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Mathematics 19 Online
OpenStudy (anonymous):

Jasper is using the following data samples to make a claim about the house values in his neighborhood: House Value A $150,000 B $175,000 C $200,000 D $167,000 E $2,500,000 Based on the data, should Jasper use the mean or the median to make an inference about the house values in his neighborhood? He should use the mean because it is in the center of the data. He should use the median because it is in the center of the data.He should use the median because there is an outlier that affects the mean. He should use the mean because there are no outliers that affect the mean.

OpenStudy (anonymous):

@missmysterious

OpenStudy (anonymous):

@Astrophysics

OpenStudy (astrophysics):

Well first, lets differentiate what the mean is and what median is.

OpenStudy (anonymous):

okay

OpenStudy (astrophysics):

Mean is the average, where median is the measurement, where the data is arranged from lowest to highest, but when you have a bunch of data, and you set it up from lowest to highest the middle number is the median.

OpenStudy (anonymous):

638400 is the mean

OpenStudy (anonymous):

right?

OpenStudy (astrophysics):

I haven't done the calculation but I believe you, so do you think it's wise to use median or mean for this?

OpenStudy (astrophysics):

Well for average you add up all the numbers and divide by the amount of numbers given.

OpenStudy (anonymous):

yeah that's what i did

OpenStudy (anonymous):

the median would be house c which is 200,000

OpenStudy (astrophysics):

Not quite

OpenStudy (anonymous):

cause it's the middle number right?

OpenStudy (astrophysics):

Remember you have to arrange them lowest to highest :)

OpenStudy (anonymous):

oh yeah damn

OpenStudy (astrophysics):

You're getting it though, it's good!

OpenStudy (anonymous):

then it would be B

OpenStudy (anonymous):

right?

OpenStudy (astrophysics):

Well this is not my expertise, but first lets eliminate a few options.

OpenStudy (anonymous):

okay what d we do first

OpenStudy (astrophysics):

Given the definition of what I said above what can we eliminate?

OpenStudy (astrophysics):

We should be able to eliminate 2 options

OpenStudy (anonymous):

i think we should be able to eliminate the mean because it is not the center of the data

OpenStudy (anonymous):

and if we eliminate that then we should be able to eliminate this one : He should use the mean because there are no outliers that affect the mean.

OpenStudy (anonymous):

is this correct isthere a way to find out?

OpenStudy (astrophysics):

There's a catch to this question, and I think you got it, because if we use mean we would get a weird answer right? Due to the high number of the last house, so yes you seem to be correct, we can say B!

OpenStudy (anonymous):

okay thanks for the help i have one more question i need help with are u able to help me?

OpenStudy (astrophysics):

Well actually it could be C

OpenStudy (astrophysics):

Lets just eliminate A and D because we know it's not the mean.

OpenStudy (anonymous):

ok

OpenStudy (astrophysics):

And B doesn't seem to be specific enough

OpenStudy (anonymous):

that's true

OpenStudy (astrophysics):

Mhm, I think it could be C, I'm sorry man, I haven't dealt with mean/ median and such in a long time haha.

OpenStudy (anonymous):

but for c is there an outlier that affects the mean?

OpenStudy (anonymous):

ha me either

OpenStudy (astrophysics):

Yeah I think they mean the last house, look how expensive it is, that would be the outlier right, and that would screw up the mean.

OpenStudy (anonymous):

yeah that's true

OpenStudy (anonymous):

so then it's c right?

OpenStudy (astrophysics):

I would say so, lets hope for the best!

OpenStudy (anonymous):

ok can you help me with one more

OpenStudy (astrophysics):

Well I was just about to go, unless it's a quick one :p

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