Gaussian distribution with absolute value
0
$begingroup$
I am doing my homework about continuous random variable and Im struggling with this problem :
Given a Gaussian random variable $T(85,10)$, find $c$ satisfying $mathbb{P}[|T| < c] = 0.9$.
Could you help me with this question? Thanks a lot in advance for your help!
probability absolute-value
edited Jan 16 at 16:10
gt6989b
34k22455
asked Jan 16 at 16:02
Nguyên ChươngNguyên Chương
51
$endgroup$
add a comment |
0
$begingroup$
I am doing my homework about continuous random variable and Im struggling with this problem :
Given a Gaussian random variable $T(85,10)$, find $c$ satisfying $mathbb{P}[|T| < c] = 0.9$.
Could you help me with this question? Thanks a lot in advance for your help!
probability absolute-value
edited Jan 16 at 16:10
gt6989b
34k22455
asked Jan 16 at 16:02
Nguyên ChươngNguyên Chương
51
$endgroup$
add a comment |
0
0
0
$begingroup$
I am doing my homework about continuous random variable and Im struggling with this problem :
Given a Gaussian random variable $T(85,10)$, find $c$ satisfying $mathbb{P}[|T| < c] = 0.9$.
Could you help me with this question? Thanks a lot in advance for your help!
probability absolute-value
edited Jan 16 at 16:10
gt6989b
34k22455
asked Jan 16 at 16:02
Nguyên ChươngNguyên Chương
51
$endgroup$
I am doing my homework about continuous random variable and Im struggling with this problem :
Given a Gaussian random variable $T(85,10)$, find $c$ satisfying $mathbb{P}[|T| < c] = 0.9$.
Could you help me with this question? Thanks a lot in advance for your help!
probability absolute-value
probability absolute-value
edited Jan 16 at 16:10
gt6989b
34k22455
asked Jan 16 at 16:02
Nguyên ChươngNguyên Chương
51
edited Jan 16 at 16:10
gt6989b
34k22455
asked Jan 16 at 16:02
Nguyên ChươngNguyên Chương
51
edited Jan 16 at 16:10
gt6989b
34k22455
edited Jan 16 at 16:10
gt6989b
34k22455
edited Jan 16 at 16:10
gt6989b
34k22455
34k22455
asked Jan 16 at 16:02
Nguyên ChươngNguyên Chương
51
asked Jan 16 at 16:02
Nguyên ChươngNguyên Chương
51
asked Jan 16 at 16:02
Nguyên ChươngNguyên Chương
51
51
add a comment |
add a comment |
1 Answer
1
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0
$begingroup$
HINT
- Can you find some transformation $X = (T-a)/b$ so that $X sim mathcal{N}(0,1)$ and $b>0$?
- Then $T = bX+a$ and your expression becomes
$$
begin{split}
0.9 &= mathbb{P}[|T| < c] \
&= mathbb{P}[-c < T < c] \
&= mathbb{P}[-c < bX+a < c] \
&= mathbb{P}left[frac{-c-a}{b} < X < frac{c-a}{b}right] \
&= Phileft(frac{c-a}{b}right) - Phileft(frac{-c-a}{b}right),
end{split}
$$
where $Phi$ is the standard normal CDF...
answered Jan 16 at 16:07
gt6989bgt6989b
34k22455
$endgroup$
$begingroup$
oh wow, I get it. thank you so much!
$endgroup$
– Nguyên Chương
Jan 16 at 16:15
$begingroup$
oops!! I still dont know how to find c, can you give me a hint ? At first, I thought that : because the probability is 0.9 then c must be greater than 85(because if c < 85 then P[|T|<c] < 0.5) , so -c < -85, then P[|T|<c] would approximately be P[T < c], which is 0.9, and then I can find c, but I'm not very sure that is the solution.
$endgroup$
– Nguyên Chương
Jan 16 at 16:40
$begingroup$
@NguyênChương do the first part, what are the values of $a$ and $b$?
$endgroup$
– gt6989b
Jan 16 at 17:24
$begingroup$
T = 10X + 85, is that right ?
$endgroup$
– Nguyên Chương
Jan 16 at 18:36
$begingroup$
@NguyênChương $b = sqrt{10}$ and $a$ is correct. Now plug them in, what do you get?
$endgroup$
– gt6989b
Jan 16 at 21:53
|
show 1 more comment
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1 Answer
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1 Answer
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0
$begingroup$
HINT
- Can you find some transformation $X = (T-a)/b$ so that $X sim mathcal{N}(0,1)$ and $b>0$?
