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statistics....we meet again.
LOL. Are you not so good with them?
not so good is an overstatement.
*understatement. I suck at it.
Oh, ok. thats fine, no worries. Ill go beg someone else. :D
Thanks for looking,tho.
see if this helps :)
Hmm... could you explain it step by step?
you find the median of your data.
have u done that
one second - let me do it.
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 sub-median 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 sub-median computations. That is, to find the sub-medians, you're only looking at the values that haven't yet been used
let me read all that.
why are you sorry?
too much to read :)
could u simplify it?
divide the data into quarters, you then find the medians of these two halves.
hmmm...could u show me the steps by using the DRAW button?
im not surehttp://www.purplemath.com/modules/boxwhisk.htm
Hmm. Okay. Thanks.
what's your question?
I have to make a box and whisker plot for the data shown at http://assets.openstudy.com/updates/attachments/5148e6dee4b05e69bfab6578-help123please.-1363732200597-screenshot20130319at5.28.53pm.png
I also would like to know how to make box and whisker plots by myself.
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
the median is 68.
oohhhh divide it into 2?
so 4 sets of data in the lower quartile and 4 in the upper?
yes the median is 68
show me step by step how to make it using the DRAW button please?
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
Uh-huh, I see.
the median of 61, 64, 65, 68 is 64.5, so Q1 = 64.5
Let me see. One second.
what's your IQR
i don't know how to get that...
IQR = Q3 - Q1
it's basically the distance from each endpoint of the box
now find the min and max
The min and max?
yes, the smallest and largest data values
oh... U mean min = 61 max = 73
good, now determine if the min is an outlier or not
you do this by seeing if the min is above median - 1.5*IQR
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
I think it was Tukey
I'm not sure how or why he got 1.5, but it works
and people agree on it
and the iqr is?
you just found it
1.5 x 6 = 9
subtract that from the median 68
Hmm. Ok. WOW, lots of steps :/ LOL.
it's not so bad once you get used to them
68 - 9 is not 56
o right sorry.
so that's your lowest threshold
i punched the wrong numbers into the calculator.
basically that's the smallest the min can be
sorry bout that.
any smaller than 59, and the min is an outlier
so we do NOT have an outlier for the min because the min (which is 61) is greater than the threshold of 59
now compute median + 1.5*IQR
Mhmm. Okay. One second - will log off for like 2 mins,k?
did you find median + 1.5*IQR
lets see. the median was?
68. so 68-9... 59?
you already did that one
its not the answer?
no I wanted you to find median + 1.5*IQR
yeah but the IQR is 6, not 9
good, is the max smaller or larger than 77?
The max is 73, so its larger.
you mean, the max is smaller than 77?
so does that make the max an outlier?
so there are no outliers
so you have enough info to plot the box and whisker plot
The basic format (the chances of your box and whisker plot looking exactly like this are very slim) is this |dw:1363736245038:dw|
this is if you don't have any outliers
Okay. I think I understand a bit better- thanks!