What is $P(x̃ > 72)$ given the following dist?
$begingroup$
It can be shown that if $x_1, x_2, dots, x_n$ are $N(mu, sigma^2)$, then $x̃$ is $N(mu, frac{sigma^2}{n})$
Here we define $x̃=frac{x_1+x_2+dots+x_n}{n}$ (the average).
The question is:
If $x_1, x_2, dots, x_{25}$ are $N(65, sigma^2 = 5^2)$, then find $P(x̃ > 72)$.
It is implied that all $x_i$ are iid (independently identically distributed).
How do I calculate this probability?
Would it be as follows:
$x̃$ is $N(65, 5^2/25)=N(65,1)$
Thus, $P(x̃ > 72) = 1 - P(x̃ leq 72) = 1 - phi(frac{72-65}{1})=1-phi(7)approx 0$
where $phi$ is how I calculate the probability using the CDF.
I don't think this is right though. If the mean is about $65$, then I do atleast except some $70+$ grades.
probability statistics probability-distributions normal-distribution
asked Dec 2 '18 at 15:53
K Split XK Split X
4,28421132
$endgroup$
add a comment |
$begingroup$
It can be shown that if $x_1, x_2, dots, x_n$ are $N(mu, sigma^2)$, then $x̃$ is $N(mu, frac{sigma^2}{n})$
Here we define $x̃=frac{x_1+x_2+dots+x_n}{n}$ (the average).
The question is:
If $x_1, x_2, dots, x_{25}$ are $N(65, sigma^2 = 5^2)$, then find $P(x̃ > 72)$.
It is implied that all $x_i$ are iid (independently identically distributed).
How do I calculate this probability?
Would it be as follows:
$x̃$ is $N(65, 5^2/25)=N(65,1)$
Thus, $P(x̃ > 72) = 1 - P(x̃ leq 72) = 1 - phi(frac{72-65}{1})=1-phi(7)approx 0$
where $phi$ is how I calculate the probability using the CDF.
I don't think this is right though. If the mean is about $65$, then I do atleast except some $70+$ grades.
probability statistics probability-distributions normal-distribution
asked Dec 2 '18 at 15:53
K Split XK Split X
4,28421132
$endgroup$
$begingroup$
It looks right to me. The pdf of $tilde x$ is more concentrated around the mean than $x_i$.
$endgroup$
– callculus
Dec 2 '18 at 16:10
$begingroup$
You've made everything completely right, but take care when $sigma^2/nneq 1$, then $$Z=frac{sqrt{n}}{sigma}Big(tilde{x}-muBig)$$ has a $N(0,1)$-distribution. The reason for that is $mathbb{E}(aX+b)=amathbb{E}(X)+b$ and $mathrm{Var}(aX+b)=a^2mathrm{Var}(X)$
$endgroup$
– Fakemistake
Dec 2 '18 at 18:50
add a comment |
$begingroup$
It can be shown that if $x_1, x_2, dots, x_n$ are $N(mu, sigma^2)$, then $x̃$ is $N(mu, frac{sigma^2}{n})$
Here we define $x̃=frac{x_1+x_2+dots+x_n}{n}$ (the average).
The question is:
If $x_1, x_2, dots, x_{25}$ are $N(65, sigma^2 = 5^2)$, then find $P(x̃ > 72)$.
It is implied that all $x_i$ are iid (independently identically distributed).
How do I calculate this probability?
Would it be as follows:
$x̃$ is $N(65, 5^2/25)=N(65,1)$
Thus, $P(x̃ > 72) = 1 - P(x̃ leq 72) = 1 - phi(frac{72-65}{1})=1-phi(7)approx 0$
where $phi$ is how I calculate the probability using the CDF.
I don't think this is right though. If the mean is about $65$, then I do atleast except some $70+$ grades.
probability statistics probability-distributions normal-distribution
asked Dec 2 '18 at 15:53
K Split XK Split X
4,28421132
$endgroup$
It can be shown that if $x_1, x_2, dots, x_n$ are $N(mu, sigma^2)$, then $x̃$ is $N(mu, frac{sigma^2}{n})$
Here we define $x̃=frac{x_1+x_2+dots+x_n}{n}$ (the average).