- Then $T = bX+a$ and your expression becomes
$$
begin{split}
0.9 &= mathbb{P}[|T| < c] \
&= mathbb{P}[-c < T < c] \
&= mathbb{P}[-c < bX+a < c] \
&= mathbb{P}left[frac{-c-a}{b} < X < frac{c-a}{b}right] \
&= Phileft(frac{c-a}{b}right) - Phileft(frac{-c-a}{b}right),
end{split}
$$
where $Phi$ is the standard normal CDF...
answered Jan 16 at 16:07
gt6989bgt6989b
34k22455
$endgroup$
$begingroup$
oh wow, I get it. thank you so much!
$endgroup$
– Nguyên Chương
Jan 16 at 16:15
$begingroup$
oops!! I still dont know how to find c, can you give me a hint ? At first, I thought that : because the probability is 0.9 then c must be greater than 85(because if c < 85 then P[|T|<c] < 0.5) , so -c < -85, then P[|T|<c] would approximately be P[T < c], which is 0.9, and then I can find c, but I'm not very sure that is the solution.
$endgroup$
– Nguyên Chương
Jan 16 at 16:40
$begingroup$
@NguyênChương do the first part, what are the values of $a$ and $b$?
$endgroup$
– gt6989b
Jan 16 at 17:24
$begingroup$
T = 10X + 85, is that right ?
$endgroup$
– Nguyên Chương
Jan 16 at 18:36
$begingroup$
@NguyênChương $b = sqrt{10}$ and $a$ is correct. Now plug them in, what do you get?
$endgroup$
– gt6989b
Jan 16 at 21:53
|
show 1 more comment
0
$begingroup$
HINT
- Can you find some transformation $X = (T-a)/b$ so that $X sim mathcal{N}(0,1)$ and $b>0$?
- Then $T = bX+a$ and your expression becomes
$$
begin{split}
0.9 &= mathbb{P}[|T| < c] \
&= mathbb{P}[-c < T < c] \
&= mathbb{P}[-c < bX+a < c] \
&= mathbb{P}left[frac{-c-a}{b} < X < frac{c-a}{b}right] \
&= Phileft(frac{c-a}{b}right) - Phileft(frac{-c-a}{b}right),
end{split}
$$
where $Phi$ is the standard normal CDF...
answered Jan 16 at 16:07
gt6989bgt6989b
34k22455
$endgroup$
$begingroup$
oh wow, I get it. thank you so much!
$endgroup$
– Nguyên Chương
Jan 16 at 16:15
$begingroup$
oops!! I still dont know how to find c, can you give me a hint ? At first, I thought that : because the probability is 0.9 then c must be greater than 85(because if c < 85 then P[|T|<c] < 0.5) , so -c < -85, then P[|T|<c] would approximately be P[T < c], which is 0.9, and then I can find c, but I'm not very sure that is the solution.
$endgroup$
– Nguyên Chương
Jan 16 at 16:40
$begingroup$
@NguyênChương do the first part, what are the values of $a$ and $b$?
$endgroup$
– gt6989b
Jan 16 at 17:24
$begingroup$
T = 10X + 85, is that right ?
$endgroup$
– Nguyên Chương
Jan 16 at 18:36
$begingroup$
@NguyênChương $b = sqrt{10}$ and $a$ is correct. Now plug them in, what do you get?
$endgroup$
– gt6989b
Jan 16 at 21:53
|
show 1 more comment
0
0
0
$begingroup$
HINT
- Can you find some transformation $X = (T-a)/b$ so that $X sim mathcal{N}(0,1)$ and $b>0$?
- Then $T = bX+a$ and your expression becomes
$$
begin{split}
0.9 &= mathbb{P}[|T| < c] \
&= mathbb{P}[-c < T < c] \
&= mathbb{P}[-c < bX+a < c] \
&= mathbb{P}left[frac{-c-a}{b} < X < frac{c-a}{b}right] \
&= Phileft(frac{c-a}{b}right) - Phileft(frac{-c-a}{b}right),
end{split}
$$
where $Phi$ is the standard normal CDF...
answered Jan 16 at 16:07
gt6989bgt6989b
34k22455
$endgroup$
HINT
- Can you find some transformation $X = (T-a)/b$ so that $X sim mathcal{N}(0,1)$ and $b>0$?
