A random number generator picks a number from 1 to 9 in a uniform manner. Find the 90th percentile. (Round your answer to one decimal place.) My set up is (k-9)(1/8)=0.90 why is this wrong
why did you use that "set up?"
in my book it says to use it
(base)(height)=0.90
first, what do you know about percentile rank in your own words
isn't it like how many # were chosen...90%?
so you're being asked for the number that ranks 90th percentile
right
how would you look for that number
that's where I am getting confused because my set up is wrong? would it be like 90%..of 9
let us list your numbers
is this how you have it set up? 1, 2, 3, 4, 5, 6, 7, 8, 9
okay yes
there are two ways to get percentile ranks one is: \(\sf \huge \frac{B\times (0.5E)}{n} \times 100 \) B: scores below x E: number of scores equal x n: total number the other is: \(\sf \huge \frac{B}{n} \times 100 \)
if you're going to use the second one, the set up should be \(\sf \large \frac{B}{9} \times 100 = 90 \) remember that your B will be lower than the actual score you're looking for, but let us focus on properly solving for B first
so it would be 8.1?
8?
I thought it ment to keep the .1
round off is the same as estimation and you need to learn about place values.
9.1 rounded off to 1 decimal means you focus on the first decimal if it is between 0-4 then it becomes 0 if it is between 5-9 then you add 1 to the number that preceded it 9.1 = 9 9.5 = 10 9.67 = 10 etcetera etcetera
let us check if our answer makes sense http://www.wolframalpha.com/input/?i=8%2F9
I understand that, but the question says "Round your answer to one decimal place" so therefore the answer would be like "9.4" keeping the 4
8 out 9 makes almost 90%
no
when I practice another problem with numbers 2-9 and asking for the 80th percentile, the answer is 7.6 ??
the .6 is kept
you're right you keep it
so youre saying when it is smaller than .4 it is rounded to 0
but 1, 2, 3, 4, 5, 6, 7, 8, 9 tell me where is 8.1
so you're saying the answer is 8
in scores 1 - 9 and if they are just 1, 2 .... 9 then yes unless someone actually has a score of 8.1 then it is 8.1
so I tried 8.1 and 8 and it is saying it is wrong
how about the first formula
the one that I used? nope
I made a typo \(\huge \frac{(B + 0.5E )}{n} \times 100 \)
so 8.1 and 8.0 are wrong huh?
yes
did you put 8.0 with point?
no I just did 8
try 8.0 because of the 10th place value
did it work?
wait but I only have one try left so I would like to be really sure!
the book says 8.2??
I don't know what you just did in the end, kropot but I do want to know why 8.2 because I want to learn if there's any other methods of solving this.
I am really not sure. They give me an example of "random number generator picks a number from 2 to 9 in a uniform manner. Find the 80th percentile(Round your answer to one decimal place . and the answer is 7.6 I am trying to work backwards from the answer but I still do not know
(7.6-2)/(9-2)=0.8
using same method, (k-1)/(9-1)=0.9 k=?
k=8.2
ohohoh I see
thank you so much for this explanation!!!!
what was the equation you used tho?
i looked at the example they gave you
basically what they are doing is
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