Mixing
Find Z-score given probability
I have a problem solving this exercise. I have this:
$P(0 le Z le z_2) = 0.3$
$P(Z le z_1) = 0.3$
$P(z_1 le Z le z_2) = 0.8$
I need to find the $z$ values for each given probability. I already solved the first and the second like this:
I calculated the inverse standard normal distribution (with LibreOffice Calc) and I found that $z_2$ is $0.841$
I calculated the inverse standard normal distribution (with LibreOffice Calc) and I found that $z_1$ is $-0.524$
How can I find $z_1$ and $z_2$ of the third point of the exercise?
probability normal-distribution
asked Jan 30 '17 at 20:36
Cricco95
11113
add a comment |
I have a problem solving this exercise. I have this:
$P(0 le Z le z_2) = 0.3$
$P(Z le z_1) = 0.3$
$P(z_1 le Z le z_2) = 0.8$
I need to find the $z$ values for each given probability. I already solved the first and the second like this:
I calculated the inverse standard normal distribution (with LibreOffice Calc) and I found that $z_2$ is $0.841$
I calculated the inverse standard normal distribution (with LibreOffice Calc) and I found that $z_1$ is $-0.524$
How can I find $z_1$ and $z_2$ of the third point of the exercise?
probability normal-distribution
asked Jan 30 '17 at 20:36
Cricco95
11113
If $z_1$ and $z_2$ in the third line are supposed to be different from $z_1$ and $z_2$ in the first two lines, then you don't have enough information to find them.
– Michael Hardy
Jan 30 '17 at 21:36
add a comment |
I have a problem solving this exercise. I have this:
$P(0 le Z le z_2) = 0.3$
$P(Z le z_1) = 0.3$
$P(z_1 le Z le z_2) = 0.8$
I need to find the $z$ values for each given probability. I already solved the first and the second like this:
I calculated the inverse standard normal distribution (with LibreOffice Calc) and I found that $z_2$ is $0.841$
I calculated the inverse standard normal distribution (with LibreOffice Calc) and I found that $z_1$ is $-0.524$
How can I find $z_1$ and $z_2$ of the third point of the exercise?
probability normal-distribution
asked Jan 30 '17 at 20:36
Cricco95
11113
I have a problem solving this exercise. I have this:
$P(0 le Z le z_2) = 0.3$
$P(Z le z_1) = 0.3$
$P(z_1 le Z le z_2) = 0.8$
I need to find the $z$ values for each given probability. I already solved the first and the second like this:
I calculated the inverse standard normal distribution (with LibreOffice Calc) and I found that $z_2$ is $0.841$
I calculated the inverse standard normal distribution (with LibreOffice Calc) and I found that $z_1$ is $-0.524$
How can I find $z_1$ and $z_2$ of the third point of the exercise?
probability normal-distribution
probability normal-distribution
asked Jan 30 '17 at 20:36
Cricco95
11113
asked Jan 30 '17 at 20:36
Cricco95
11113
asked Jan 30 '17 at 20:36
Cricco95
11113
asked Jan 30 '17 at 20:36
Cricco95
11113
asked Jan 30 '17 at 20:36
Cricco95
11113
11113
If $z_1$ and $z_2$ in the third line are supposed to be different from $z_1$ and $z_2$ in the first two lines, then you don't have enough information to find them.
– Michael Hardy
Jan 30 '17 at 21:36
add a comment |
If $z_1$ and $z_2$ in the third line are supposed to be different from $z_1$ and $z_2$ in the first two lines, then you don't have enough information to find them.
– Michael Hardy
Jan 30 '17 at 21:36
If $z_1$ and $z_2$ in the third line are supposed to be different from $z_1$ and $z_2$ in the first two lines, then you don't have enough information to find them.
– Michael Hardy
Jan 30 '17 at 21:36
If $z_1$ and $z_2$ in the third line are supposed to be different from $z_1$ and $z_2$ in the first two lines, then you don't have enough information to find them.
