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December 15th, 2016, 10:54 AM   #1
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Chebychev inequality

Hi people,

I got the following question
A random variable X has mean 3 and variance 1. Use Chebychev's inequality to obtain an upper bound for:
a. P(|X-3|≥1),
b. P(|X-3|≥2).
I tried to answer A
kq=1 that means k=1/q, therefore k²=1²/q² since the variance is 1
k²=1²/1² k²=1
Therefore the upper bound is
P(|X-3|≥1)<1
P(|X-3|1<1)>0
Somehow it feels wrong, can anyone help me?
Thanks
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December 15th, 2016, 11:03 AM   #2
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$P[|X-\mu| \geq k \sigma] \leq \dfrac {1}{k^2}$

$\sigma=1$

$\mu=3$

$P[|X-3| \geq k] \leq \dfrac{1}{k^2}$

$P[|X-3|\geq 1] \leq \dfrac{1}{1^2}=1$

$P[|X-3| \geq 2] \leq \dfrac{1}{2^2} = \dfrac 1 4$
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December 15th, 2016, 12:02 PM   #3
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Ok, Thanks so I was right.
One more question, can you maybe help me again?
Fifty-five percent of all students on a campus are in Humanities. If 150 students are selected at random, what is the probability that the proportion of Humanities students in this particular sample will be between 45% and 55%.
I did the following
150*0.55=82.5
150*0.45=67.5
P(67.5<X<82.5)
σ²=np(1-p)=37.125
Here comes where I´m not quite sure
15=kσ k=15/σ k=15/√σ² k=15/√37.25 k=0.1654
Is it correct? I thought I have to use 15 since thats the distnace form 82.5 to 67.5 and therefore the interval I´m interested in.
Thanks again
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