Estimating Probability with Chebyshev's inequality
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I am working on a problem and am a bit stuck.
The problem:
Average height = 1.7, Standard Deviation = .25
Estimate the Probability that a person has height X such that |X - 1.7| $ge$ .3
Use Chebyshev's Inequality
What I have done so far:
P(|X - $mu| ge$ b) $le$ $Var[X]over b^2$
Var[X] = (SD[X])$^2$ = (.25)$^2$ = .0625
P(|X - 1.7| $ge$ .3) $le$ $.0625over (.3)^2$
I am a little bit unsure how to proceed here. I know that we need to separate this into bounds like...
P(X $ge$ x or X $le$ x) = $1over sqrt 2 pi$ $int_{-infty}^x$ e$^{-x^2/2}$ dx + $int_{x}^{infty}$ e$^{-x^2/2}$ dx $le$ some percent?
But I am a little bit confused.
probability probability-theory probability-distributions chebyshev-function
asked Dec 3 '18 at 22:50
EthanEthan
10513
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|
show 4 more comments
$begingroup$
I am working on a problem and am a bit stuck.
The problem:
Average height = 1.7, Standard Deviation = .25
Estimate the Probability that a person has height X such that |X - 1.7| $ge$ .3
Use Chebyshev's Inequality
What I have done so far:
P(|X - $mu| ge$ b) $le$ $Var[X]over b^2$
Var[X] = (SD[X])$^2$ = (.25)$^2$ = .0625
P(|X - 1.7| $ge$ .3) $le$ $.0625over (.3)^2$
I am a little bit unsure how to proceed here. I know that we need to separate this into bounds like...
P(X $ge$ x or X $le$ x) = $1over sqrt 2 pi$ $int_{-infty}^x$ e$^{-x^2/2}$ dx + $int_{x}^{infty}$ e$^{-x^2/2}$ dx $le$ some percent?
But I am a little bit confused.
probability probability-theory probability-distributions chebyshev-function
asked Dec 3 '18 at 22:50
EthanEthan
10513
$endgroup$
1
$begingroup$
Your last line with the probability, it should be $leq$, but I don’t understand your question. You found an upper bound on the probability that you wish to seek. Also, in your last line with the integrals, you’re working with a standard normal distribution. The probability that $X$ will either be bigger or smaller than $x$ is simply $frac{1}{2}$ (the standard normal distribution is symmetric about the vertical axis).
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– Live Free or π Hard
Dec 4 '18 at 3:04
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The problem is to estimate the probability.
$endgroup$
– Ethan
Dec 4 '18 at 15:32
1
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@ Ethan The only thing that Chebyshev’s Inequality gives is an upper bound. That’s the best you can do in terms of giving an “estimated” probability. Unless you added “Use Chebyshev’s Inequality” yourself. Then, if you know the distribution of $X$, you can derive the exact probability.
$endgroup$
– Live Free or π Hard
Dec 4 '18 at 15:56
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So in this case, the last line is as far as we can go?
$endgroup$
– Ethan
Dec 4 '18 at 17:27
1
$begingroup$
Ok, thank you that makes sense.
$endgroup$
– Ethan
Dec 4 '18 at 21:30
|
show 4 more comments
$begingroup$
I am working on a problem and am a bit stuck.
The problem:
Average height = 1.7, Standard Deviation = .25
Estimate the Probability that a person has height X such that |X - 1.7| $ge$ .3
Use Chebyshev's Inequality
What I have done so far:
P(|X - $mu| ge$ b) $le$ $Var[X]over b^2$
Var[X] = (SD[X])$^2$ = (.25)$^2$ = .0625
P(|X - 1.7| $ge$ .3) $le$ $.0625over (.3)^2$
I am a little bit unsure how to proceed here. I know that we need to separate this into bounds like...
P(X $ge$ x or X $le$ x) = $1over sqrt 2 pi$ $int_{-infty}^x$ e$^{-x^2/2}$ dx + $int_{x}^{infty}$ e$^{-x^2/2}$ dx $le$ some percent?
But I am a little bit confused.
probability probability-theory probability-distributions chebyshev-function
asked Dec 3 '18 at 22:50
EthanEthan
10513
$endgroup$
I am working on a problem and am a bit stuck.
The problem:
Average height = 1.7, Standard Deviation = .25
Estimate the Probability that a person has height X such that |X - 1.7| $ge$ .3
Use Chebyshev's Inequality
What I have done so far:
P(|X - $mu| ge$ b) $le$ $Var[X]over b^2$
Var[X] = (SD[X])$^2$ = (.25)$^2$ = .0625
P(|X - 1.7| $ge$ .3) $le$ $.0625over (.3)^2$
I am a little bit unsure how to proceed here. I know that we need to separate this into bounds like...
P(X $ge$ x or X $le$ x) = $1over sqrt 2 pi$ $int_{-infty}^x$ e$^{-x^2/2}$ dx + $int_{x}^{infty}$ e$^{-x^2/2}$ dx $le$ some percent?
But I am a little bit confused.
probability probability-theory probability-distributions chebyshev-function
probability probability-theory probability-distributions chebyshev-function
asked Dec 3 '18 at 22:50
EthanEthan
10513
asked Dec 3 '18 at 22:50
EthanEthan
10513
edited Dec 4 '18 at 4:24
Ethan
asked Dec 3 '18 at 22:50
EthanEthan
10513
asked Dec 3 '18 at 22:50
EthanEthan
10513
asked Dec 3 '18 at 22:50
EthanEthan
10513
10513
1
$begingroup$
Your last line with the probability, it should be $leq$, but I don’t understand your question. You found an upper bound on the probability that you wish to seek. Also, in your last line with the integrals, you’re working with a standard normal distribution. The probability that $X$ will either be bigger or smaller than $x$ is simply $frac{1}{2}$ (the standard normal distribution is symmetric about the vertical axis).
