I have a problem solving this exercise. I have this: P(0<=Z<=z_2)=0.3 P(Z<=z_1)=0.3 P(z_1<=Z<=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?

limunom623

limunom623

Answered question

2022-11-02

I have a problem solving this exercise. I have this:
1. P ( 0 Z z 2 ) = 0.3
2. P ( Z z 1 ) = 0.3
3. P ( z 1 Z 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:
1. I calculated the inverse standard normal distribution (with LibreOffice Calc) and I found that z 2 is 0.841
2. 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?

Answer & Explanation

Regan Holloway

Regan Holloway

Beginner2022-11-03Added 17 answers

By using 1. and 2. , there is no solution
P ( z 1 Z z 2 ) = P ( Z z 2 ) P ( Z z 1 ) + P ( Z 0 ) P ( Z 0 )
= [ P ( Z z 2 ) P ( Z 0 ) ] P ( Z z 1 ) + P ( Z 0 )
= P ( 0 Z z 2 ) P ( Z z 1 ) + P ( Z 0 )
We have that P ( Z 0 ) = 0.5, P ( 0 Z z 2 ) = 0.3 and P ( z 1 Z z 2 ) = 0.8, but we do not have
0.8 0.3 0.3 + 0.5
If we solve independently of question 1. and 2.

Let ( z 1 , z 2 ) R 2 , we have z 1 < z 2 and
P ( z 1 Z z 2 ) = Φ ( z 2 ) Φ ( z 1 ) = 0.8.
Obviously, Φ ( z 1 ) < 0.2 because Φ ( z 2 ) < 1, which imposes that z 1 < Φ 1 ( 0.2 ) 0.81462
Then, we have
z 2 = Φ 1 ( 0.8 + Φ ( z 1 ) ) .
Finally, the space of solutions is
{ ( z 1 , z 2 ) R 2 | z 1 < Φ 1 ( 0.2 ) , z 2 = Φ 1 ( 0.8 + Φ ( z 1 ) ) }

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