Markov's and Chebyshev's inequalities

**cursedbg** · May 28th 2011, 06:11 AM

1.The max weight an elevator can carry is 700 kg.
Let the probability distribution of human weight be Unit{71,72,...,90}.
8 people (who's weight is independent) get on the elevator.
A)Find the upper limit to the probability that the total weight will exceed the maximum permitted using Markov's inequality.
B)Find the upper limit to the probability that the total weight will exceed the maximum permitted using Chebyshev's inequality.

2.Let $X_n$ be the number of tails after $n$ tries,
P(tails)=p.
Find n that assures the probability of | $X_n$ /n-p|< 0.1 is at least 0.95.

As far as I know in the first question according to Markov's inequality it shold go something like that:
P(X $\geqslant$ 700)=E(X)/700=80.5/700=0.115,is it correct?
Please correct me if I'm wrong and help me solve the rest.
Thanks in advance

**CaptainBlack** · May 28th 2011, 11:10 PM

Originally Posted by cursedbg

1.The max weight an elevator can carry is 700 kg.
Let the probability distribution of human weight be Unit{71,72,...,90}.
8 people (who's weight is independent) get on the elevator.
A)Find the upper limit to the probability that the total weight will exceed the maximum permitted using Markov's inequality.
B)Find the upper limit to the probability that the total weight will exceed the maximum permitted using Chebyshev's inequality.

2.Let $X_n$ be the number of tails after $n$ tries,
P(tails)=p.
Find n that assures the probability of | $X_n$ /n-p|< 0.1 is at least 0.95.

As far as I know in the first question according to Markov's inequality it shold go something like that:
P(X $\geqslant$ 700)=E(X)/700=80.5/700=0.115,is it correct?
Please correct me if I'm wrong and help me solve the rest.
Thanks in advance

1A is not correct, you should be using the mean weight of eight people not of one. To see that your answer is not reasonable you need only realise that the mean weight of eight people is $644$ kg, so $644/700 \approx 0.92$

For 1B you will need the standard deviation [Math]\sigma_1[/tex] of the weight distribution (I will assume you are to use the two-sided Chebyshev Inequality). Now the standard deviation of the sum of the weights of 8 people is $\sigma_8=\sigma_1\sqrt{8}$ . Then the Chebyshev Inequality tells you that:

$P(|X-644|\ge k\sigma_8)\le \frac{1}{k^2}$

AS we are interested in the probability of the total weight exceeding $700$ kg we put:

$k \sigma_8 =700-644=56$

CB

**mahefo** · May 30th 2011, 06:08 AM

Originally Posted by CaptainBlack

1A is not correct, you should be using the mean weight of eight people not of one. To see that your answer is not reasonable you need only realise that the mean weight of eight people is $644$ kg, so $644/700 \approx 0.92$

For 1B you will need the standard deviation [Math]\sigma_1[/tex] of the weight distribution (I will assume you are to use the two-sided Chebyshev Inequality). Now the standard deviation of the sum of the weights of 8 people is $\sigma_8=\sigma_1\sqrt{8}$ . Then the Chebyshev Inequality tells you that:

$P(|X-644|\ge k\sigma_8)\le \frac{1}{k}$

AS we are interested in the probability of the total weight exceeding $700$ kg we put:

$k \sigma_8 =700-644=56$

CB

1B. I think X is symmetric so the Chebyshev's inequation can represent
$2P\{X-644 \ge k\sigma_X \}=P\{|X-644|\ge k\sigma_X\}\le\frac{1}{k^2}$
Chose $k\sigma_X=56$ . From this we obtain the upper limit $\frac{1}{2k^2}$ .

**mahefo** · May 30th 2011, 06:18 AM

Originally Posted by cursedbg

2.Let $X_n$ be the number of tails after $n$ tries,
P(tails)=p.
Find n that assures the probability of | $X_n$ /n-p|< 0.1 is at least 0.95.

With note that $E(X_n/n)=p, Var (X_n/n)=pq/n$ , we use Chebyshev's inequation
$P\{|\frac{X_n/n-p}{\sqrt{pq/n}}|\ge \frac{0.1}{\sqrt{pq/n}} \}$ $\le \frac{pq/n}{0.01}$
and let the upper limit equal 0.95

**CaptainBlack** · May 30th 2011, 06:33 AM

Originally Posted by mahefo

1B. I think X is symmetric so the Chebyshev's inequation can represent
$2P\{X-644 \ge k\sigma_X \}=P\{|X-644|\ge k\sigma_X\}\le\frac{1}{k^2}$
Chose $k\sigma_X=56$ . From this we obtain the upper limit $\frac{1}{2k^2}$ .

Yes you can use the one sided inequality if you have covered it, but what you have posted is not the one sided Chebyshev Inequality, which is:

$P(X-\mu \ge k\sigma)\le \frac{1}{1+k^2}$

CB

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