Central Limit Theorem Bounds
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I am working on a Central Limit Theorem problem and am a bit stuck.
The problem:
N = 10000
Xi = X ~ Exp(1)
Yi = Y ~ Exp(2)
All independent, find P(total amount is between 7400 and 7560)
What I have done so far:
Sn = X1 + ... + x5000 + Y1 + ... + Y5000
E[xi] = 1 Var[xi] = 1
E[yi] = $1over2$ Var[yi] = $1over4$
E[Sn] = 5000 (1) + 5000 ($1over2$) = 7500
Var[Sn] = 5000 (1) + 5000 ($1over4$) = 6250
Central Limit Theorem:
$Sn - 7500 over sqrt(6250)$ ~ N(0,1)
But here is where I am stuck.
I know that that we need to put this in between the bounds of a and b (ie. 7400 and 7560) but how exactly do we do this? I know we need to use the Z - Table for some values but we are not exactly using a confidence interval? Do we just literally place it between these two bounds?
probability central-limit-theorem
asked Dec 1 at 20:47
Ethan
10012
|
show 4 more comments
up vote
0
down vote
favorite
I am working on a Central Limit Theorem problem and am a bit stuck.
The problem:
N = 10000
Xi = X ~ Exp(1)
Yi = Y ~ Exp(2)
All independent, find P(total amount is between 7400 and 7560)
What I have done so far:
Sn = X1 + ... + x5000 + Y1 + ... + Y5000
E[xi] = 1 Var[xi] = 1
E[yi] = $1over2$ Var[yi] = $1over4$
E[Sn] = 5000 (1) + 5000 ($1over2$) = 7500
Var[Sn] = 5000 (1) + 5000 ($1over4$) = 6250
Central Limit Theorem:
$Sn - 7500 over sqrt(6250)$ ~ N(0,1)
But here is where I am stuck.
I know that that we need to put this in between the bounds of a and b (ie. 7400 and 7560) but how exactly do we do this? I know we need to use the Z - Table for some values but we are not exactly using a confidence interval? Do we just literally place it between these two bounds?
probability central-limit-theorem
asked Dec 1 at 20:47
Ethan
10012
For a continuous random variable X we have $P(aleq Xleq b)=P(Xleq b)-P(Xleq a)$.
– callculus
Dec 1 at 21:26
We aren't give an confidence number though to use a z score for finding the P?
– Ethan
Dec 1 at 22:17
What do you mean by confidence number? Can you rephrase your question?
– callculus
Dec 1 at 23:12
1
I used the wiki-table. The z-value for $P(S_nleq 7400)$ is $z=frac{7400-7500}{sqrt{6250}}approx-1.265$ If you have no negative z-values at you table you take the positive one. At $z=1.26$ we have $p=0.39617$ and at $z=1.27$ we have $p=0.39796$ we calculate the arithmetic mean to obtain (approximately) $Phi(1.265)=39.71%$
– callculus
Dec 1 at 23:54
1
I used the wiki-table (cumulative). The z-value for $P(S_nleq 7400)$ is $z=frac{7400-7500}{sqrt{6250}}approx-1.265$ If you have no negative z-values at your table you take the positive one. At $z=1.26$ we have $p=0.89617$ and at $z=1.27$ we have $p=0.89796$ we calculate the arithmetic mean to obtain (approximately) $Phi(1.265)=89.71%$. Due symmetry of the normal distribution we have $Phi(-1.265)=1-Phi(1.265)=1-0.8971=10.29%$
– callculus
Dec 2 at 0:03
|
show 4 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am working on a Central Limit Theorem problem and am a bit stuck.
The problem:
N = 10000
Xi = X ~ Exp(1)
Yi = Y ~ Exp(2)
All independent, find P(total amount is between 7400 and 7560)
What I have done so far:
Sn = X1 + ... + x5000 + Y1 + ... + Y5000
E[xi] = 1 Var[xi] = 1
E[yi] = $1over2$ Var[yi] = $1over4$
E[Sn] = 5000 (1) + 5000 ($1over2$) = 7500
Var[Sn] = 5000 (1) + 5000 ($1over4$) = 6250
Central Limit Theorem:
$Sn - 7500 over sqrt(6250)$ ~ N(0,1)
But here is where I am stuck.
