Question

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I'm having trouble with these four questions.

Answer 1

For question 2 first identify the median.

Answer 2

4

Answer 3

Not quite. Median is the number in the middle of the set...

The median is 7. Now we want `to keep the mean equal to the median` so `median=mean` so since the median is 7 then the mean has to be 7...though we want to add one more number to make this true...this number will name it as `x`. Now we find the mean by adding all the numbers and then dividing it. So we would add up all the numbers (including the variable) and divide by the amount of numbers (with the variable included it is 8) so we have the equation.

\(\huge\bf{\frac{3+4+6+7+9+9+11+x}{8}=7}\)
Simplify ;)

Answer 4

Oh I see what I did wrong. The formula I was using was referring to the fourth term; not a number. Alright. Can you give me thirty minutes?

Answer 5

I actually got to go in a bit :(

Answer 6

I got 6.125

@mathmate

You think you can help me with these since he's gone?

Answer 7

For #2, the formula he gives sounds logical.
However, .... can you tell me what is the median on the left-hand side (neglecting the denominator)?
\(\Large\bf{\frac{3+4+6+7\color{red}{+}9+9+11+x}{8}=\color{red}{7}}\)

Answer 8

Im sorry Im not sure what you're saying.

@mathmate

Answer 9

I am continuing what @563blackghost was suggesting to you.
Since the new data set has 8 numbers, you need to know the median of the 8 numbers
A={3,4,6,7,9,9,11,x} before you can complete the calculations for finding x.
So what is the median of data set A defined above?

Answer 10

8
@mathmate

Answer 11

Exactly!
Now continue with the previous equation and solve for x.
We know the median is 8, so we try to impose that the mean is also eight by the following equation. All you need to do is to solve for x to find the number you're supposed to add.
\(\Large\bf{\frac{3+4+6+7+9+9+11+\color{blue}{x}}{8}=\color{red}{8}}\)
If the number comes out equal to or greater than 11, you can put it at the end. If not, you need to place it to maintain the ascending order of the numbers, and CHECK if the median is still 8.

Answer 12

Hm, alright this is what I got.

49x/8=8

6.125

Answer 13

@Loser66

Answer 14

@Kikuo
Do you realize that 49x means 49 times x?

The solution works like this:
\(\frac{49+x}{8}=8\)
cross multiply
\(49+x=8\times 8\)
x=64-49=11

Answer 15

15

Answer 16

Yes, 15 is correct! for #2.

Answer 17

Alright, what do we do for question two?

@mathmate

Answer 18

For number 3, read the question carefully. If you have already read the question, reread the question. Often rereading the questions 3 times or more will give you more insight.
If you're still stuck, reread question 2 which you have already solved.
After that,
Hint: the data set used in Q3 is the same as that of Q2.
Consider that whatever number x you add that is greater than 8 will not change the median (8). Use the answer from #2 to help you solve #3. The two are related.

Answer 19

I apologize, but I haven't been able to figure it out yet lol.

Or, more so, I know I keep getting the incorrect answer.

Answer 20

are u there

Answer 21

Example:
The mean and median of data set below are the same if x=5
{1 3 x}
If x=5, data set becomes {1,3,5}, mean (1+3+5)/3=9/3= 3, median = 3

What values of x would make the mean GREATER than the median?

Answer 22

Example:
The mean and median of data set below are the same if x=5
{1 3 x}
If x=5, data set becomes {1,3,5}, mean (1+3+5)/3=9/3= 3, median = 3

What values of x would make the mean GREATER than the median?
Try x=5, mean = 3 as before, mean NOT greater than median.
try x=8, mean=(1+3+8)/3=12/3=4 > median=3
try x=11, mean=(1+3+11)/3=15/3=5 > median=3
try x=100, mean=(1+3+100)/3=104/3=34.666... > ,median=3

So `What values of x would make the mean GREATER than the median?`

Answer 23

@mathmate

What I don't understand is how are they related when the question asks directly to use the data set given to us?

Answer 24

@Kikuo
`What I don't understand is how are they related when the question asks directly to use the data set given to us?`
I am assuming 'they' refer to the two questions or the two data sets.

Questions 2 and 3 are related by simple coincidence.
It "turns out" that the data sets of both questions are identical.

In the general case, to answer #3, you would have to repeat what you did in #2 and then continue.

In this particular case where they are identical, you can just take advantage of what you've done in #2 and apply the result to #3.

Answer 25

Question how did you figure out the median was 8 for #2 and #3.

I was using the formula to find the median when the numbers in the set form an even number
((n/2)th term+(n+1/2)th term)/2

I got 2th term+4.5th term which shouldn't be a choice

Even if I rounded it would be (4+9)/2 which is 6.5 rounded 7

How did you get 8?