$endgroup$
– Live Free or π Hard
Dec 4 '18 at 3:04
$begingroup$
The problem is to estimate the probability.
$endgroup$
– Ethan
Dec 4 '18 at 15:32
1
$begingroup$
@ Ethan The only thing that Chebyshev’s Inequality gives is an upper bound. That’s the best you can do in terms of giving an “estimated” probability. Unless you added “Use Chebyshev’s Inequality” yourself. Then, if you know the distribution of $X$, you can derive the exact probability.
$endgroup$
– Live Free or π Hard
Dec 4 '18 at 15:56
$begingroup$
So in this case, the last line is as far as we can go?
$endgroup$
– Ethan
Dec 4 '18 at 17:27
1
$begingroup$
Ok, thank you that makes sense.
$endgroup$
– Ethan
Dec 4 '18 at 21:30
|
show 4 more comments
1
$begingroup$
Your last line with the probability, it should be $leq$, but I don’t understand your question. You found an upper bound on the probability that you wish to seek. Also, in your last line with the integrals, you’re working with a standard normal distribution. The probability that $X$ will either be bigger or smaller than $x$ is simply $frac{1}{2}$ (the standard normal distribution is symmetric about the vertical axis).
$endgroup$
– Live Free or π Hard
Dec 4 '18 at 3:04
$begingroup$
The problem is to estimate the probability.
$endgroup$
– Ethan
Dec 4 '18 at 15:32
1
$begingroup$
@ Ethan The only thing that Chebyshev’s Inequality gives is an upper bound. That’s the best you can do in terms of giving an “estimated” probability. Unless you added “Use Chebyshev’s Inequality” yourself. Then, if you know the distribution of $X$, you can derive the exact probability.
$endgroup$
– Live Free or π Hard
Dec 4 '18 at 15:56
$begingroup$
So in this case, the last line is as far as we can go?
$endgroup$
– Ethan
Dec 4 '18 at 17:27
1
$begingroup$
Ok, thank you that makes sense.
$endgroup$
– Ethan
Dec 4 '18 at 21:30
1
1
$begingroup$
Your last line with the probability, it should be $leq$, but I don’t understand your question. You found an upper bound on the probability that you wish to seek. Also, in your last line with the integrals, you’re working with a standard normal distribution. The probability that $X$ will either be bigger or smaller than $x$ is simply $frac{1}{2}$ (the standard normal distribution is symmetric about the vertical axis).
$endgroup$
– Live Free or π Hard
Dec 4 '18 at 3:04
$begingroup$
Your last line with the probability, it should be $leq$, but I don’t understand your question. You found an upper bound on the probability that you wish to seek. Also, in your last line with the integrals, you’re working with a standard normal distribution. The probability that $X$ will either be bigger or smaller than $x$ is simply $frac{1}{2}$ (the standard normal distribution is symmetric about the vertical axis).
$endgroup$
– Live Free or π Hard
Dec 4 '18 at 3:04
$begingroup$
The problem is to estimate the probability.
$endgroup$
– Ethan
Dec 4 '18 at 15:32
$begingroup$
The problem is to estimate the probability.
$endgroup$
– Ethan
Dec 4 '18 at 15:32
1
1
$begingroup$
@ Ethan The only thing that Chebyshev’s Inequality gives is an upper bound. That’s the best you can do in terms of giving an “estimated” probability. Unless you added “Use Chebyshev’s Inequality” yourself. Then, if you know the distribution of $X$, you can derive the exact probability.
$endgroup$
– Live Free or π Hard
Dec 4 '18 at 15:56
$begingroup$
@ Ethan The only thing that Chebyshev’s Inequality gives is an upper bound. That’s the best you can do in terms of giving an “estimated” probability. Unless you added “Use Chebyshev’s Inequality” yourself. Then, if you know the distribution of $X$, you can derive the exact probability.
$endgroup$
– Live Free or π Hard
Dec 4 '18 at 15:56
$begingroup$
So in this case, the last line is as far as we can go?
$endgroup$
– Ethan
Dec 4 '18 at 17:27
$begingroup$
So in this case, the last line is as far as we can go?
$endgroup$
– Ethan
Dec 4 '18 at 17:27
1
1
$begingroup$
Ok, thank you that makes sense.
$endgroup$
– Ethan
Dec 4 '18 at 21:30
$begingroup$
Ok, thank you that makes sense.
$endgroup$
– Ethan
Dec 4 '18 at 21:30
|
show 4 more comments
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$begingroup$
Your last line with the probability, it should be $leq$, but I don’t understand your question. You found an upper bound on the probability that you wish to seek. Also, in your last line with the integrals, you’re working with a standard normal distribution. The probability that $X$ will either be bigger or smaller than $x$ is simply $frac{1}{2}$ (the standard normal distribution is symmetric about the vertical axis).
$endgroup$
– Live Free or π Hard
Dec 4 '18 at 3:04
$begingroup$
The problem is to estimate the probability.
$endgroup$
– Ethan
Dec 4 '18 at 15:32
1
$begingroup$
@ Ethan The only thing that Chebyshev’s Inequality gives is an upper bound. That’s the best you can do in terms of giving an “estimated” probability. Unless you added “Use Chebyshev’s Inequality” yourself. Then, if you know the distribution of $X$, you can derive the exact probability.
$endgroup$
– Live Free or π Hard
Dec 4 '18 at 15:56
$begingroup$
So in this case, the last line is as far as we can go?
$endgroup$
– Ethan
Dec 4 '18 at 17:27
1
$begingroup$
Ok, thank you that makes sense.
$endgroup$
– Ethan
Dec 4 '18 at 21:30