I know that that we need to put this in between the bounds of a and b (ie. 7400 and 7560) but how exactly do we do this? I know we need to use the Z - Table for some values but we are not exactly using a confidence interval? Do we just literally place it between these two bounds?
probability central-limit-theorem
asked Dec 1 at 20:47
Ethan
10012
I am working on a Central Limit Theorem problem and am a bit stuck.
The problem:
N = 10000
Xi = X ~ Exp(1)
Yi = Y ~ Exp(2)
All independent, find P(total amount is between 7400 and 7560)
What I have done so far:
Sn = X1 + ... + x5000 + Y1 + ... + Y5000
E[xi] = 1 Var[xi] = 1
E[yi] = $1over2$ Var[yi] = $1over4$
E[Sn] = 5000 (1) + 5000 ($1over2$) = 7500
Var[Sn] = 5000 (1) + 5000 ($1over4$) = 6250
Central Limit Theorem:
$Sn - 7500 over sqrt(6250)$ ~ N(0,1)
But here is where I am stuck.
I know that that we need to put this in between the bounds of a and b (ie. 7400 and 7560) but how exactly do we do this? I know we need to use the Z - Table for some values but we are not exactly using a confidence interval? Do we just literally place it between these two bounds?
probability central-limit-theorem
probability central-limit-theorem
asked Dec 1 at 20:47
Ethan
10012
asked Dec 1 at 20:47
Ethan
10012
asked Dec 1 at 20:47
Ethan
10012
asked Dec 1 at 20:47
Ethan
10012
asked Dec 1 at 20:47
Ethan
10012
10012
For a continuous random variable X we have $P(aleq Xleq b)=P(Xleq b)-P(Xleq a)$.
– callculus
Dec 1 at 21:26
We aren't give an confidence number though to use a z score for finding the P?
– Ethan
Dec 1 at 22:17
What do you mean by confidence number? Can you rephrase your question?
– callculus
Dec 1 at 23:12
1
I used the wiki-table. The z-value for $P(S_nleq 7400)$ is $z=frac{7400-7500}{sqrt{6250}}approx-1.265$ If you have no negative z-values at you table you take the positive one. At $z=1.26$ we have $p=0.39617$ and at $z=1.27$ we have $p=0.39796$ we calculate the arithmetic mean to obtain (approximately) $Phi(1.265)=39.71%$
– callculus
Dec 1 at 23:54
1
I used the wiki-table (cumulative). The z-value for $P(S_nleq 7400)$ is $z=frac{7400-7500}{sqrt{6250}}approx-1.265$ If you have no negative z-values at your table you take the positive one. At $z=1.26$ we have $p=0.89617$ and at $z=1.27$ we have $p=0.89796$ we calculate the arithmetic mean to obtain (approximately) $Phi(1.265)=89.71%$. Due symmetry of the normal distribution we have $Phi(-1.265)=1-Phi(1.265)=1-0.8971=10.29%$
– callculus
Dec 2 at 0:03
|
show 4 more comments
For a continuous random variable X we have $P(aleq Xleq b)=P(Xleq b)-P(Xleq a)$.
– callculus
Dec 1 at 21:26
We aren't give an confidence number though to use a z score for finding the P?
– Ethan
Dec 1 at 22:17
What do you mean by confidence number? Can you rephrase your question?
– callculus
Dec 1 at 23:12
1
I used the wiki-table. The z-value for $P(S_nleq 7400)$ is $z=frac{7400-7500}{sqrt{6250}}approx-1.265$ If you have no negative z-values at you table you take the positive one. At $z=1.26$ we have $p=0.39617$ and at $z=1.27$ we have $p=0.39796$ we calculate the arithmetic mean to obtain (approximately) $Phi(1.265)=39.71%$
– callculus
Dec 1 at 23:54
1
I used the wiki-table (cumulative). The z-value for $P(S_nleq 7400)$ is $z=frac{7400-7500}{sqrt{6250}}approx-1.265$ If you have no negative z-values at your table you take the positive one. At $z=1.26$ we have $p=0.89617$ and at $z=1.27$ we have $p=0.89796$ we calculate the arithmetic mean to obtain (approximately) $Phi(1.265)=89.71%$. Due symmetry of the normal distribution we have $Phi(-1.265)=1-Phi(1.265)=1-0.8971=10.29%$
– callculus
Dec 2 at 0:03
For a continuous random variable X we have $P(aleq Xleq b)=P(Xleq b)-P(Xleq a)$.
