Statistics distribution of numbers and effects of mean,median, and mode

[DODGEVIPER13]

Homework Statement

The number 15 is added to each of the biggest 150 numbers in a distribution of 301 numbers in other words n=301

A. How does this addition affect the median of the distribution
B. how does this addition affect the mode of the distribution
C. How does this addition affect the mean of the distribution

Homework Equations

The Attempt at a Solution

Ok so for part a it will be no change if I consider numbers from 1 to 301 for simplicity since it could be random numbers. I will go 1-150 numbers on a line then add the middle number 151 and the another 150 numbers 152-301. There fore an addition of 15 to the top end won't effect the median. Mode is what I'm not sure about I know it is the number repeated the most but I can't see how this would be affected by the addition of 15 to higher numbers. I would assume it would decrease it? As for the mean obviously if you increase your higher numbers when you average it or take the mean it will be higher. More specifically 15(150)/301 to get the approximate amount it goes up by. So all I really need help on is mode.

 

[DODGEVIPER13]

Sorry for messy writing it was done on my ipad. I want to say my professor said there were a couple of answers to this. I just can't remember this part

 

[haruspex]

Yes, the mode is the complicated case. What will happen if the existing mode is in the top 150? What if it isn't?

 

[DODGEVIPER13]

If the mode is in the top and you add 15 it should increase the actual mode value but not the size of the mode for example if it were Unimodal it would stay as such so no change. In the lower 150 it would stay the same in both regards. I know this isn't right what am I doing wrong.

 

[haruspex]

If the mode is in the top and you add 15 it should increase the actual mode value but not the size of the mode for example if it were Unimodal it would stay as such so no change.

Yes.

In the lower 150 it would stay the same in both regards.

Sounds right to me. Why do you think that's wrong?

 

[DODGEVIPER13]

Sweet thanks yah that's what I figured

 

[Dick]

Sweet thanks yah that's what I figured

Suppose you are given nine numbers 1,1,1,2,2,2,2,3,4 and you add 15 to the largest 4 numbers?

 

[haruspex]

Suppose you are given nine numbers 1,1,1,2,2,2,2,3,4 and you add 15 to the largest 4 numbers?

I thought about that, but it seems to me you cannot do so - there are not '4 largest numbers'. There are 2 largest or six largest, etc. Therefore the stated conditions rule out a case like this.

 

[Dick]

I thought about that, but it seems to me you cannot do so - there are not '4 largest numbers'. There are 2 largest or six largest, etc. Therefore the stated conditions rule out a case like this.

Maybe. I'll consult my lawyer on the question phrasing. :)

 

[DODGEVIPER13]

Ok so is I add 15 to only 4 of those numbers I get 16,1,1,17,2,2,2,18,19 that would effectivley decrase the mode.

 

[Dick]

Ok so is I add 15 to only 4 of those numbers I get 16,1,1,17,2,2,2,18,19 that would effectivley decrase the mode.

I meant add 15 to the 'largest' four numbers in the list. 2,2,3,4 making them 17,17,18,19. Then sure, the mode changes from 2 to 1. But haruspex makes a good point that I'm probably reading the question wrong. He thinks the problem means that there are 301 distinct numbers in the set, some of which are repeated. I think that's probably right.

 

[DODGEVIPER13]

Ok well thanks for the help anyways

 

[haruspex]

haruspex ... thinks the problem means that there are 301 distinct numbers in the set, some of which are repeated.

Actually, I didn't. My point was that you can't add 15 to the largest 150 unless there is a largest 150, i.e. the 151st must be smaller than the 150th. If it does mean 301 different values then the answer for the mean in the OP is wrong (because n is the number of values, not the number of numbers).

 

[Dick]

Actually, I didn't. My point was that you can't add 15 to the largest 150 unless there is a largest 150, i.e. the 151st must be smaller than the 150th. If it does mean 301 different values then the answer for the mean in the OP is wrong (because n is the number of values, not the number of numbers).

True, but the mean will still certainly go up. I think what I thought you meant is probably more correct than a distribution with special properties. It just says 'a distribution of 301 numbers', it's really not clear whether the numbers are the values or the individual data points. I'll call my lawyer back. I still like the interpretation I attributed to you. At least it's unambiguous. But then again if the professor said there were multiple answers, maybe it's supposed to be confusing.