anonymous
  • anonymous
Display the data in the chart attached below by making a box and whisker plot. Explain how you made it.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
anonymous
  • anonymous
@rajathsbhat
anonymous
  • anonymous
statistics....we meet again.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
LOL. Are you not so good with them?
anonymous
  • anonymous
not so good is an overstatement.
anonymous
  • anonymous
*understatement. I suck at it.
anonymous
  • anonymous
Oh, ok. thats fine, no worries. Ill go beg someone else. :D
anonymous
  • anonymous
Thanks for looking,tho.
pooja195
  • pooja195
http://www.purplemath.com/modules/boxwhisk3.htm
pooja195
  • pooja195
see if this helps :)
anonymous
  • anonymous
Hmm... could you explain it step by step?
pooja195
  • pooja195
hold on
anonymous
  • anonymous
ok. ty:)
pooja195
  • pooja195
you find the median of your data.
pooja195
  • pooja195
have u done that
anonymous
  • anonymous
one second - let me do it.
anonymous
  • anonymous
Median: 68
pooja195
  • pooja195
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
anonymous
  • anonymous
hmmm.
anonymous
  • anonymous
let me read all that.
pooja195
  • pooja195
srry
anonymous
  • anonymous
why are you sorry?
pooja195
  • pooja195
too much to read :)
anonymous
  • anonymous
indeed. lol.
anonymous
  • anonymous
could u simplify it?
pooja195
  • pooja195
divide the data into quarters, you then find the medians of these two halves.
anonymous
  • anonymous
hmmm...could u show me the steps by using the DRAW button?
anonymous
  • anonymous
Please.
pooja195
  • pooja195
im not surehttp://www.purplemath.com/modules/boxwhisk.htm
anonymous
  • anonymous
Hmm. Okay. Thanks.
anonymous
  • anonymous
Lets see.....
anonymous
  • anonymous
@eigenschmeigen
pooja195
  • pooja195
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
what's your question?
anonymous
  • anonymous
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
anonymous
  • anonymous
I also would like to know how to make box and whisker plots by myself.
jim_thompson5910
  • jim_thompson5910
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
anonymous
  • anonymous
the median is 68.
anonymous
  • anonymous
ok.....hmmm...
anonymous
  • anonymous
oohhhh divide it into 2?
anonymous
  • anonymous
8/2=4
anonymous
  • anonymous
so 4 sets of data in the lower quartile and 4 in the upper?
jim_thompson5910
  • jim_thompson5910
yes the median is 68
anonymous
  • anonymous
Okay.
anonymous
  • anonymous
show me step by step how to make it using the DRAW button please?
jim_thompson5910
  • jim_thompson5910
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
anonymous
  • anonymous
Uh-huh, I see.
jim_thompson5910
  • jim_thompson5910
the median of 61, 64, 65, 68 is 64.5, so Q1 = 64.5
anonymous
  • anonymous
Okay.
jim_thompson5910
  • jim_thompson5910
what's Q3?
anonymous
  • anonymous
Let me see. One second.
jim_thompson5910
  • jim_thompson5910
ok
anonymous
  • anonymous
70.5
jim_thompson5910
  • jim_thompson5910
good
jim_thompson5910
  • jim_thompson5910
what's your IQR
anonymous
  • anonymous
i don't know how to get that...
jim_thompson5910
  • jim_thompson5910
IQR = Q3 - Q1
jim_thompson5910
  • jim_thompson5910
it's basically the distance from each endpoint of the box
anonymous
  • anonymous
Oh,ok.
anonymous
  • anonymous
70.5-64.5?
jim_thompson5910
  • jim_thompson5910
yep
anonymous
  • anonymous
thats 6.
jim_thompson5910
  • jim_thompson5910
now find the min and max
anonymous
  • anonymous
The min and max?
jim_thompson5910
  • jim_thompson5910
yes, the smallest and largest data values
anonymous
  • anonymous
oh... U mean min = 61 max = 73
jim_thompson5910
  • jim_thompson5910
good, now determine if the min is an outlier or not
jim_thompson5910
  • jim_thompson5910
you do this by seeing if the min is above median - 1.5*IQR
anonymous
  • anonymous
Why 1.5?
