Computing Probability Using Central Limit Theorem
$begingroup$
$X_1,X_2...X_{100}$ are IID random variables with variance equal to 1. The mean of the random variables is not known. We estimate the mean by taking the sample average $hat{mu}=frac{sum_{i=1}^{100} X_i}{100} $. Use the central limit theorem to find the probability that the estimated mean and the true mean differ by less than $0.2$.
My solution:
Need to find
$P(-0.2<hat{mu}-mu<0.2)$
which is equivalent to finding:
$P(frac{-0.2}{1/sqrt(100)}<frac{hat{mu}-mu}{1sqrt(100}<frac{0.2}{1/sqrt(100)})$
By Central Limit the expression in the middle of the inequality is approximately standard normal.
Thus the Probability is just
$phi(2)-phi(-2)approx0.95$. Is this correct?
I'm not sure. Thanks.
probability probability-theory statistics
asked Dec 8 '18 at 8:39
kembkemb
700313
$endgroup$
add a comment |
$begingroup$
$X_1,X_2...X_{100}$ are IID random variables with variance equal to 1. The mean of the random variables is not known. We estimate the mean by taking the sample average $hat{mu}=frac{sum_{i=1}^{100} X_i}{100} $. Use the central limit theorem to find the probability that the estimated mean and the true mean differ by less than $0.2$.
My solution:
Need to find
$P(-0.2<hat{mu}-mu<0.2)$
which is equivalent to finding:
$P(frac{-0.2}{1/sqrt(100)}<frac{hat{mu}-mu}{1sqrt(100}<frac{0.2}{1/sqrt(100)})$
By Central Limit the expression in the middle of the inequality is approximately standard normal.
Thus the Probability is just
$phi(2)-phi(-2)approx0.95$. Is this correct?
I'm not sure. Thanks.
probability probability-theory statistics
asked Dec 8 '18 at 8:39
kembkemb
700313
$endgroup$
1
$begingroup$
Well, the percentage of values within $2$ Standard Deviations is not exactly $95%$ (it’s closer to $95.45%$), but otherwise it’s correct.
$endgroup$
– Kyky
Dec 8 '18 at 9:01
$begingroup$
@Kyky Thanks, I didn't know how to do the approximately equals to symbol in Latex
$endgroup$
– kemb
Dec 8 '18 at 9:10
1
$begingroup$
In LaTeXapproxgives $approx$
$endgroup$
– Henry
Dec 8 '18 at 9:20
$begingroup$
@Henry Thanks I will edit
$endgroup$
– kemb
Dec 8 '18 at 9:20
$begingroup$
It has been edited
$endgroup$
– kemb
Dec 8 '18 at 9:21
add a comment |
$begingroup$
$X_1,X_2...X_{100}$ are IID random variables with variance equal to 1. The mean of the random variables is not known. We estimate the mean by taking the sample average $hat{mu}=frac{sum_{i=1}^{100} X_i}{100} $. Use the central limit theorem to find the probability that the estimated mean and the true mean differ by less than $0.2$.
My solution:
Need to find
$P(-0.2<hat{mu}-mu<0.2)$
which is equivalent to finding:
$P(frac{-0.2}{1/sqrt(100)}<frac{hat{mu}-mu}{1sqrt(100}<frac{0.2}{1/sqrt(100)})$
By Central Limit the expression in the middle of the inequality is approximately standard normal.
Thus the Probability is just
$phi(2)-phi(-2)approx0.95$. Is this correct?
I'm not sure. Thanks.
probability probability-theory statistics
asked Dec 8 '18 at 8:39
kembkemb
700313
$endgroup$
$X_1,X_2...X_{100}$ are IID random variables with variance equal to 1. The mean of the random variables is not known. We estimate the mean by taking the sample average $hat{mu}=frac{sum_{i=1}^{100} X_i}{100} $. Use the central limit theorem to find the probability that the estimated mean and the true mean differ by less than $0.2$.
My solution:
Need to find
$P(-0.2<hat{mu}-mu<0.2)$
which is equivalent to finding:
$P(frac{-0.2}{1/sqrt(100)}<frac{hat{mu}-mu}{1sqrt(100}<frac{0.2}{1/sqrt(100)})$
By Central Limit the expression in the middle of the inequality is approximately standard normal.