- Then $T = bX+a$ and your expression becomes
$$
begin{split}
0.9 &= mathbb{P}[|T| < c] \
&= mathbb{P}[-c < T < c] \
&= mathbb{P}[-c < bX+a < c] \
&= mathbb{P}left[frac{-c-a}{b} < X < frac{c-a}{b}right] \
&= Phileft(frac{c-a}{b}right) - Phileft(frac{-c-a}{b}right),
end{split}
$$
where $Phi$ is the standard normal CDF...
answered Jan 16 at 16:07
gt6989bgt6989b
34k22455
answered Jan 16 at 16:07
gt6989bgt6989b
34k22455
answered Jan 16 at 16:07
gt6989bgt6989b
34k22455
answered Jan 16 at 16:07
gt6989bgt6989b
34k22455
34k22455
$begingroup$
oh wow, I get it. thank you so much!
$endgroup$
– Nguyên Chương
Jan 16 at 16:15
$begingroup$
oops!! I still dont know how to find c, can you give me a hint ? At first, I thought that : because the probability is 0.9 then c must be greater than 85(because if c < 85 then P[|T|<c] < 0.5) , so -c < -85, then P[|T|<c] would approximately be P[T < c], which is 0.9, and then I can find c, but I'm not very sure that is the solution.
$endgroup$
– Nguyên Chương
Jan 16 at 16:40
$begingroup$
@NguyênChương do the first part, what are the values of $a$ and $b$?
$endgroup$
– gt6989b
Jan 16 at 17:24
$begingroup$
T = 10X + 85, is that right ?
$endgroup$
– Nguyên Chương
Jan 16 at 18:36
$begingroup$
@NguyênChương $b = sqrt{10}$ and $a$ is correct. Now plug them in, what do you get?
$endgroup$
– gt6989b
Jan 16 at 21:53
|
show 1 more comment
$begingroup$
oh wow, I get it. thank you so much!
$endgroup$
– Nguyên Chương
Jan 16 at 16:15
$begingroup$
oops!! I still dont know how to find c, can you give me a hint ? At first, I thought that : because the probability is 0.9 then c must be greater than 85(because if c < 85 then P[|T|<c] < 0.5) , so -c < -85, then P[|T|<c] would approximately be P[T < c], which is 0.9, and then I can find c, but I'm not very sure that is the solution.
$endgroup$
– Nguyên Chương
Jan 16 at 16:40
$begingroup$
@NguyênChương do the first part, what are the values of $a$ and $b$?
$endgroup$
– gt6989b
Jan 16 at 17:24
$begingroup$
T = 10X + 85, is that right ?
$endgroup$
– Nguyên Chương
Jan 16 at 18:36
$begingroup$
@NguyênChương $b = sqrt{10}$ and $a$ is correct. Now plug them in, what do you get?
$endgroup$
– gt6989b
Jan 16 at 21:53
$begingroup$
oh wow, I get it. thank you so much!
$endgroup$
– Nguyên Chương
Jan 16 at 16:15
$begingroup$
oh wow, I get it. thank you so much!
$endgroup$
– Nguyên Chương
Jan 16 at 16:15
$begingroup$
oops!! I still dont know how to find c, can you give me a hint ? At first, I thought that : because the probability is 0.9 then c must be greater than 85(because if c < 85 then P[|T|<c] < 0.5) , so -c < -85, then P[|T|<c] would approximately be P[T < c], which is 0.9, and then I can find c, but I'm not very sure that is the solution.
$endgroup$
– Nguyên Chương
Jan 16 at 16:40
$begingroup$
oops!! I still dont know how to find c, can you give me a hint ? At first, I thought that : because the probability is 0.9 then c must be greater than 85(because if c < 85 then P[|T|<c] < 0.5) , so -c < -85, then P[|T|<c] would approximately be P[T < c], which is 0.9, and then I can find c, but I'm not very sure that is the solution.
$endgroup$
– Nguyên Chương
Jan 16 at 16:40
$begingroup$
@NguyênChương do the first part, what are the values of $a$ and $b$?
$endgroup$
– gt6989b
Jan 16 at 17:24
$begingroup$
@NguyênChương do the first part, what are the values of $a$ and $b$?
$endgroup$
– gt6989b
Jan 16 at 17:24
$begingroup$
T = 10X + 85, is that right ?
$endgroup$
– Nguyên Chương
Jan 16 at 18:36
$begingroup$
T = 10X + 85, is that right ?
$endgroup$
– Nguyên Chương
Jan 16 at 18:36
$begingroup$
@NguyênChương $b = sqrt{10}$ and $a$ is correct. Now plug them in, what do you get?
$endgroup$
– gt6989b
Jan 16 at 21:53
$begingroup$
@NguyênChương $b = sqrt{10}$ and $a$ is correct. Now plug them in, what do you get?
$endgroup$
– gt6989b
Jan 16 at 21:53
|
show 1 more comment
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