– callculus
Dec 1 at 21:26
For a continuous random variable X we have $P(aleq Xleq b)=P(Xleq b)-P(Xleq a)$.
– callculus
Dec 1 at 21:26
We aren't give an confidence number though to use a z score for finding the P?
– Ethan
Dec 1 at 22:17
We aren't give an confidence number though to use a z score for finding the P?
– Ethan
Dec 1 at 22:17
What do you mean by confidence number? Can you rephrase your question?
– callculus
Dec 1 at 23:12
What do you mean by confidence number? Can you rephrase your question?
– callculus
Dec 1 at 23:12
1
1
I used the wiki-table. The z-value for $P(S_nleq 7400)$ is $z=frac{7400-7500}{sqrt{6250}}approx-1.265$ If you have no negative z-values at you table you take the positive one. At $z=1.26$ we have $p=0.39617$ and at $z=1.27$ we have $p=0.39796$ we calculate the arithmetic mean to obtain (approximately) $Phi(1.265)=39.71%$
– callculus
Dec 1 at 23:54
I used the wiki-table. The z-value for $P(S_nleq 7400)$ is $z=frac{7400-7500}{sqrt{6250}}approx-1.265$ If you have no negative z-values at you table you take the positive one. At $z=1.26$ we have $p=0.39617$ and at $z=1.27$ we have $p=0.39796$ we calculate the arithmetic mean to obtain (approximately) $Phi(1.265)=39.71%$
– callculus
Dec 1 at 23:54
1
1
I used the wiki-table (cumulative). The z-value for $P(S_nleq 7400)$ is $z=frac{7400-7500}{sqrt{6250}}approx-1.265$ If you have no negative z-values at your table you take the positive one. At $z=1.26$ we have $p=0.89617$ and at $z=1.27$ we have $p=0.89796$ we calculate the arithmetic mean to obtain (approximately) $Phi(1.265)=89.71%$. Due symmetry of the normal distribution we have $Phi(-1.265)=1-Phi(1.265)=1-0.8971=10.29%$
– callculus
Dec 2 at 0:03
I used the wiki-table (cumulative). The z-value for $P(S_nleq 7400)$ is $z=frac{7400-7500}{sqrt{6250}}approx-1.265$ If you have no negative z-values at your table you take the positive one. At $z=1.26$ we have $p=0.89617$ and at $z=1.27$ we have $p=0.89796$ we calculate the arithmetic mean to obtain (approximately) $Phi(1.265)=89.71%$. Due symmetry of the normal distribution we have $Phi(-1.265)=1-Phi(1.265)=1-0.8971=10.29%$
– callculus
Dec 2 at 0:03
|
show 4 more comments
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For a continuous random variable X we have $P(aleq Xleq b)=P(Xleq b)-P(Xleq a)$.
– callculus
Dec 1 at 21:26
We aren't give an confidence number though to use a z score for finding the P?
– Ethan
Dec 1 at 22:17
What do you mean by confidence number? Can you rephrase your question?
– callculus
Dec 1 at 23:12
1
I used the wiki-table. The z-value for $P(S_nleq 7400)$ is $z=frac{7400-7500}{sqrt{6250}}approx-1.265$ If you have no negative z-values at you table you take the positive one. At $z=1.26$ we have $p=0.39617$ and at $z=1.27$ we have $p=0.39796$ we calculate the arithmetic mean to obtain (approximately) $Phi(1.265)=39.71%$
– callculus
Dec 1 at 23:54
1
I used the wiki-table (cumulative). The z-value for $P(S_nleq 7400)$ is $z=frac{7400-7500}{sqrt{6250}}approx-1.265$ If you have no negative z-values at your table you take the positive one. At $z=1.26$ we have $p=0.89617$ and at $z=1.27$ we have $p=0.89796$ we calculate the arithmetic mean to obtain (approximately) $Phi(1.265)=89.71%$. Due symmetry of the normal distribution we have $Phi(-1.265)=1-Phi(1.265)=1-0.8971=10.29%$
– callculus
Dec 2 at 0:03