@mathmate

Answer 26

For Q#2 & Q#3:
Both relate to an \(initial\) data set of A={3,4,6,7,9,9,11}
8 refers to a data set of 8 numbers, since we are supposed to add one number to the set of 7.
original set: A={3,4,6,\(\color{red}{7}\),9,9,11} n=7, the middle number is the fourth, namely \(\color{red}{7}\).

A'={3,4,6,\(\color{blue}{7,9}\),9,11,X}
However, if we add one number (X) to it, assuming it is greater than or equal to 9, then the median would be the 4.5th number, or the mean between 7 & 9, namely (7+9)/2=\(\color{blue}{8}\).

Answer 27

Oh I understand!

Alright back to 3.

I tried to set it up the same way we setup 2 but it doesnt work.

@mathmate

Answer 28

For Q#3, we know that data set A is the same as the one in Q#2, so we can say that the median of A' is also 8.
The question asks
`Identify a number (X) you could add to the data set to make the mean larger than the median`.
In #2, we have identified X as being 15.
If the number (X) to be added to data set A were 15, would the mean be larger than the median (8)?
What if X=14?
What if X=16?

Answer 29

By the way, are you aware of the fact that
'adding numbers outside of the middle number (or middle two numbers will NOT affect the value of the median, only the mean'?

For example,
4,5,6 has a median of 5, and a mean of 5
but
4,5,9 still has a median of 5 (middle number), but a mean of (4+5+9)/3=6.

Answer 30

* Changing numbers .... not Adding numbers....

Answer 31

@mathmate

Surely 16 would work right?

Since the median wont change?

Answer 32

Exactly, would 17 work? How about 18, 19, 20...?
So what would a general answer be? (now that you have identified 16 as a possible answer).

Answer 33

Yes, they would, correct?

@mathmate

Answer 34

Anything higher than 15.

Answer 35

Exactly! Well done!
Will you be working on #4 & #5?
They are based on the concepts you gathered from working with #2 and #3.
Give them a try, and I think you can figure things out.
If not, tag me. I should be on and off.

Answer 36

Alright give me 10 mins or so : )

Answer 37

'adding numbers outside of the middle number (or middle two numbers will NOT affect the value of the median, only the mean'?

Out of curosity could you give me an example where the mean would change if the values were changed

@mathmate

Answer 38

Here are some examples where the mean would change by changing the "outside" numbers.
{3,4,5} median = 4, mean = 4
{3,4,8} median = 4, mean = (3+4+8)/3=5
{3,4,101} median = 4, mean=(3+4+101)/3=36
{-2,4,4} median = 4, mean = (-1+4+4)/3=2

Answer 39

I apologize lol I meant median

@mathmate

Im working on the problems now also

Sorry it took so long had to help

Answer 40

No problem, I thought so too.

To change the median, the change must be either
1. change of the median value
2. adding a number.

Examples:
P={4,5,6,7} median=mean=5.5
P'={4,5,7,7} median=6, mean=23/4=5.75
P"={4,5,6,7,8} median=6, mean=6

Answer 41

Hm, I'm a bit confused. Sorry this is a bit irrelevant lol but I try to make sure I understand everything.

I thought you said adding a number wouldnt change the median?

Answer 42

@mathmate

Answer 43

Yes, you're right. That's what I wrote initially. On proof-reading, I realized that the statement was not correct. I then edited it to read "changing" a number.
Read the post immediately following the one that says "adding..."
I put
`* Changing numbers .... not Adding numbers....`
Sorry for the confustion.

So
changing a number outside of the middle number or two middle numbers will not change the median. But adding a number will change the median in most cases.

Answer 44

Got it!

A bit confused by this part.

changing a number outside of the middle number or two middle numbers will not change the median.

This sentence confused me.

So if we have a set like

24578

doing this 24678 would change it

doing this

24589

wouldn't


@mathmate

Answer 45

Yes, that's what it means. Numbers that you change OUTSIDE of the median number(s) will not change the median, but the mean is affected. That's exactly why the median is a better measure of central tendency than the mean when the numbers are skewed.

Say you're required to make a data set of 5 numbers with median 20 and mean 16.
You can start off with {20,20,20,20,20}, n=5 which has both median and mean 20.

The sum of the 5 numbers equals 5*20=100
For a mean of 16, the total of the numbers is 16n=16(5)=80.
So you change the "outside" numbers without touching the middle 10 so that the total is 75, try {10,10,20,20,20}. sum=10+10+20+20+20=80, so the mean is 16, while the median stays at 10.

Answer 46

This is how I setup #4.

12468/14=10

@mathmate

Answer 47

not sure what you meant!
Is your data set {1,2,4,6,8} ?
If it is, the median is 4, and the mean is 4.25. Not quite what #4 wants.

