Mathematics
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OpenStudy (anonymous):
the math test scores were: 50,65,70,72,72,78,80,82,84,84,85,86,88,88,90,94,96,98,98,99 . find the percentile rank for a score of 84 on the test
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OpenStudy (anonymous):
@Hero could u help me
OpenStudy (anonymous):
could u help
hero (hero):
Hint:
\(\large P = \frac{100(r - .5)}{n}\)
r = order of rank
n = number of students
hero (hero):
@annej
OpenStudy (anonymous):
ok i got 422.5
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OpenStudy (anonymous):
is that correct
OpenStudy (anonymous):
when u say order of rank what do you mean
OpenStudy (anonymous):
@Hero
hero (hero):
You can't get a percentile of 422.5
OpenStudy (anonymous):
oh ok jk then
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hero (hero):
It has to be between 0.1% - 99%
hero (hero):
Order of rank will be from lowest to greatest
OpenStudy (anonymous):
so the median
hero (hero):
1. 50
2. 65
3. 70
4. 72
4. 72
6. 78
7. 80
...
hero (hero):
the students who scored 84 have a rank of 9, therefore r = 9 and n = 20, so: \[ \large P = \frac{100(9 - .5)}{20}\]
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hero (hero):
Ha, it's 42.5%
OpenStudy (anonymous):
where did the 9 come from ???
hero (hero):
9th ranked score
hero (hero):
I'm curious as to how you calculated it
hero (hero):
I used the given formula. What did you use?
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OpenStudy (anonymous):
i used what my teacher gave me
hero (hero):
Which is...?
OpenStudy (anonymous):
so if im trying to find the percentile score of 86 with the same data set , the rank would change
nxk/100
hero (hero):
And what is k?
OpenStudy (anonymous):
the required percentile
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OpenStudy (anonymous):
@Hero
OpenStudy (anonymous):
ok so will the rank always be 9
hero (hero):
No, the rank will not always be 9
hero (hero):
Did you read what I posted above?
OpenStudy (anonymous):
yes but i still dont get how u dont 9 because when i count the number lower that 84 i get 8
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OpenStudy (anonymous):
do i still count 84
even if it repeats
OpenStudy (anonymous):
@Hero
hero (hero):
Well....If you look at what I posted above, you'll see what happens when numbers repeat.
OpenStudy (anonymous):
ok so if it where the perceptional rank of a score of 86 would the answer be 57.5
OpenStudy (anonymous):
@Hero
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OpenStudy (anonymous):
plz just answer this last question
hero (hero):
Yes, you have it figured out
OpenStudy (anonymous):
ok thank you