The question is:
If $x_1, x_2, dots, x_{25}$ are $N(65, sigma^2 = 5^2)$, then find $P(x̃ > 72)$.
It is implied that all $x_i$ are iid (independently identically distributed).
How do I calculate this probability?
Would it be as follows:
$x̃$ is $N(65, 5^2/25)=N(65,1)$
Thus, $P(x̃ > 72) = 1 - P(x̃ leq 72) = 1 - phi(frac{72-65}{1})=1-phi(7)approx 0$
where $phi$ is how I calculate the probability using the CDF.
I don't think this is right though. If the mean is about $65$, then I do atleast except some $70+$ grades.
probability statistics probability-distributions normal-distribution
probability statistics probability-distributions normal-distribution
asked Dec 2 '18 at 15:53
K Split XK Split X
4,28421132
asked Dec 2 '18 at 15:53
K Split XK Split X
4,28421132
asked Dec 2 '18 at 15:53
K Split XK Split X
4,28421132
asked Dec 2 '18 at 15:53
K Split XK Split X
4,28421132
asked Dec 2 '18 at 15:53
K Split XK Split X
4,28421132
4,28421132
$begingroup$
It looks right to me. The pdf of $tilde x$ is more concentrated around the mean than $x_i$.
$endgroup$
– callculus
Dec 2 '18 at 16:10
$begingroup$
You've made everything completely right, but take care when $sigma^2/nneq 1$, then $$Z=frac{sqrt{n}}{sigma}Big(tilde{x}-muBig)$$ has a $N(0,1)$-distribution. The reason for that is $mathbb{E}(aX+b)=amathbb{E}(X)+b$ and $mathrm{Var}(aX+b)=a^2mathrm{Var}(X)$
$endgroup$
– Fakemistake
Dec 2 '18 at 18:50
add a comment |
$begingroup$
It looks right to me. The pdf of $tilde x$ is more concentrated around the mean than $x_i$.
$endgroup$
– callculus
Dec 2 '18 at 16:10
$begingroup$
You've made everything completely right, but take care when $sigma^2/nneq 1$, then $$Z=frac{sqrt{n}}{sigma}Big(tilde{x}-muBig)$$ has a $N(0,1)$-distribution. The reason for that is $mathbb{E}(aX+b)=amathbb{E}(X)+b$ and $mathrm{Var}(aX+b)=a^2mathrm{Var}(X)$
$endgroup$
– Fakemistake
Dec 2 '18 at 18:50
$begingroup$
It looks right to me. The pdf of $tilde x$ is more concentrated around the mean than $x_i$.
$endgroup$
– callculus
Dec 2 '18 at 16:10
$begingroup$
It looks right to me. The pdf of $tilde x$ is more concentrated around the mean than $x_i$.
$endgroup$
– callculus
Dec 2 '18 at 16:10
$begingroup$
You've made everything completely right, but take care when $sigma^2/nneq 1$, then $$Z=frac{sqrt{n}}{sigma}Big(tilde{x}-muBig)$$ has a $N(0,1)$-distribution. The reason for that is $mathbb{E}(aX+b)=amathbb{E}(X)+b$ and $mathrm{Var}(aX+b)=a^2mathrm{Var}(X)$
$endgroup$
– Fakemistake
Dec 2 '18 at 18:50
$begingroup$
You've made everything completely right, but take care when $sigma^2/nneq 1$, then $$Z=frac{sqrt{n}}{sigma}Big(tilde{x}-muBig)$$ has a $N(0,1)$-distribution. The reason for that is $mathbb{E}(aX+b)=amathbb{E}(X)+b$ and $mathrm{Var}(aX+b)=a^2mathrm{Var}(X)$
$endgroup$
– Fakemistake
Dec 2 '18 at 18:50
add a comment |
1 Answer
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$begingroup$
The random variable $X_i$ is distributed as $X_isim mathcal N(65,25)$. We can see at the graph how the pdf looks like:
It can be seen that $P(X_i> 72) gg 0$
What is the variance of $frac1nsum_{i=1}^n X_i$, where $X_i$ are iid?