– Michael Hardy
Jan 30 '17 at 21:36
add a comment |
1 Answer
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By using 1. and 2. , there is no solution
$$P(z_1 le Z le z_2)=P( Z le z_2)-P( Z le z_1)+P( Z le 0)-P( Z le 0)$$
$$= [P( Z le z_2)-P( Z le 0)]-P( Z le z_1)+P( Z le 0)$$
$$= P(0 le Z le z_2)-P( Z le z_1)+P( Z le 0)$$
We have that $P( Z le 0)=0.5$,$P(0 le Z le z_2)=0.3$ and $P(z_1 le Z le z_2)=0.8$, but we do not have $$0.8 ne 0.3-0.3+0.5$$
If we solve independently of question 1. and 2.
Let $(z_1,z_2) in mathbb{R^2}$ , we have $z_1 < z_2$ and
$$P(z_1 le Z le z_2) = Phi(z_2)-Phi(z_1)=0.8.$$
Obviously, $Phi(z_1)<0.2$ because $Phi(z_2)<1$, which imposes that $z_1<Phi^{-1}({0.2}) approx-0.81462$
Then, we have $$z_2=Phi^{-1}(0.8+Phi(z_1)).$$
Finally , the space of solutions is $${(z_1,z_2)in mathbb{R^2}|z_1<Phi^{-1}({0.2}),z_2=Phi^{-1}(0.8+Phi(z_1)) }$$
edited Jan 30 '17 at 21:36
Michael Hardy
1
answered Jan 30 '17 at 21:16
Canardini
2,4751519
add a comment |
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1 Answer
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By using 1. and 2. , there is no solution
$$P(z_1 le Z le z_2)=P( Z le z_2)-P( Z le z_1)+P( Z le 0)-P( Z le 0)$$
$$= [P( Z le z_2)-P( Z le 0)]-P( Z le z_1)+P( Z le 0)$$
$$= P(0 le Z le z_2)-P( Z le z_1)+P( Z le 0)$$
We have that $P( Z le 0)=0.5$,$P(0 le Z le z_2)=0.3$ and $P(z_1 le Z le z_2)=0.8$, but we do not have $$0.8 ne 0.3-0.3+0.5$$
If we solve independently of question 1. and 2.
Let $(z_1,z_2) in mathbb{R^2}$ , we have $z_1 < z_2$ and
$$P(z_1 le Z le z_2) = Phi(z_2)-Phi(z_1)=0.8.$$
Obviously, $Phi(z_1)<0.2$ because $Phi(z_2)<1$, which imposes that $z_1<Phi^{-1}({0.2}) approx-0.81462$
Then, we have $$z_2=Phi^{-1}(0.8+Phi(z_1)).$$
Finally , the space of solutions is $${(z_1,z_2)in mathbb{R^2}|z_1<Phi^{-1}({0.2}),z_2=Phi^{-1}(0.8+Phi(z_1)) }$$
edited Jan 30 '17 at 21:36
Michael Hardy
1
answered Jan 30 '17 at 21:16
Canardini
2,4751519
add a comment |
By using 1. and 2. , there is no solution
$$P(z_1 le Z le z_2)=P( Z le z_2)-P( Z le z_1)+P( Z le 0)-P( Z le 0)$$
$$= [P( Z le z_2)-P( Z le 0)]-P( Z le z_1)+P( Z le 0)$$
$$= P(0 le Z le z_2)-P( Z le z_1)+P( Z le 0)$$
We have that $P( Z le 0)=0.5$,$P(0 le Z le z_2)=0.3$ and $P(z_1 le Z le z_2)=0.8$, but we do not have $$0.8 ne 0.3-0.3+0.5$$
If we solve independently of question 1. and 2.