jim_thompson5910
  • jim_thompson5910
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
jim_thompson5910
  • jim_thompson5910
I think it was Tukey
anonymous
  • anonymous
Oh.
jim_thompson5910
  • jim_thompson5910
I'm not sure how or why he got 1.5, but it works
jim_thompson5910
  • jim_thompson5910
and people agree on it
anonymous
  • anonymous
and the iqr is?
jim_thompson5910
  • jim_thompson5910
you just found it
anonymous
  • anonymous
o sorry....6
jim_thompson5910
  • jim_thompson5910
yep
anonymous
  • anonymous
1.5 x 6 = 9
jim_thompson5910
  • jim_thompson5910
subtract that from the median 68
anonymous
  • anonymous
Hmm. Ok. WOW, lots of steps :/ LOL.
jim_thompson5910
  • jim_thompson5910
it's not so bad once you get used to them
anonymous
  • anonymous
56.
jim_thompson5910
  • jim_thompson5910
no
jim_thompson5910
  • jim_thompson5910
68 - 9 is not 56
jim_thompson5910
  • jim_thompson5910
it's 59
anonymous
  • anonymous
o right sorry.
jim_thompson5910
  • jim_thompson5910
so that's your lowest threshold
anonymous
  • anonymous
i punched the wrong numbers into the calculator.
jim_thompson5910
  • jim_thompson5910
basically that's the smallest the min can be
anonymous
  • anonymous
sorry bout that.
jim_thompson5910
  • jim_thompson5910
any smaller than 59, and the min is an outlier
anonymous
  • anonymous
Ah,ok.
jim_thompson5910
  • jim_thompson5910
so we do NOT have an outlier for the min because the min (which is 61) is greater than the threshold of 59
jim_thompson5910
  • jim_thompson5910
now compute median + 1.5*IQR
anonymous
  • anonymous
Mhmm. Okay. One second - will log off for like 2 mins,k?
anonymous
  • anonymous
ill brb
jim_thompson5910
  • jim_thompson5910
ok
anonymous
  • anonymous
im back
jim_thompson5910
  • jim_thompson5910
ok
jim_thompson5910
  • jim_thompson5910
did you find median + 1.5*IQR
anonymous
  • anonymous
lets see. the median was?
jim_thompson5910
  • jim_thompson5910
68
anonymous
  • anonymous
68. so 68-9... 59?
jim_thompson5910
  • jim_thompson5910
you already did that one
anonymous
  • anonymous
its not the answer?
jim_thompson5910
  • jim_thompson5910
no I wanted you to find median + 1.5*IQR
anonymous
  • anonymous
Oh...Ok.
anonymous
  • anonymous
68+1.5*9
jim_thompson5910
  • jim_thompson5910
yeah but the IQR is 6, not 9
anonymous
  • anonymous
right,sorry
anonymous
  • anonymous
77
jim_thompson5910
  • jim_thompson5910
good, is the max smaller or larger than 77?
anonymous
  • anonymous
The max is 73, so its larger.
jim_thompson5910
  • jim_thompson5910
you mean, the max is smaller than 77?
anonymous
  • anonymous
Right.
jim_thompson5910
  • jim_thompson5910
so does that make the max an outlier?
anonymous
  • anonymous
no.
anonymous
  • anonymous
wait..
jim_thompson5910
  • jim_thompson5910
so there are no outliers
jim_thompson5910
  • jim_thompson5910
you're correct
anonymous
  • anonymous
nvm.
anonymous
  • anonymous
oh,okay.
jim_thompson5910
  • jim_thompson5910
so you have enough info to plot the box and whisker plot
anonymous
  • anonymous
Alright,thanks!
jim_thompson5910
  • jim_thompson5910
The basic format (the chances of your box and whisker plot looking exactly like this are very slim) is this |dw:1363736245038:dw|
anonymous
  • anonymous
Yes.
jim_thompson5910
  • jim_thompson5910
|dw:1363736267701:dw|
jim_thompson5910
  • jim_thompson5910
this is if you don't have any outliers
anonymous
  • anonymous
Okay. I think I understand a bit better- thanks!
jim_thompson5910
  • jim_thompson5910
yw

Looking for something else?

Not the answer you are looking for? Search for more explanations.