$Varleft( frac1nsum_{i=1}^n X_iright)=frac1{n^2}cdot Varleft(sum_{i=1}^n X_iright)=frac{1}{n^2}cdot left( Var(X_1)+Var(X_2)+ldots + Var(X_n) right)$
$=frac{1}{n^2}cdot left(ncdot Var(X_1) right)=frac{sigma^2}{n}=frac{25}{25}=1$. Therefore $frac1{25}sum_{i=1}^{25} X_i=overline X_{25}sim mathcal{N}(65, 1)$
The graph of the pdf is
It can be seen that $P(overline X_{25}> 72)approx 0$.
The pdf of $overline X_{25}$ is more concentrated around the mean than $X_i$. It confirms your result.
answered Dec 2 '18 at 17:18
callculuscallculus
18.1k31427
$endgroup$
$begingroup$
Thank you foir the graph, it helped alot!
$endgroup$
– K Split X
Dec 5 '18 at 21:28
$begingroup$
@KSplitX You´re welcome. I´m glad that it helped.
$endgroup$
– callculus
Dec 7 '18 at 20:33
add a comment |
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$begingroup$
The random variable $X_i$ is distributed as $X_isim mathcal N(65,25)$. We can see at the graph how the pdf looks like:
It can be seen that $P(X_i> 72) gg 0$
What is the variance of $frac1nsum_{i=1}^n X_i$, where $X_i$ are iid?
$Varleft( frac1nsum_{i=1}^n X_iright)=frac1{n^2}cdot Varleft(sum_{i=1}^n X_iright)=frac{1}{n^2}cdot left( Var(X_1)+Var(X_2)+ldots + Var(X_n) right)$
$=frac{1}{n^2}cdot left(ncdot Var(X_1) right)=frac{sigma^2}{n}=frac{25}{25}=1$. Therefore $frac1{25}sum_{i=1}^{25} X_i=overline X_{25}sim mathcal{N}(65, 1)$
The graph of the pdf is
It can be seen that $P(overline X_{25}> 72)approx 0$.
The pdf of $overline X_{25}$ is more concentrated around the mean than $X_i$. It confirms your result.
answered Dec 2 '18 at 17:18
callculuscallculus
18.1k31427
$endgroup$
$begingroup$
Thank you foir the graph, it helped alot!
$endgroup$
– K Split X
Dec 5 '18 at 21:28
$begingroup$
@KSplitX You´re welcome. I´m glad that it helped.
$endgroup$
– callculus
Dec 7 '18 at 20:33
add a comment |
$begingroup$
The random variable $X_i$ is distributed as $X_isim mathcal N(65,25)$. We can see at the graph how the pdf looks like:
It can be seen that $P(X_i> 72) gg 0$
What is the variance of $frac1nsum_{i=1}^n X_i$, where $X_i$ are iid?
$Varleft( frac1nsum_{i=1}^n X_iright)=frac1{n^2}cdot Varleft(sum_{i=1}^n X_iright)=frac{1}{n^2}cdot left( Var(X_1)+Var(X_2)+ldots + Var(X_n) right)$
$=frac{1}{n^2}cdot left(ncdot Var(X_1) right)=frac{sigma^2}{n}=frac{25}{25}=1$. Therefore $frac1{25}sum_{i=1}^{25} X_i=overline X_{25}sim mathcal{N}(65, 1)$
The graph of the pdf is
It can be seen that $P(overline X_{25}> 72)approx 0$.
The pdf of $overline X_{25}$ is more concentrated around the mean than $X_i$. It confirms your result.
answered Dec 2 '18 at 17:18
callculuscallculus
18.1k31427
$endgroup$
$begingroup$
Thank you foir the graph, it helped alot!
$endgroup$
– K Split X
Dec 5 '18 at 21:28
$begingroup$
@KSplitX You´re welcome. I´m glad that it helped.
$endgroup$
– callculus
Dec 7 '18 at 20:33
add a comment |
$begingroup$
The random variable $X_i$ is distributed as $X_isim mathcal N(65,25)$. We can see at the graph how the pdf looks like:
It can be seen that $P(X_i> 72) gg 0$
What is the variance of $frac1nsum_{i=1}^n X_i$, where $X_i$ are iid?