Thus the Probability is just
$phi(2)-phi(-2)approx0.95$. Is this correct?
I'm not sure. Thanks.
probability probability-theory statistics
probability probability-theory statistics
asked Dec 8 '18 at 8:39
kembkemb
700313
asked Dec 8 '18 at 8:39
kembkemb
700313
edited Dec 8 '18 at 9:21
kemb
asked Dec 8 '18 at 8:39
kembkemb
700313
asked Dec 8 '18 at 8:39
kembkemb
700313
asked Dec 8 '18 at 8:39
kembkemb
700313
700313
1
$begingroup$
Well, the percentage of values within $2$ Standard Deviations is not exactly $95%$ (it’s closer to $95.45%$), but otherwise it’s correct.
$endgroup$
– Kyky
Dec 8 '18 at 9:01
$begingroup$
@Kyky Thanks, I didn't know how to do the approximately equals to symbol in Latex
$endgroup$
– kemb
Dec 8 '18 at 9:10
1
$begingroup$
In LaTeXapproxgives $approx$
$endgroup$
– Henry
Dec 8 '18 at 9:20
$begingroup$
@Henry Thanks I will edit
$endgroup$
– kemb
Dec 8 '18 at 9:20
$begingroup$
It has been edited
$endgroup$
– kemb
Dec 8 '18 at 9:21
add a comment |
1
$begingroup$
Well, the percentage of values within $2$ Standard Deviations is not exactly $95%$ (it’s closer to $95.45%$), but otherwise it’s correct.
$endgroup$
– Kyky
Dec 8 '18 at 9:01
$begingroup$
@Kyky Thanks, I didn't know how to do the approximately equals to symbol in Latex
$endgroup$
– kemb
Dec 8 '18 at 9:10
1
$begingroup$
In LaTeXapproxgives $approx$
$endgroup$
– Henry
Dec 8 '18 at 9:20
$begingroup$
@Henry Thanks I will edit
$endgroup$
– kemb
Dec 8 '18 at 9:20
$begingroup$
It has been edited
$endgroup$
– kemb
Dec 8 '18 at 9:21
1
1
$begingroup$
Well, the percentage of values within $2$ Standard Deviations is not exactly $95%$ (it’s closer to $95.45%$), but otherwise it’s correct.
$endgroup$
– Kyky
Dec 8 '18 at 9:01
$begingroup$
Well, the percentage of values within $2$ Standard Deviations is not exactly $95%$ (it’s closer to $95.45%$), but otherwise it’s correct.
$endgroup$
– Kyky
Dec 8 '18 at 9:01
$begingroup$
@Kyky Thanks, I didn't know how to do the approximately equals to symbol in Latex
$endgroup$
– kemb
Dec 8 '18 at 9:10
$begingroup$
@Kyky Thanks, I didn't know how to do the approximately equals to symbol in Latex
$endgroup$
– kemb
Dec 8 '18 at 9:10
1
1
$begingroup$
In LaTeX
approx gives $approx$$endgroup$
– Henry
Dec 8 '18 at 9:20
$begingroup$
In LaTeX
approx gives $approx$$endgroup$
– Henry
Dec 8 '18 at 9:20
$begingroup$
@Henry Thanks I will edit
$endgroup$
– kemb
Dec 8 '18 at 9:20
$begingroup$
@Henry Thanks I will edit
$endgroup$
– kemb
Dec 8 '18 at 9:20
$begingroup$
It has been edited
$endgroup$
– kemb
Dec 8 '18 at 9:21
$begingroup$
It has been edited
$endgroup$
– kemb
Dec 8 '18 at 9:21
add a comment |
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$begingroup$
Well, the percentage of values within $2$ Standard Deviations is not exactly $95%$ (it’s closer to $95.45%$), but otherwise it’s correct.
$endgroup$
– Kyky
Dec 8 '18 at 9:01
$begingroup$
@Kyky Thanks, I didn't know how to do the approximately equals to symbol in Latex
$endgroup$
– kemb
Dec 8 '18 at 9:10
1
$begingroup$
In LaTeX
approxgives $approx$$endgroup$
– Henry
Dec 8 '18 at 9:20
$begingroup$
@Henry Thanks I will edit
$endgroup$
– kemb
Dec 8 '18 at 9:20
$begingroup$
It has been edited
$endgroup$
– kemb
Dec 8 '18 at 9:21