Mixing
Normal Distribution Quantiles - Is my solution correct?
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Probability and statistics exam coming up and I have no way to check if my answers are correct, so bear with me, please :) Could someone please check if the below solution is correct? Thank you so much!
X is a normally distributed random variable with expected value 3 and standard deviation 2. Determine the 0.95-quantile for X.
In other words, what we have is, E(X)=$mu$=3 and $sigma$=2
Using that for Normal distribution z$_p$ = $mu$ + $sigma$z$_p$.
We obtain z$_p$=3 +2*0.8289 (from the Standard Normal Distribution
Table) = 4.6578
probability statistics normal-distribution
asked Feb 2 at 16:41
VRTVRT
957
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show 2 more comments
$begingroup$
Probability and statistics exam coming up and I have no way to check if my answers are correct, so bear with me, please :) Could someone please check if the below solution is correct? Thank you so much!
X is a normally distributed random variable with expected value 3 and standard deviation 2. Determine the 0.95-quantile for X.
In other words, what we have is, E(X)=$mu$=3 and $sigma$=2
Using that for Normal distribution z$_p$ = $mu$ + $sigma$z$_p$.
We obtain z$_p$=3 +2*0.8289 (from the Standard Normal Distribution
Table) = 4.6578
probability statistics normal-distribution
asked Feb 2 at 16:41
VRTVRT
957
$endgroup$
$begingroup$
You have an expression $z_p=mu+sigma z_p$. Fix it. The final answer looks correct, although I haven't checked the numbers.
$endgroup$
– herb steinberg
Feb 2 at 17:40
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I dont know how to insert the monkey tail symbol above the first zp :/ But thank you tof rchecking!
$endgroup$
– VRT
Feb 2 at 17:49
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@VRT I don´t know how do you get the value of $z_{0.95}$. You should read of the table $z_{0.95}=1.645$ But your equation is right.
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– callculus
Feb 2 at 17:58
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@callculus It comes from $Phi(0.95) approx 0.8289$ when $Phi^{-1}(0.95) approx 1.645$ should be used
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– Henry
Feb 2 at 18:19
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In R statistical softwareqnorm(.95, 3, 2)returns 6.289707. You can come very close with printed tables. $P((X-3)/2 < (c-3)/2) = P(Z < (c-3)/2)) = .95,$ where Z is standard normal. So from a printed table you can get that $(c-3)/2 = 1.645.$ Solve for $c.$ // If you don't mind, I rather 'bear' with you than 'bare'.
$endgroup$
– BruceET
Feb 3 at 3:16
|
show 2 more comments
$begingroup$
Probability and statistics exam coming up and I have no way to check if my answers are correct, so bear with me, please :) Could someone please check if the below solution is correct? Thank you so much!
X is a normally distributed random variable with expected value 3 and standard deviation 2. Determine the 0.95-quantile for X.
In other words, what we have is, E(X)=$mu$=3 and $sigma$=2
Using that for Normal distribution z$_p$ = $mu$ + $sigma$z$_p$.
We obtain z$_p$=3 +2*0.8289 (from the Standard Normal Distribution
Table) = 4.6578
probability statistics normal-distribution
asked Feb 2 at 16:41
VRTVRT
957
$endgroup$
Probability and statistics exam coming up and I have no way to check if my answers are correct, so bear with me, please :) Could someone please check if the below solution is correct? Thank you so much!
X is a normally distributed random variable with expected value 3 and standard deviation 2. Determine the 0.95-quantile for X.
In other words, what we have is, E(X)=$mu$=3 and $sigma$=2
Using that for Normal distribution z$_p$ = $mu$ + $sigma$z$_p$.
We obtain z$_p$=3 +2*0.8289 (from the Standard Normal Distribution
Table) = 4.6578
probability statistics normal-distribution
probability statistics normal-distribution
asked Feb 2 at 16:41
VRTVRT
957
asked Feb 2 at 16:41
VRTVRT
957
edited Feb 3 at 12:21
VRT
asked Feb 2 at 16:41
VRTVRT
957
asked Feb 2 at 16:41
VRTVRT
957
asked Feb 2 at 16:41
VRTVRT
957
957
$begingroup$
You have an expression $z_p=mu+sigma z_p$. Fix it. The final answer looks correct, although I haven't checked the numbers.
$endgroup$
– herb steinberg
Feb 2 at 17:40
$begingroup$
I dont know how to insert the monkey tail symbol above the first zp :/ But thank you tof rchecking!
$endgroup$
– VRT
Feb 2 at 17:49
$begingroup$
@VRT I don´t know how do you get the value of $z_{0.95}$. You should read of the table $z_{0.95}=1.645$ But your equation is right.
$endgroup$
– callculus
Feb 2 at 17:58
$begingroup$
@callculus It comes from $Phi(0.95) approx 0.8289$ when $Phi^{-1}(0.95) approx 1.645$ should be used
$endgroup$
– Henry
Feb 2 at 18:19
$begingroup$
In R statistical softwareqnorm(.95, 3, 2)returns 6.289707. You can come very close with printed tables. $P((X-3)/2 < (c-3)/2) = P(Z < (c-3)/2)) = .95,$ where Z is standard normal. So from a printed table you can get that $(c-3)/2 = 1.645.$ Solve for $c.$ // If you don't mind, I rather 'bear' with you than 'bare'.
