odd number of people is easy, but what about even numbers? Like in this video, it said 11 salespeople, but what about 10 salespeople? Maybe you can do a video about an even amount of people or things?(10 votes)
- It certainly does make the question a lot harder.

Just for fun, lets try to solve it anyway.

Instead of 11 salespersons, lets have 10!*min, _, _, _, A ,B, _, _, _ ,max*

We know that*(A + B)/2 = 6*, so*A + B = 12*

We also know that the range in 4, so*max - min = 4*

To test the claim that at least one salesperson sold more than 10 cars, lets make the max = 11.

As,*max - min = 4*,*min = 7**7, _, _, _, A ,B, _, _, _ ,11*

The lowest possible value of A is 7 (*7, 7, 7, 7, A ,B, _, _, _ ,11*).

If*A = 7*, then*B = 5*because we knew*A + B =12*

But B cannot be less than A, therefore the statement that at least one salesperson sold more than 10 cars is**false**.

The key lies in the fact that if there is an even number of items in the data set, then the median is found by taking the mean (average) of the two middlemost numbers.

I hope this helps.(17 votes)

- We can have 10 as maximum, and actually we can even have 5 10 in total.

e.g. 6 6 6 6 6 6 10 10 10 10 10

Median is 6

Range is 4(13 votes)- Wrong! The problem says that at least one of the salespersons sold more than 10 cars. You do not have any more than 10 cars sold in your own data.(3 votes)

- Sal's drawing skills on point 🚘😎(6 votes)
- Can't we use this method? We know the median equals (min+max)/2 so (min+max)/2 = 6 => min + max = 12

and we know the range = max - min. and the range is 4. so max = 4 + min. so we have two equations. min + max =12 and max = 4 + min. If we substitute in place of max we will have. min + 4 + min =12 =>2min=8 => min = 4 and max = 8. now we have the exact number. isn't this a more efficient way?(5 votes) - This video doesn't really help me. I still don't get it, I have a general math contest on March 24,2018. Someone,please, give me a simpler explanation. and lease have it by March 10,2018, because I NEED to get a place, and know what a range it. Sal,No offense, just some advice, I can't understand what your saying. Thank God for the words and script,LOL. Please accept my advice and help me!

-Curious About-to-be Champion.(5 votes)- Simple, the range is the difference between the largest and smallest values.

Hope this helps! If you have any further questions or need help, please ask! :)

Also, good luck on your contest!(1 vote)

- Hey diddle diddle

the mean is in the middle

you add then divide for the mean

the mode is the one you see the most

and the range is the difference between

difference between= biggest minus the smaller number(3 votes) - Okayy , but what if i assumed that the maximum number would be six ? I feel like that the answer shouldve been i don't know because 6 could be repeated more than once.(2 votes)
- If you assume the maximum is 6, the answer would still be false because no one would have sold more than 10 cars if the max is 6. No matter how many 6s there are, the range would have to be one of these choices: 2-6, 3-7, 4-8, 5-9, or 6-10. There is no reason to assume something that is not given. These ranges show the minimum number of possible cars sold has to be 2 and the maximum number sold has to be 10. IF it is 2, then all numbers after the median have to be a 6, and IF it is 10, all numbers before the median have to be 6.(3 votes)

- I don't know what to ask i'm frustrated to even think.(3 votes)
- no u cant have over 10 people(2 votes)

