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December 15th, 2016, 11:54 AM  #1 
Newbie Joined: Dec 2016 From: Trinidad and Tobago Posts: 5 Thanks: 0  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(X3≥1), b. P(X3≥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(X3≥1)<1 P(X31<1)>0 Somehow it feels wrong, can anyone help me? Thanks 
December 15th, 2016, 12:03 PM  #2 
Senior Member Joined: Sep 2015 From: CA Posts: 586 Thanks: 316 
$P[X\mu \geq k \sigma] \leq \dfrac {1}{k^2}$ $\sigma=1$ $\mu=3$ $P[X3 \geq k] \leq \dfrac{1}{k^2}$ $P[X3\geq 1] \leq \dfrac{1}{1^2}=1$ $P[X3 \geq 2] \leq \dfrac{1}{2^2} = \dfrac 1 4$ 
December 15th, 2016, 01:02 PM  #3 
Newbie Joined: Dec 2016 From: Trinidad and Tobago Posts: 5 Thanks: 0 
Ok, Thanks so I was right. One more question, can you maybe help me again? Fiftyfive 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(1p)=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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