# Chebyshevs theorem : find k so that at most 10%

1. Aug 9, 2010

### idioteque

the test scores for a large statistics class have an unknown distribution with a mean of 70 and a standard deviation of 10

find k so that at most 10% of the scores are more than k standard deviations above the mean.

I'm a bit confused by the question it self.
does the question means :
1-1/k^2 = 0.1
k = 1.05

or

1-1/k^2 = 0.2
k = 1.12

or

1-1/k^2 = 0.8
k = square root 5 = 2.23

Last edited: Aug 9, 2010
2. Aug 9, 2010

### idioteque

to avoid being flamed for homework type of question, I added my own opinion towards this question. your help is very much appreciated.

3. Aug 9, 2010

### SW VandeCarr

I misread your question. The minimal value for k given "at most" 10% k SDs above the mean would be $$1-1/k^2=0.8$$ so $$k=2.236$$. This assumes your distribution is perfectly symmetrical.

Last edited: Aug 9, 2010
4. Aug 10, 2010

### idioteque

'at most' 10%.

why take the other remaining 80%?
it asked for 'at most' 10%.
doesn't this mean anything not greater than 10%?
if it asked 'at least' 10% then anything greater than 10%

5. Aug 10, 2010

### SW VandeCarr

80% of the test scores are within 2.236 SD of the mean. 10% are at least 2.236 SD above the mean, 10% at least 2.236 SD below the mean. This is the minimal value of k. If you want to be sure that at most 10% are more than k SD above the mean, take any arbitrary value of k greater than 2.236. That's the way I read it anyway.

Last edited: Aug 10, 2010
6. Aug 11, 2010

### idioteque

ok thank you. I got it.
the keyword is above the mean.
the value of k before the last 10% of each side(20%). so 80% is in between the 20%
is how I interpreted it as it is.
correct?

7. Aug 11, 2010

Correct.