Answer:
a) [tex] P(-1<Z<1)= P(Z<1) -P(Z<-1)= 0.841-0.159= 0.682[/tex]
b) [tex] P(-2<Z<2)= P(Z<2) -P(Z<-2)= 0.977-0.0228= 0.954[/tex]
c) [tex] P(-3<Z<3)= P(Z<3) -P(Z<-3)= 0.999-0.0013= 0.998[/tex]
The results are on the fogure attached.
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
For this case we want to find this probability:
[tex] P(-1<Z<1)[/tex]
And we can find this probability with this difference:
[tex] P(-1<Z<1)= P(Z<1) -P(Z<-1)[/tex]
And if we find the probability using the normal standard distribution or excel we got:
[tex] P(-1<Z<1)= P(Z<1) -P(Z<-1)= 0.841-0.159= 0.682[/tex]
Part b
For this case we want to find this probability:
[tex] P(-2<Z<2)[/tex]
And we can find this probability with this difference:
[tex] P(-2<Z<2)= P(Z<2) -P(Z<-2)[/tex]
And if we find the probability using the normal standard distribution or excel we got:
[tex] P(-2<Z<2)= P(Z<2) -P(Z<-2)= 0.977-0.0228= 0.954[/tex]
Part c
For this case we want to find this probability:
[tex] P(-3<Z<3)[/tex]
And we can find this probability with this difference:
[tex] P(-3<Z<3)= P(Z<3) -P(Z<-3)[/tex]
And if we find the probability using the normal standard distribution or excel we got:
[tex] P(-3<Z<3)= P(Z<3) -P(Z<-3)= 0.999-0.0013= 0.998[/tex]
