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Display the data in the chart attached below by making a box and whisker plot.
Explain how you made it.
 one year ago
 one year ago
Display the data in the chart attached below by making a box and whisker plot. Explain how you made it.
 one year ago
 one year ago

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rajathsbhatBest ResponseYou've already chosen the best response.0
statistics....we meet again.
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
LOL. Are you not so good with them?
 one year ago

rajathsbhatBest ResponseYou've already chosen the best response.0
not so good is an overstatement.
 one year ago

rajathsbhatBest ResponseYou've already chosen the best response.0
*understatement. I suck at it.
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
Oh, ok. thats fine, no worries. Ill go beg someone else. :D
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
Thanks for looking,tho.
 one year ago

pooja195Best ResponseYou've already chosen the best response.0
http://www.purplemath.com/modules/boxwhisk3.htm
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
Hmm... could you explain it step by step?
 one year ago

pooja195Best ResponseYou've already chosen the best response.0
you find the median of your data.
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
one second  let me do it.
 one year ago

pooja195Best ResponseYou've already chosen the best response.0
goodTo divide the data into quarters, you then find the medians of these two halves. Note: If you have an even number of values, so the first median was the average of the two middle values, then you include the middle values in your submedian computations. If you have an odd number of values, so the first median was an actual data point, then you do not include that value in your submedian computations. That is, to find the submedians, you're only looking at the values that haven't yet been used
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
let me read all that.
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
why are you sorry?
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
could u simplify it?
 one year ago

pooja195Best ResponseYou've already chosen the best response.0
divide the data into quarters, you then find the medians of these two halves.
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
hmmm...could u show me the steps by using the DRAW button?
 one year ago

pooja195Best ResponseYou've already chosen the best response.0
im not surehttp://www.purplemath.com/modules/boxwhisk.htm
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
Hmm. Okay. Thanks.
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
@eigenschmeigen
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
what's your question?
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
I have to make a box and whisker plot for the data shown at http://assets.openstudy.com/updates/attachments/5148e6dee4b05e69bfab6578help123please.1363732200597screenshot20130319at5.28.53pm.png
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
I also would like to know how to make box and whisker plots by myself.
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
Find the median first Then split the data into two groups (the median is the cut off point) and find the median of each group. The first quartile is the median of the lower half, the third quartile is the median of the upper half
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
the median is 68.
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
oohhhh divide it into 2?
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
so 4 sets of data in the lower quartile and 4 in the upper?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
yes the median is 68
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
show me step by step how to make it using the DRAW button please?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
so the sorted number set 61, 64, 65, 68, 68, 70, 71, 73 turns into 61, 64, 65, 68  68, 70, 71, 73 when you split the data in half along the median
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
the median of 61, 64, 65, 68 is 64.5, so Q1 = 64.5
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
Let me see. One second.
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
what's your IQR
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
i don't know how to get that...
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
IQR = Q3  Q1
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
it's basically the distance from each endpoint of the box
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
now find the min and max
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
The min and max?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
yes, the smallest and largest data values
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
oh... U mean min = 61 max = 73
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
good, now determine if the min is an outlier or not
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
you do this by seeing if the min is above median  1.5*IQR
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
just a number that some statistician a long time ago determined was the best number to see if outliers exist on a box and whisker plot
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
I think it was Tukey
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
I'm not sure how or why he got 1.5, but it works
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
and people agree on it
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
and the iqr is?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
you just found it
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
subtract that from the median 68
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
Hmm. Ok. WOW, lots of steps :/ LOL.
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
it's not so bad once you get used to them
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
68  9 is not 56
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
so that's your lowest threshold
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
i punched the wrong numbers into the calculator.
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
basically that's the smallest the min can be
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
sorry bout that.
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
any smaller than 59, and the min is an outlier
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
so we do NOT have an outlier for the min because the min (which is 61) is greater than the threshold of 59
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
now compute median + 1.5*IQR
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
Mhmm. Okay. One second  will log off for like 2 mins,k?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
did you find median + 1.5*IQR
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
lets see. the median was?
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
68. so 689... 59?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
you already did that one
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
its not the answer?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
no I wanted you to find median + 1.5*IQR
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
yeah but the IQR is 6, not 9
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
good, is the max smaller or larger than 77?
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
The max is 73, so its larger.
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
you mean, the max is smaller than 77?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
so does that make the max an outlier?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
so there are no outliers
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
you're correct
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
so you have enough info to plot the box and whisker plot
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
Alright,thanks!
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
The basic format (the chances of your box and whisker plot looking exactly like this are very slim) is this dw:1363736245038:dw
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
dw:1363736267701:dw
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.2
this is if you don't have any outliers
 one year ago

help123please.Best ResponseYou've already chosen the best response.0
Okay. I think I understand a bit better thanks!
 one year ago
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