Answer 48

I must have to use x somewhere in all this

Is it 1, 2, 4, 6, 8/x=x?

Answer 49

@mathmate

Answer 50

There is no X required. You make up all the five numbers in the data set.

Answer 51

Start with making any 5 number set that gives the required median. Then change the "outside" values to match the required mean, similar to what we've seen before.

Answer 52

1, 4, 10, 13, 15

Done

Answer 53

Hm, let's see
{1, 4,10, 13, 15}
median is definitely 10, checks.
mean=(1+4+10+13+15)/5=43/5=8.6
Question wants mean = 14.
What you need to do about it is with a mean of 14 out of 5 numbers, the total equals 14*5=70.
So change the "outside" numbers so that the total adds up to 70. You're missing 70-43=27, and that can be made up by adding numbers to the first and last two numbers. Leave the middle number (median=10) untouched.

Answer 54

Why did multiplying 14 by 5 give us our answer?

Answer 55

@mathmate

Answer 56

1, 27, 10, 13, 15

Does that work?

Answer 57

14*5 because the mean is 14, and there are 5 numbers.
For example,
{14,14,14,14,14} would give you a mean of 14, but not median of 10.

{1, 27, 10, 13, 15 } does not work, because after reordering, you get
{1, 10, 13, 15, 27 } and the median becomes 13. In any case the mean is (1+10+13+15+27)/5=66/5=13.2 not quite 14.

I gave you an example before where you would start with any set of numbers, with 10 in the middle (to make sure median is 10), say
{10,10,10,10,10}
that makes a median of 10, and a mean of 50/5=10.
The sum of the numbers is 50 (instead of 70), so you need to have bigger numbers.
We know we cannot increase the first two numbers, because otherwise the order will change and 10 will no longer be the median.
So you can increase the last two numbers by 20 in any way you want (10,10), (5,10), etc. to make a total of 70, which also makes a mean of 14.
Am I going too fast?

Answer 58

(sorry, I was afk for chores)

Answer 59

9, 9, 10, 13, 29?
How is that

Answer 60

@mathmate

Answer 61

Hmm, let's see!
{9, 9, 10, 13, 29} is ordered correctly (ascending).
Now the middle number is 10, the required median, good!
Let's check the mean: (9+9+10+13+29)/5=70/5=14 !
Excellent, so you have a five-number data set with median 10 and mean 14.
Check with the question to make sure that's what is needed!
Well done!

Answer 62

FINALLY ON THE LAST QUESTION

Answer 63

Good! It's an essay. So I'll give you some time to work on it.
If in doubt, review what we discussed. The concept is quite related to what we did.
Will be glad to review what you did, when you finish.

Answer 64

All I could figure out so far is that the mean is the average number.

But I can't figure out why it would be so much larger than the middle price.

Answer 65

I know the median is the middle number but I can't figure out why its relevant to this.


I know it is.

Answer 66

The idea is the same as Q3. What you did in Q3 is basically what the data set for house prices look like in California. If you can explain why the data set in Q3 has a higher mean than the median, then it's exactly the same reason for Q4.
Does that make sense?

Answer 67

The best explanation I have is...

"The median stays the same as long as the middle number(s) stay the same regardless of how many numbers are added. The mean is the summation of all values in a data set, or in this case the prices of the houses in California, which is irrelevant to the middle number in a data set. As prices accumulate, causing values to be added to the data set, the mean increases, while the middle number can stay the same."

Answer 68

{9, 9, 10, 13, 29}
Think of the above numbers as the price of a 4 bed-room house in L.A. in $100,000.
For example, 9= a $900,000 house,
29=a $2.9 million house, etc.
What can you say that causes the median price to be $1 million, but the mean price is $1.4 million?

Answer 69

Hm, I would probably say the same thing haha.

The mean is the average and the median is the middle?

I must be missing something.

Answer 70

Yes, mean=average, and median=middle number.
Look at the numbers and see if you find some outstanding number that cause the mean to 'jump' higher than the median.

Sorry, I have to go now, so I'll leave you work on this. tag me when you finish. I will look at it when I come back.

Answer 71

The mean could be much higher because the prices of some houses are much higher than the prices of others?

Answer 72

Oh you know what, I actually learned this and it slipped my mind.

The concept of resistance says the median isn't vulnerable to extreme values while the mean is. While the mean might change as extreme values are added to a data set, the change will likely be little compared to the change of the mean, as an extreme value can easily change the product of the mean.

If we have a data set of 4, 4, 4, 4 and one of the values on the end change to 700, the median stays resistant to the extreme values while the mean is greatly affected.

Answer 73

Yes, that's exactly the idea the question wants you to express, and it makes it much easier if you have already learned the concept before.
On the other hand, try to express the concept in your own words, and do not cite verbatim from your notes.
Well done! :)