Mathematics
6 Online
OpenStudy (anonymous):
@ganeshie8
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OpenStudy (anonymous):
like this one
ganeshie8 (ganeshie8):
Notice that July is left skewed, and the August is right skeewed
ganeshie8 (ganeshie8):
we dont know how many observations are outliers
ganeshie8 (ganeshie8):
so there is no way to tell which mean would be higher
OpenStudy (anonymous):
right so there's no way to tell the mean?
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OpenStudy (anonymous):
oh haha
ganeshie8 (ganeshie8):
yep, pick the last option and try ur luck :)
ganeshie8 (ganeshie8):
after u submit, do let me knw what the grader says,...
OpenStudy (anonymous):
Could you help me with a few more?
OpenStudy (anonymous):
will do :)
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ganeshie8 (ganeshie8):
yeah sure
OpenStudy (anonymous):
ganeshie8 (ganeshie8):
check for outliers in Rome and Newyork
ganeshie8 (ganeshie8):
here is the rule :
if you have outliers, then use median
if you dont have outliers, then use mean
OpenStudy (anonymous):
A?
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ganeshie8 (ganeshie8):
Rome has no outliers => mean
Newyork has outliers => median
ganeshie8 (ganeshie8):
C
OpenStudy (anonymous):
wait but isnt the minimum in rome an outlier?
OpenStudy (anonymous):
or not..idk what i was thinking
ganeshie8 (ganeshie8):
Range for Rome in box plot :
\((3 -1.5*10,~ 13+1.5*10) = (-12, ~28)\)
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OpenStudy (anonymous):
ahhh -12...i was getting 12
ganeshie8 (ganeshie8):
since the minimum and maximum \((0, ~16)\) lies within \((-12, ~28)\),
there are no outliers in Rome
OpenStudy (anonymous):
i see...i have 2 more...?
ganeshie8 (ganeshie8):
sure
OpenStudy (anonymous):
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OpenStudy (anonymous):
what is standard deviation?
ganeshie8 (ganeshie8):
its a quantity that measures the amount of spread
ganeshie8 (ganeshie8):
i see no outliers in both box plots
so i would use standard deviation for both
ganeshie8 (ganeshie8):
but i could be wrong on this one...
next question ?
OpenStudy (anonymous):
im still confused
if standard dev. means the spread and theyre asking for spread, wouldnt it be standard dev?
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ganeshie8 (ganeshie8):
here is the rule :
outliers => use median/IQR
no ouliers => use mean/standard deviation
ganeshie8 (ganeshie8):
both `IQR` and `standard deviation` quantify the amount of spread in a dataset
ganeshie8 (ganeshie8):
but you need to use `stadard deviation` as you dont have any outliers
OpenStudy (anonymous):
but the bb game has an outlier causing the skew..no?
OpenStudy (anonymous):
wouldnt it?
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ganeshie8 (ganeshie8):
its slightly skewed, but there are no outliers
OpenStudy (anonymous):
so neither has outliers therefore its D?
ganeshie8 (ganeshie8):
Yep !
OpenStudy (anonymous):
last one! @ganeshie8
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OpenStudy (anonymous):
i know B and C are wrong
OpenStudy (anonymous):
i just cant decide whether its A or D since they both sorta make sense
ganeshie8 (ganeshie8):
i think its B
OpenStudy (anonymous):
okay wait the females has an outlier but the males doesnt
ganeshie8 (ganeshie8):
i think ur box plots are not showing outliers separately
so the skewed males can have outliers
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OpenStudy (anonymous):
im super confused about this last one
ganeshie8 (ganeshie8):
yeah usually the boxplots need to show dots to indicate outliers,
but ur plots are not showing the outliers separately
ganeshie8 (ganeshie8):
So that means the males have a significant outlier as the plot is skewed to right
OpenStudy (anonymous):
if you do 1.5(24)+25 it gives the upper range to be 61 so there arent any outliers
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OpenStudy (anonymous):
yea and the outlier follows the outlier test on that problem
ganeshie8 (ganeshie8):
what test ?
OpenStudy (anonymous):
15-1.5(21-15)=9 so the lower range is 6 thats why theres an outlier on the problem you showed me
OpenStudy (anonymous):
lower range is 9*
ganeshie8 (ganeshie8):
oh yes you're right :
Q1 = 15
IQR = 5
left boundary = 15 - 1.5*5 > 0
so there are outliers
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ganeshie8 (ganeshie8):
one sec, leme work the present problem again :)
OpenStudy (anonymous):
okay
OpenStudy (anonymous):
does A make sense?
ganeshie8 (ganeshie8):
A makes perfect sense ! and yes you're right both plots have any outliers
ganeshie8 (ganeshie8):
so the earlier discussion on question#7 is incorrect
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OpenStudy (anonymous):
so what would it be for 7?
ganeshie8 (ganeshie8):
Movies : No outliers => standard deviation
Games : outliers => IQR
ganeshie8 (ganeshie8):
it has to be B
OpenStudy (anonymous):
so it is B
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ganeshie8 (ganeshie8):
yes
OpenStudy (anonymous):
okay now what about 4?
ganeshie8 (ganeshie8):
we concluded right, the females worked less
OpenStudy (anonymous):
so A?
ganeshie8 (ganeshie8):
oh you're gona ask why its not D
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OpenStudy (anonymous):
sure, why isnt it D? :)
ganeshie8 (ganeshie8):
lol idk both A and D are true about the shown box plots
ganeshie8 (ganeshie8):
but the question is not asking "which of the below options are true"
ganeshie8 (ganeshie8):
its asking which of the below "affect the spread and center"
OpenStudy (anonymous):
right it asked about affect
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ganeshie8 (ganeshie8):
I'd go wid A - i feel it -_-
OpenStudy (anonymous):
You did good :) it'll be an A i hope
ganeshie8 (ganeshie8):
stats is probably the toughest topic in alg1/2 i guess... good luck wid the test !
OpenStudy (anonymous):
no kidding haha, thanks
OpenStudy (anonymous):
when data is skewed in a box plot do you use mean or median?
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ganeshie8 (ganeshie8):
median,
one sec i have a nice table showing when to use mean/median/IQR/standard deviations
OpenStudy (anonymous):
|dw:1399959631358:dw|