How do I calculate the value of K using Tchebycheff's theorem for this data?

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Homework Statement


Hello all,

I have a one number stat:

Xi
2
4
5
6
6
8
10

Tchebycheff :
It ask me to find the value of K.

Interval of confidence: 89%

I know 1-1/k2
I have no idea how to calculate the value of K.
Do I have to somehow use the interval of confidence?

I checked my book and notes and there is nothing stating how to calculate K.
I just need to know the formula. Thanks.
 
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masterchiefo said:

Homework Statement


Hello all,

I have a one number stat:

Xi
2
4
5
6
6
8
10

Tchebycheff :
It ask me to find the value of K.

Interval of confidence: 89%

I know 1-1/k2
I have no idea how to calculate the value of K.
Do I have to somehow use the interval of confidence?

I checked my book and notes and there is nothing stating how to calculate K.
I just need to know the formula. Thanks.

(1) Your question makes no sense. 89% confidence interval for what? What do you mean by K?
(2) You are required to show your work before getting any help.
 
Ray Vickson said:
(1) Your question makes no sense. 89% confidence interval for what? What do you mean by K?
(2) You are required to show your work before getting any help.
This is all I have, no other info from teacher.
trust me I would love to show any work...but...I have no idea what is K. or what is the relation with 89% confidence interval. it probably have no relation and its just there to mess with my head.

All I have is this on my paper:
Fill the blank:

Xi
2
4
5
6
6
8
10

Tchebycheff :
Interval of confidence: 89%
K=
 
You titled this "Tychebycheff". Are you saying you do not know what Tchebycheff's theorem is? If so did you try looking it up?

Tchebycheff's theorem says that if X is a random variable with mean \mu and standard deviation \sigma then the probability that |x- \mu|\ge k\sigma is less than or equal to 1/k^2. What are the mean and standard deviation for this data? What are the upper and lower numbers such that 89% of the data is closer to the mean than those?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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