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 December 15th, 2016, 10:54 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(|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
 December 15th, 2016, 11:03 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,780 Thanks: 919 $P[|X-\mu| \geq k \sigma] \leq \dfrac {1}{k^2}$ $\sigma=1$ $\mu=3$ $P[|X-3| \geq k] \leq \dfrac{1}{k^2}$ $P[|X-3|\geq 1] \leq \dfrac{1}{1^2}=1$ $P[|X-3| \geq 2] \leq \dfrac{1}{2^2} = \dfrac 1 4$
 December 15th, 2016, 12:02 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? 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

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