Let $(z_1,z_2) in mathbb{R^2}$ , we have $z_1 < z_2$ and
$$P(z_1 le Z le z_2) = Phi(z_2)-Phi(z_1)=0.8.$$
Obviously, $Phi(z_1)<0.2$ because $Phi(z_2)<1$, which imposes that $z_1<Phi^{-1}({0.2}) approx-0.81462$
Then, we have $$z_2=Phi^{-1}(0.8+Phi(z_1)).$$
Finally , the space of solutions is $${(z_1,z_2)in mathbb{R^2}|z_1<Phi^{-1}({0.2}),z_2=Phi^{-1}(0.8+Phi(z_1)) }$$
edited Jan 30 '17 at 21:36
Michael Hardy
1
answered Jan 30 '17 at 21:16
Canardini
2,4751519
add a comment |
By using 1. and 2. , there is no solution
$$P(z_1 le Z le z_2)=P( Z le z_2)-P( Z le z_1)+P( Z le 0)-P( Z le 0)$$
$$= [P( Z le z_2)-P( Z le 0)]-P( Z le z_1)+P( Z le 0)$$
$$= P(0 le Z le z_2)-P( Z le z_1)+P( Z le 0)$$
We have that $P( Z le 0)=0.5$,$P(0 le Z le z_2)=0.3$ and $P(z_1 le Z le z_2)=0.8$, but we do not have $$0.8 ne 0.3-0.3+0.5$$
If we solve independently of question 1. and 2.
Let $(z_1,z_2) in mathbb{R^2}$ , we have $z_1 < z_2$ and
$$P(z_1 le Z le z_2) = Phi(z_2)-Phi(z_1)=0.8.$$
Obviously, $Phi(z_1)<0.2$ because $Phi(z_2)<1$, which imposes that $z_1<Phi^{-1}({0.2}) approx-0.81462$
Then, we have $$z_2=Phi^{-1}(0.8+Phi(z_1)).$$
Finally , the space of solutions is $${(z_1,z_2)in mathbb{R^2}|z_1<Phi^{-1}({0.2}),z_2=Phi^{-1}(0.8+Phi(z_1)) }$$
edited Jan 30 '17 at 21:36
Michael Hardy
1
answered Jan 30 '17 at 21:16
Canardini
2,4751519
By using 1. and 2. , there is no solution
$$P(z_1 le Z le z_2)=P( Z le z_2)-P( Z le z_1)+P( Z le 0)-P( Z le 0)$$
$$= [P( Z le z_2)-P( Z le 0)]-P( Z le z_1)+P( Z le 0)$$
$$= P(0 le Z le z_2)-P( Z le z_1)+P( Z le 0)$$
We have that $P( Z le 0)=0.5$,$P(0 le Z le z_2)=0.3$ and $P(z_1 le Z le z_2)=0.8$, but we do not have $$0.8 ne 0.3-0.3+0.5$$
If we solve independently of question 1. and 2.
Let $(z_1,z_2) in mathbb{R^2}$ , we have $z_1 < z_2$ and
$$P(z_1 le Z le z_2) = Phi(z_2)-Phi(z_1)=0.8.$$
Obviously, $Phi(z_1)<0.2$ because $Phi(z_2)<1$, which imposes that $z_1<Phi^{-1}({0.2}) approx-0.81462$
Then, we have $$z_2=Phi^{-1}(0.8+Phi(z_1)).$$
Finally , the space of solutions is $${(z_1,z_2)in mathbb{R^2}|z_1<Phi^{-1}({0.2}),z_2=Phi^{-1}(0.8+Phi(z_1)) }$$
edited Jan 30 '17 at 21:36
Michael Hardy
1
answered Jan 30 '17 at 21:16
Canardini
2,4751519
edited Jan 30 '17 at 21:36
Michael Hardy
1
edited Jan 30 '17 at 21:36
Michael Hardy
1
edited Jan 30 '17 at 21:36
Michael Hardy
1
1
answered Jan 30 '17 at 21:16
Canardini
2,4751519
answered Jan 30 '17 at 21:16
Canardini
2,4751519
answered Jan 30 '17 at 21:16
Canardini
2,4751519
2,4751519
add a comment |
add a comment |
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If $z_1$ and $z_2$ in the third line are supposed to be different from $z_1$ and $z_2$ in the first two lines, then you don't have enough information to find them.
– Michael Hardy
Jan 30 '17 at 21:36