Respuesta :
Answer:
a) [tex] P(0 <z<2.0) = P(Z<2.0)-P(Z<0)[/tex]
And we can use the following excel code to find the probability:
"=NORM.DIST(2,0,1,TRUE)-NORM.DIST(0,0,1,TRUE)"
[tex] P(0 <z<2.0) = P(Z<2.0)-P(Z<0)=0.47725[/tex]
b) [tex] P(Z>1.5) =1-P(Z<1.5) [/tex]
And we can use the following code and we got:
"=1-NORM.DIST(1.5,0,1,TRUE)"
[tex] P(Z>1.5) =1-P(Z<1.5)=0.0668[/tex]
c) [tex] P(z<-1.75)[/tex]
And we can use the following code and we got:
"=NORM.DIST(-1.75,0,1,TRUE)"
[tex] P(z<-1.75)=0.0400[/tex]
d) [tex] P(-2.78 <z<1.66) = P(Z<1.66)-P(Z<-2.78)[/tex]
And we can use the following excel code to find the probability:
"=NORM.DIST(1.66,0,1,TRUE)-NORM.DIST(-2.78,0,1,TRUE)"
[tex] P(-2.78 <z<1.66) = P(Z<1.66)-P(Z<-2.78)=0.949[/tex]
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
Part a
We want this probability:
[tex] P(0 <z<2.0) = P(Z<2.0)-P(Z<0)[/tex]
And we can use the following excel code to find the probability:
"=NORM.DIST(2,0,1,TRUE)-NORM.DIST(0,0,1,TRUE)"
[tex] P(0 <z<2.0) = P(Z<2.0)-P(Z<0)=0.47725[/tex]
Part b
For this case we want this probability:
[tex] P(Z>1.5)[/tex]
And we can use the complement rule and we have:
[tex] P(Z>1.5) =1-P(Z<1.5) [/tex]
And we can use the following code and we got:
"=1-NORM.DIST(1.5,0,1,TRUE)"
[tex] P(Z>1.5) =1-P(Z<1.5)=0.0668[/tex]
Part c
We want this probability:
[tex] P(z<-1.75)[/tex]
And we can use the following code and we got:
"=NORM.DIST(-1.75,0,1,TRUE)"
[tex] P(z<-1.75)=0.0400[/tex]
Part d
We want this probability:
[tex] P(-2.78 <z<1.66) = P(Z<1.66)-P(Z<-2.78)[/tex]
And we can use the following excel code to find the probability:
"=NORM.DIST(1.66,0,1,TRUE)-NORM.DIST(-2.78,0,1,TRUE)"
[tex] P(-2.78 <z<1.66) = P(Z<1.66)-P(Z<-2.78)=0.949[/tex]
The area under the standard normal distribution curve for the interval between z = 0 and z = 2.0 is; 0.47725
How to use the normal distribution table?
A) P(0 < z < 2)
From online p-value from two z-score calculator, we have;
p-value = (0.97725 - 0.5000) = 0.47725
B) P(z > 1.5)
From online p-value from z-score calculator, we have;
p-value = 0.0668
C) P(z < 1.75)
From online p-value from z-score calculator, we have;
p-value = 1 - 0.040059
p-value = 0.9599
D) P(-2.78 < z < 1.6)
From online p-value from two z-score calculator, we have;
p-value = 0.94882
Read more about Normal Distribution table at; https://brainly.com/question/4079902
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