Respuesta :
Answer:
[tex] P(0<Z<z) =P(Z<z)-P(Z<0) = P(Z<z)-0.5= 0.4750[/tex]
And solving for z we have
[tex]P(Z<z)= 0.4750+0.5= 0.9750[/tex]
And we can find the value for z with the following excel code:
"=NORM.INV(0.975,0,1)"
And we got z =1.96
[tex] P(Z>z)= 0.1314[/tex]
And we can use the complement rule and we got:
[tex] P(Z>z) = 1-P(Z<z) = 0.1314[/tex]
[tex] P(Z<z)= 1-0.1314= 0.8686[/tex]
And we can find the value for z with the following excel code:
"=NORM.INV(0.8686,0,1)"
And we got z =1.120
[tex] P(Z<z)= 0.67[/tex]
And we can find the value for z with the following excel code:
"=NORM.INV(0.67,0,1)"
And we got z =0.440
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
We want this probability:
[tex] P(0<Z<z) =P(Z<z)-P(Z<0) = P(Z<z)-0.5= 0.4750[/tex]
And solving for z we have
[tex]P(Z<z)= 0.4750+0.5= 0.9750[/tex]
And we can find the value for z with the following excel code:
"=NORM.INV(0.975,0,1)"
And we got z =1.96
For the next part we want to calculate:
[tex] P(Z>z)= 0.1314[/tex]
And we can use the complement rule and we got:
[tex] P(Z>z) = 1-P(Z<z) = 0.1314[/tex]
[tex] P(Z<z)= 1-0.1314= 0.8686[/tex]
And we can find the value for z with the following excel code:
"=NORM.INV(0.8686,0,1)"
And we got z =1.120
For the next part we want to calculate:
[tex] P(Z<z)= 0.67[/tex]
And we can find the value for z with the following excel code:
"=NORM.INV(0.67,0,1)"
And we got z =0.440