## Video transcript

- [Voiceover] Let's say that
we run a car dealership, and we have 11 salespeople, and we record the number of cars each of the 11 salespeople
sold in the past week. So that's one of them,
two, three, four, five, six, seven, eight, nine, 10, 11. So we record how much each of them sold. So maybe one person sold,
maybe they sold five cars. Maybe the next person sold seven cars. Maybe the next person sold 10 cars. And so that's how we would do it. So we're going to record all of that. But then we're told that
the median car sold is six. So if we ordered all of these numbers, so let's just assume
that we ordered them all from least to greatest. So maybe, and I'm just
making up numbers here, maybe this is four. This is four. Maybe this is five,
five, five, six, seven, seven, eight, nine, 10. The median car sold, meaning
that the middle number here is going to be six. Now we don't know whether
all of these other numbers are the actual numbers. I just made those up. But we know that the middle
number needs to be six. So let me just fill out
only the middle number. So the middle number, if
you have 11 data points, the middle number's going
to have five on either side. So that's that one right over there. One, two, three, four, five. One, two, three, four, five. So we know that the median,
we know the median-- Actually, let me do that a purple color. The median is going to be six. Now the other thing we know is that the range of cars sold is four. Now let's remind ourselves, the range is the maximum
number of cars sold minus the minimum number of cars sold, and if we have sorted all of these out, if we take-- If this is our max right over here, because we've sorted them all out, and this is our min right over here, our range is going to be
our max minus our min. So let me write that down. So you take your min and your max. You have your range. Range is equal to the
maximum minus the minimum. And they're telling us that that is four. This range is four. So the maximum minus the minimum, the difference between the number of cars that the most productive salesperson sold and the number of cars that the least productive
salesperson sold, that difference is going to be four. So given all of those assumptions, giving all of that information, I'm now going to give you a statement, and our challenge is if we assume everything I just said is true, is this statement going to be true, or is this statement going to be false, or do we not know? So let me write this down. So is this going to be true, false, or do we not know? Do we not have enough information to say it's for sure going to be true, for sure going to be false, or for sure we don't know? And so I encourage you
to pause the video now and try it out. All right, so let's work
through it together. This is kind of a fun puzzle here, where we've been given some clues, and then we have a statement, and we figure out can
we make this statement. Can we say it's true or can we
say for sure that it's false, or do we just not know? So a couple of ways to tackle it. One way to tackle it is, well look, one of the salespeople
must have sold six cars. If the median sold is six, and there's 11 salespeople here, the middle number here,
that if we order it, one of them must have sold six cars. If we had an even number here, then that would be the
average of the two middles, but if we have an odd number here, that is literally the
middle number of cars sold. So someone sold six cars. So if we assume-- We're trying to find a
world where someone sold more than 10 cars. So if someone sold six cars, is there a way, with a range of four, that someone sold more than 10? Well let's just think about it. Let's just assume that the min is six. We don't know that the min is six, but let's just try it out. If the minimum is six,
if the minimum is six, what's the maximum going to be? Well remember, the range is equal to-- We know that the range is four. The range is equal to the
maximum minus the minimum, maximum minus the minimum, and so the maximum in this
case would be 10, would be 10. The max would be equal
to, would be equal to 10. If the min is equal to
six, and instead of colons I'm going to write an equals sign, if the min is equal to six, then the max at most can be equal to 10. And we can't take a higher min, because we know that the six is going to be one of the values. We could try a lower minimum value. We could say the minimum
value is five or four or three or two or one, but then the maximum
would go down even more, because remember, the
maximum is no more than four larger than the minimum. So if we assume six, and we know that one of the salespeople sold six cars, then the maximum that
any of the salespeople could have sold is 10. And so the statement, at
least one of the salespeople sold more than 10 cars,
that's got to be false. That's got to be false. Now there's another way that
you could think about it. You could assume that
someone sold more than 10. You could assume. In the last example, or the
last way of thinking about it, we assumed that the min was six, but now let's just assume, let's just assume that someone sold-- Let's just try it out. Let's see if it's possible. Let's just assume that the max is 11, that someone sold more than 10 cars. If the max is equal to 11, what's the min going to need to be? Well we just have to remind ourselves, range is equal to max minus min. So four is equal to the maximum, 11, minus the minimum. So 11 minus what is equal to four? Well 11 minus seven is equal to four. So if we assume that the maximum is 11, then the minimum is going to have to be, the minimum is going to
have to be equal to seven. Now can the minimum be seven
when your median is six? No. If your median is six,
that means you have-- That means if you have an
odd number of data points, that means one of the data points is six, and if you had an even
number of data points, that means that the middle two are going to average out to be six, which means you even have data points that are less than six. So your minimum can't be seven. You're going to have a value
that is at least as low, at least as low as six. So your assumption can't
be true, can't be true, so once again, the assumption
based on the statement can't be true. This statement is going to be false.