Thread: I need help with some Chebyshev's Theorum problems..

1. I need help with some Chebyshev's Theorum problems..

Hi everyone.. I am doing an at-home university course on Statistics for Business and Economics, I have been out of school for a number of years now and I need a little help.

I have just finished reading and learning Chebyshev's Theorum, z-Scores, and Empirical Rule.. i understand them, and did the first 5 problems without a hitch.
Heres what ive run in to that i dont get:
Of 1154 adults, average hours of sleep per night is 6.9, standard deviation is 1.2
a) use chebyshevs to find percentage that sleep between 4.5 and 9.3 - i worked that one out to be 75%
b) use chebyshevs to find % thats between 3.3 and 9.9..... heres where i am stuck, 6.9 - 3.3 = 3.6, 6.9 - 9.9 = -3.0
3.6 and -3.0 are not.. that same number of standard deviations apart. None of the methods or notes in the chapter tell me how to use this. All of the previous problems had the same number of SD's apart.. such as -2.4 and 2.4 in the a) part of the question.

Orginally i just skipped that question and went to the next one, but i have a problem with it as well. Mean = 3, SD = 1
a) whats the percent between 2 and 3? <-- same as above, not the same number of SDs apart?
b) whats the percent between 1 and 4? <-- again.
c) what percent is more than 4?

Thanks in advance for any help you can offer me... please go easy on me, like i said ive been outta school for a while

2. Originally Posted by shatrocious
Hi everyone.. I am doing an at-home university course on Statistics for Business and Economics, I have been out of school for a number of years now and I need a little help.

I have just finished reading and learning Chebyshev's Theorum, z-Scores, and Empirical Rule.. i understand them, and did the first 5 problems without a hitch.
Heres what ive run in to that i dont get:
Of 1154 adults, average hours of sleep per night is 6.9, standard deviation is 1.2
a) use chebyshevs to find percentage that sleep between 4.5 and 9.3 - i worked that one out to be 75%
b) use chebyshevs to find % thats between 3.3 and 9.9..... heres where i am stuck, 6.9 - 3.3 = 3.6, 6.9 - 9.9 = -3.0
3.6 and -3.0 are not.. that same number of standard deviations apart. None of the methods or notes in the chapter tell me how to use this. All of the previous problems had the same number of SD's apart.. such as -2.4 and 2.4 in the a) part of the question.

From Chebyshev's inequality we have:

P(|x-mu|>=3.6)<=1/(3.6/1.2)^2

and

P(|x-mu|>=3.0)<=1/(3.0/1.2)^2

so:

P(|x-mu|<3.6)>1-1/(3.6/1.2)^2=0.8889

P(|x-mu|<3.0)>1-1/(3.0/1.2)^2=0.84

so we take the smaller of these:

P(-3<x-mu<3.6)>0.84

RonL

3. so.. chebyshevs inequality is ... an addition to the chebyshev theorum? my text mentions NOTHING about that (thusfar)
so i would always work them both out and use the smaller of the two?

sigh, i dont know if i can do this on my own..

4. Originally Posted by shatrocious
so.. chebyshevs inequality is ... an addition to the chebyshev theorum? my text mentions NOTHING about that (thusfar)
so i would always work them both out and use the smaller of the two?

sigh, i dont know if i can do this on my own..
Please tell us what you have been given under the name Chebyshev's theorem. The only variant I can find on the inequality is (and this may
be more appropriate to your problem) is the one sided Chebyshev inequality:

P(x-mu>= k1 sigma)<=1/(1+k1^2), for k1>0

Which will have a dual form:

P(x-mu<= -k2 sigma)<=1/(1+k2^2), for k2>0

Then in this case we would have:

P(x-mu>=3.6)<=0.1

and

P(x-mu<=-3.0)<=0.138

So:

P(-0.3<x-mu<3.6)>1-0.238 = 0.762

(though I doubt that this inequality is tight)

RonL