$endgroup$
– BruceET
Feb 3 at 3:16
|
show 2 more comments
$begingroup$
You have an expression $z_p=mu+sigma z_p$. Fix it. The final answer looks correct, although I haven't checked the numbers.
$endgroup$
– herb steinberg
Feb 2 at 17:40
$begingroup$
I dont know how to insert the monkey tail symbol above the first zp :/ But thank you tof rchecking!
$endgroup$
– VRT
Feb 2 at 17:49
$begingroup$
@VRT I don´t know how do you get the value of $z_{0.95}$. You should read of the table $z_{0.95}=1.645$ But your equation is right.
$endgroup$
– callculus
Feb 2 at 17:58
$begingroup$
@callculus It comes from $Phi(0.95) approx 0.8289$ when $Phi^{-1}(0.95) approx 1.645$ should be used
$endgroup$
– Henry
Feb 2 at 18:19
$begingroup$
In R statistical softwareqnorm(.95, 3, 2)returns 6.289707. You can come very close with printed tables. $P((X-3)/2 < (c-3)/2) = P(Z < (c-3)/2)) = .95,$ where Z is standard normal. So from a printed table you can get that $(c-3)/2 = 1.645.$ Solve for $c.$ // If you don't mind, I rather 'bear' with you than 'bare'.
$endgroup$
– BruceET
Feb 3 at 3:16
$begingroup$
You have an expression $z_p=mu+sigma z_p$. Fix it. The final answer looks correct, although I haven't checked the numbers.
$endgroup$
– herb steinberg
Feb 2 at 17:40
$begingroup$
You have an expression $z_p=mu+sigma z_p$. Fix it. The final answer looks correct, although I haven't checked the numbers.
$endgroup$
– herb steinberg
Feb 2 at 17:40
$begingroup$
I dont know how to insert the monkey tail symbol above the first zp :/ But thank you tof rchecking!
$endgroup$
– VRT
Feb 2 at 17:49
$begingroup$
I dont know how to insert the monkey tail symbol above the first zp :/ But thank you tof rchecking!
$endgroup$
– VRT
Feb 2 at 17:49
$begingroup$
@VRT I don´t know how do you get the value of $z_{0.95}$. You should read of the table $z_{0.95}=1.645$ But your equation is right.
$endgroup$
– callculus
Feb 2 at 17:58
$begingroup$
@VRT I don´t know how do you get the value of $z_{0.95}$. You should read of the table $z_{0.95}=1.645$ But your equation is right.
$endgroup$
– callculus
Feb 2 at 17:58
$begingroup$
@callculus It comes from $Phi(0.95) approx 0.8289$ when $Phi^{-1}(0.95) approx 1.645$ should be used
$endgroup$
– Henry
Feb 2 at 18:19
$begingroup$
@callculus It comes from $Phi(0.95) approx 0.8289$ when $Phi^{-1}(0.95) approx 1.645$ should be used
$endgroup$
– Henry
Feb 2 at 18:19
$begingroup$
In R statistical software
qnorm(.95, 3, 2) returns 6.289707. You can come very close with printed tables. $P((X-3)/2 < (c-3)/2) = P(Z < (c-3)/2)) = .95,$ where Z is standard normal. So from a printed table you can get that $(c-3)/2 = 1.645.$ Solve for $c.$ // If you don't mind, I rather 'bear' with you than 'bare'.$endgroup$
– BruceET
Feb 3 at 3:16
$begingroup$
In R statistical software
qnorm(.95, 3, 2) returns 6.289707. You can come very close with printed tables. $P((X-3)/2 < (c-3)/2) = P(Z < (c-3)/2)) = .95,$ where Z is standard normal. So from a printed table you can get that $(c-3)/2 = 1.645.$ Solve for $c.$ // If you don't mind, I rather 'bear' with you than 'bare'.$endgroup$
– BruceET
Feb 3 at 3:16
|
show 2 more comments
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$begingroup$
You have an expression $z_p=mu+sigma z_p$. Fix it. The final answer looks correct, although I haven't checked the numbers.
$endgroup$
– herb steinberg
Feb 2 at 17:40
$begingroup$
I dont know how to insert the monkey tail symbol above the first zp :/ But thank you tof rchecking!
$endgroup$
– VRT
Feb 2 at 17:49
$begingroup$
@VRT I don´t know how do you get the value of $z_{0.95}$. You should read of the table $z_{0.95}=1.645$ But your equation is right.
$endgroup$
– callculus
Feb 2 at 17:58
$begingroup$
@callculus It comes from $Phi(0.95) approx 0.8289$ when $Phi^{-1}(0.95) approx 1.645$ should be used
$endgroup$
– Henry
Feb 2 at 18:19
$begingroup$
In R statistical software
qnorm(.95, 3, 2)returns 6.289707. You can come very close with printed tables. $P((X-3)/2 < (c-3)/2) = P(Z < (c-3)/2)) = .95,$ where Z is standard normal. So from a printed table you can get that $(c-3)/2 = 1.645.$ Solve for $c.$ // If you don't mind, I rather 'bear' with you than 'bare'.$endgroup$
– BruceET
Feb 3 at 3:16