Use the Law of Large Number to compute the probability of at least 49% heads on n, as n goes to infinity.
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Q: Use the Law of Large Number to compute the probability of at least 49% heads on n, as n goes to infinity.
From the Text, there is an example that is similar to this question which is this↓
probability statistics
asked Jan 12 at 17:58
MichelleMichelle
1
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add a comment |
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Q: Use the Law of Large Number to compute the probability of at least 49% heads on n, as n goes to infinity.
From the Text, there is an example that is similar to this question which is this↓
probability statistics
asked Jan 12 at 17:58
MichelleMichelle
1
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Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close with reading the comments.
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– saulspatz
Jan 12 at 18:07
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Q: Use the Law of Large Number to compute the probability of at least 49% heads on n, as n goes to infinity.
From the Text, there is an example that is similar to this question which is this↓
probability statistics
asked Jan 12 at 17:58
MichelleMichelle
1
$endgroup$
Q: Use the Law of Large Number to compute the probability of at least 49% heads on n, as n goes to infinity.
From the Text, there is an example that is similar to this question which is this↓
probability statistics
probability statistics
asked Jan 12 at 17:58
MichelleMichelle
1
asked Jan 12 at 17:58
MichelleMichelle
1
asked Jan 12 at 17:58
MichelleMichelle
1
asked Jan 12 at 17:58
MichelleMichelle
1
asked Jan 12 at 17:58
MichelleMichelle
1
1
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Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close with reading the comments.
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– saulspatz
Jan 12 at 18:07
add a comment |
$begingroup$
Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close with reading the comments.
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– saulspatz
Jan 12 at 18:07
$begingroup$
Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close with reading the comments.
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– saulspatz
Jan 12 at 18:07
$begingroup$
Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close with reading the comments.
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– saulspatz
Jan 12 at 18:07
add a comment |
1 Answer
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The law of large numbers is that if a random variable has mean $mu$ and variance $sigma^2$, then for every $epsilon > 0$ the sample mean $X_n$ satisfies $P(|X_n - mu| ge epsilon) le frac{sigma^2}{nepsilon}$. So the probability of heads being out of the $49% - 51%$ range is less than or equal to $frac{sigma^2}{nepsilon}$. As $ntoinfty$, $frac{sigma^2}{nepsilon} to 0$, so the probability of at least $49%$ heads is $1$ as $ntoinfty$.
answered Jan 12 at 19:07
Erik ParkinsonErik Parkinson
1,0859
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Do you mean 1 as in 100% as n goes to infinity? I am confused.
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– Michelle
Jan 13 at 21:24
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Yes, as n goes to infinity the probability is 100%
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– Erik Parkinson
Jan 14 at 6:18
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The law of large numbers is that if a random variable has mean $mu$ and variance $sigma^2$, then for every $epsilon > 0$ the sample mean $X_n$ satisfies $P(|X_n - mu| ge epsilon) le frac{sigma^2}{nepsilon}$. So the probability of heads being out of the $49% - 51%$ range is less than or equal to $frac{sigma^2}{nepsilon}$. As $ntoinfty$, $frac{sigma^2}{nepsilon} to 0$, so the probability of at least $49%$ heads is $1$ as $ntoinfty$.
answered Jan 12 at 19:07
Erik ParkinsonErik Parkinson
1,0859
$endgroup$
$begingroup$
Do you mean 1 as in 100% as n goes to infinity? I am confused.
$endgroup$
– Michelle
Jan 13 at 21:24
$begingroup$
Yes, as n goes to infinity the probability is 100%
$endgroup$
– Erik Parkinson
Jan 14 at 6:18
add a comment |
$begingroup$
The law of large numbers is that if a random variable has mean $mu$ and variance $sigma^2$, then for every $epsilon > 0$ the sample mean $X_n$ satisfies $P(|X_n - mu| ge epsilon) le frac{sigma^2}{nepsilon}$. So the probability of heads being out of the $49% - 51%$ range is less than or equal to $frac{sigma^2}{nepsilon}$. As $ntoinfty$, $frac{sigma^2}{nepsilon} to 0$, so the probability of at least $49%$ heads is $1$ as $ntoinfty$.
answered Jan 12 at 19:07
Erik ParkinsonErik Parkinson
1,0859
$endgroup$
$begingroup$
Do you mean 1 as in 100% as n goes to infinity? I am confused.
$endgroup$
– Michelle
Jan 13 at 21:24
$begingroup$
Yes, as n goes to infinity the probability is 100%
$endgroup$
– Erik Parkinson
Jan 14 at 6:18
add a comment |
$begingroup$
The law of large numbers is that if a random variable has mean $mu$ and variance $sigma^2$, then for every $epsilon > 0$ the sample mean $X_n$ satisfies $P(|X_n - mu| ge epsilon) le frac{sigma^2}{nepsilon}$. So the probability of heads being out of the $49% - 51%$ range is less than or equal to $frac{sigma^2}{nepsilon}$. As $ntoinfty$, $frac{sigma^2}{nepsilon} to 0$, so the probability of at least $49%$ heads is $1$ as $ntoinfty$.
answered Jan 12 at 19:07
Erik ParkinsonErik Parkinson
1,0859
$endgroup$
The law of large numbers is that if a random variable has mean $mu$ and variance $sigma^2$, then for every $epsilon > 0$ the sample mean $X_n$ satisfies $P(|X_n - mu| ge epsilon) le frac{sigma^2}{nepsilon}$. So the probability of heads being out of the $49% - 51%$ range is less than or equal to $frac{sigma^2}{nepsilon}$. As $ntoinfty$, $frac{sigma^2}{nepsilon} to 0$, so the probability of at least $49%$ heads is $1$ as $ntoinfty$.
answered Jan 12 at 19:07
Erik ParkinsonErik Parkinson
1,0859
answered Jan 12 at 19:07
Erik ParkinsonErik Parkinson
1,0859
answered Jan 12 at 19:07
Erik ParkinsonErik Parkinson
1,0859
answered Jan 12 at 19:07
Erik ParkinsonErik Parkinson
1,0859
1,0859
$begingroup$
Do you mean 1 as in 100% as n goes to infinity? I am confused.
$endgroup$
– Michelle
Jan 13 at 21:24
$begingroup$
Yes, as n goes to infinity the probability is 100%
$endgroup$
– Erik Parkinson
Jan 14 at 6:18
add a comment |
$begingroup$
Do you mean 1 as in 100% as n goes to infinity? I am confused.
$endgroup$
– Michelle
Jan 13 at 21:24
$begingroup$
Yes, as n goes to infinity the probability is 100%
$endgroup$
– Erik Parkinson
Jan 14 at 6:18
$begingroup$
Do you mean 1 as in 100% as n goes to infinity? I am confused.
$endgroup$
– Michelle
Jan 13 at 21:24
$begingroup$
Do you mean 1 as in 100% as n goes to infinity? I am confused.
$endgroup$
– Michelle
Jan 13 at 21:24
$begingroup$
Yes, as n goes to infinity the probability is 100%
$endgroup$
– Erik Parkinson
Jan 14 at 6:18
$begingroup$
Yes, as n goes to infinity the probability is 100%
$endgroup$
– Erik Parkinson
Jan 14 at 6:18
add a comment |
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$begingroup$
Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close with reading the comments.
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– saulspatz
Jan 12 at 18:07