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Mathematics 6 Online
OpenStudy (anonymous):

@ganeshie8

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?

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 :)

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?

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 : \((3 -1.5*10,~ 13+1.5*10) = (-12, ~28)\)

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):

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?

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?

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

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

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

ganeshie8 (ganeshie8):

yep i see your point :) look at below correct problem : http://prntscr.com/3irbtz

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

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

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

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

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

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?

ganeshie8 (ganeshie8):

median, one sec i have a nice table showing when to use mean/median/IQR/standard deviations

OpenStudy (anonymous):

|dw:1399959631358:dw|

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