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

- anonymous

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- schrodinger

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- anonymous

##### 1 Attachment

- anonymous

@rajathsbhat

- anonymous

statistics....we meet again.

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- anonymous

LOL. Are you not so good with them?

- anonymous

not so good is an overstatement.

- anonymous

*understatement. I suck at it.

- anonymous

Oh, ok. thats fine, no worries. Ill go beg someone else.
:D

- anonymous

Thanks for looking,tho.

- pooja195

http://www.purplemath.com/modules/boxwhisk3.htm

- pooja195

see if this helps :)

- anonymous

Hmm... could you explain it step by step?

- pooja195

hold on

- anonymous

ok. ty:)

- pooja195

you find the median of your data.

- pooja195

have u done that

- anonymous

one second - let me do it.

- anonymous

Median: 68

- 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

hmmm.

- anonymous

let me read all that.

- pooja195

srry

- anonymous

why are you sorry?

- pooja195

too much to read :)

- anonymous

indeed. lol.

- anonymous

could u simplify it?

- pooja195

divide the data into quarters, you then find the medians of these two halves.

- anonymous

hmmm...could u show me the steps by using the DRAW button?

- anonymous

Please.

- pooja195

im not surehttp://www.purplemath.com/modules/boxwhisk.htm

- anonymous

Hmm. Okay. Thanks.

- anonymous

Lets see.....

- anonymous

@eigenschmeigen

- pooja195

@jim_thompson5910

- jim_thompson5910

what's your question?

- 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

I also would like to know how to make box and whisker plots by myself.

- 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

the median is 68.

- anonymous

ok.....hmmm...

- anonymous

oohhhh divide it into 2?

- anonymous

8/2=4

- anonymous

so 4 sets of data in the lower quartile and 4 in the upper?

- jim_thompson5910

yes the median is 68

- anonymous

Okay.

- anonymous

show me step by step how to make it using the DRAW button please?

- 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

Uh-huh, I see.

- jim_thompson5910

the median of 61, 64, 65, 68 is 64.5, so Q1 = 64.5

- anonymous

Okay.

- jim_thompson5910

what's Q3?

- anonymous

Let me see. One second.

- jim_thompson5910

ok

- anonymous

70.5

- jim_thompson5910

good

- jim_thompson5910

what's your IQR

- anonymous

i don't know how to get that...

- jim_thompson5910

IQR = Q3 - Q1

- jim_thompson5910

it's basically the distance from each endpoint of the box

- anonymous

Oh,ok.

- anonymous

70.5-64.5?

- jim_thompson5910

yep

- anonymous

thats 6.

- jim_thompson5910

now find the min and max

- anonymous

The min and max?

- jim_thompson5910

yes, the smallest and largest data values

- anonymous

oh... U mean
min = 61
max = 73

- jim_thompson5910

good, now determine if the min is an outlier or not

- jim_thompson5910

you do this by seeing if the min is above
median - 1.5*IQR

- anonymous

Why 1.5?

- 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

I think it was Tukey

- anonymous

Oh.

- jim_thompson5910

I'm not sure how or why he got 1.5, but it works

- jim_thompson5910

and people agree on it

- anonymous

and the iqr is?

- jim_thompson5910

you just found it

- anonymous

o sorry....6

- jim_thompson5910

yep

- anonymous

1.5 x 6 = 9

- jim_thompson5910

subtract that from the median 68

- anonymous

Hmm. Ok. WOW, lots of steps
:/
LOL.

- jim_thompson5910

it's not so bad once you get used to them

- anonymous

56.

- jim_thompson5910

no

- jim_thompson5910

68 - 9 is not 56

- jim_thompson5910

it's 59

- anonymous

o right sorry.

- jim_thompson5910

so that's your lowest threshold

- anonymous

i punched the wrong numbers into the calculator.

- jim_thompson5910

basically that's the smallest the min can be

- anonymous

sorry bout that.

- jim_thompson5910

any smaller than 59, and the min is an outlier

- anonymous

Ah,ok.

- 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

now compute
median + 1.5*IQR

- anonymous

Mhmm. Okay. One second - will log off for like 2 mins,k?

- anonymous

ill brb

- jim_thompson5910

ok

- anonymous

im back

- jim_thompson5910

ok

- jim_thompson5910

did you find
median + 1.5*IQR

- anonymous

lets see. the median was?

- jim_thompson5910

68

- anonymous

68. so 68-9...
59?

- jim_thompson5910

you already did that one

- anonymous

its not the answer?

- jim_thompson5910

no I wanted you to find
median + 1.5*IQR

- anonymous

Oh...Ok.

- anonymous

68+1.5*9

- jim_thompson5910

yeah but the IQR is 6, not 9

- anonymous

right,sorry

- anonymous

77

- jim_thompson5910

good, is the max smaller or larger than 77?

- anonymous

The max is 73, so its larger.

- jim_thompson5910

you mean, the max is smaller than 77?

- anonymous

Right.

- jim_thompson5910

so does that make the max an outlier?

- anonymous

no.

- anonymous

wait..

- jim_thompson5910

so there are no outliers

- jim_thompson5910

you're correct

- anonymous

nvm.

- anonymous

oh,okay.

- jim_thompson5910

so you have enough info to plot the box and whisker plot

- anonymous

Alright,thanks!

- 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

Yes.

- jim_thompson5910

|dw:1363736267701:dw|

- jim_thompson5910

this is if you don't have any outliers

- anonymous

Okay. I think I understand a bit better- thanks!

- jim_thompson5910

yw

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