$Varleft( frac1nsum_{i=1}^n X_iright)=frac1{n^2}cdot Varleft(sum_{i=1}^n X_iright)=frac{1}{n^2}cdot left( Var(X_1)+Var(X_2)+ldots + Var(X_n) right)$
$=frac{1}{n^2}cdot left(ncdot Var(X_1) right)=frac{sigma^2}{n}=frac{25}{25}=1$. Therefore $frac1{25}sum_{i=1}^{25} X_i=overline X_{25}sim mathcal{N}(65, 1)$
The graph of the pdf is
It can be seen that $P(overline X_{25}> 72)approx 0$.
The pdf of $overline X_{25}$ is more concentrated around the mean than $X_i$. It confirms your result.
answered Dec 2 '18 at 17:18
callculuscallculus
18.1k31427
$endgroup$
The random variable $X_i$ is distributed as $X_isim mathcal N(65,25)$. We can see at the graph how the pdf looks like:
It can be seen that $P(X_i> 72) gg 0$
What is the variance of $frac1nsum_{i=1}^n X_i$, where $X_i$ are iid?
$Varleft( frac1nsum_{i=1}^n X_iright)=frac1{n^2}cdot Varleft(sum_{i=1}^n X_iright)=frac{1}{n^2}cdot left( Var(X_1)+Var(X_2)+ldots + Var(X_n) right)$
$=frac{1}{n^2}cdot left(ncdot Var(X_1) right)=frac{sigma^2}{n}=frac{25}{25}=1$. Therefore $frac1{25}sum_{i=1}^{25} X_i=overline X_{25}sim mathcal{N}(65, 1)$
The graph of the pdf is
It can be seen that $P(overline X_{25}> 72)approx 0$.
The pdf of $overline X_{25}$ is more concentrated around the mean than $X_i$. It confirms your result.
answered Dec 2 '18 at 17:18
callculuscallculus
18.1k31427
answered Dec 2 '18 at 17:18
callculuscallculus
18.1k31427
answered Dec 2 '18 at 17:18
callculuscallculus
18.1k31427
answered Dec 2 '18 at 17:18
callculuscallculus
18.1k31427
18.1k31427
$begingroup$
Thank you foir the graph, it helped alot!
$endgroup$
– K Split X
Dec 5 '18 at 21:28
$begingroup$
@KSplitX You´re welcome. I´m glad that it helped.
$endgroup$
– callculus
Dec 7 '18 at 20:33
add a comment |
$begingroup$
Thank you foir the graph, it helped alot!
$endgroup$
– K Split X
Dec 5 '18 at 21:28
$begingroup$
@KSplitX You´re welcome. I´m glad that it helped.
$endgroup$
– callculus
Dec 7 '18 at 20:33
$begingroup$
Thank you foir the graph, it helped alot!
$endgroup$
– K Split X
Dec 5 '18 at 21:28
$begingroup$
Thank you foir the graph, it helped alot!
$endgroup$
– K Split X
Dec 5 '18 at 21:28
$begingroup$
@KSplitX You´re welcome. I´m glad that it helped.
$endgroup$
– callculus
Dec 7 '18 at 20:33
$begingroup$
@KSplitX You´re welcome. I´m glad that it helped.
$endgroup$
– callculus
Dec 7 '18 at 20:33
add a comment |
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$begingroup$
It looks right to me. The pdf of $tilde x$ is more concentrated around the mean than $x_i$.
$endgroup$
– callculus
Dec 2 '18 at 16:10
$begingroup$
You've made everything completely right, but take care when $sigma^2/nneq 1$, then $$Z=frac{sqrt{n}}{sigma}Big(tilde{x}-muBig)$$ has a $N(0,1)$-distribution. The reason for that is $mathbb{E}(aX+b)=amathbb{E}(X)+b$ and $mathrm{Var}(aX+b)=a^2mathrm{Var}(X)$
$endgroup$
– Fakemistake
Dec 2 '18 at 18:50