Answer:
P(45<X<81) = P(-3<z<3)=0.9974 or 99.74%
Step-by-step explanation:
Mean = 63
Standard Deviation =6
We need to find probability that X lies between 45 and 81 i.e P(45<X<81)
First we will find z-score for X=45 and X=81
For X=45 the z-score can be found using formula: [tex]z-score=\frac{X-\mu}{\sigma}[/tex]
Putting values and finding z-score
[tex]z-score=\frac{X-\mu}{\sigma}\\z-score=\frac{45-63}{6}\\z-score=-3[/tex]
For X=45, z-score is -3
For X=81 the z-score can be found using formula: [tex]z-score=\frac{X-\mu}{\sigma}[/tex]
Putting values and finding z-score
[tex]z-score=\frac{X-\mu}{\sigma}\\z-score=\frac{81-63}{6}\\z-score=3[/tex]
For X=81, z-score is 3
Now finding the probability of P(-3<z<3)
It can be solved as:
P(-3<z<3)=P(z<3)-P(z<-3)
Looking at the z-score table to find
P(z<3)= 0.9987
P(z<-3)=0.0013
So, P(-3<z<3)=P(z<3)-P(z<-3)
P(-3<z<3)=0.9987 - 0.0013
P(-3<z<3)=0.9974
or P(-3<z<3)= 99.74%
So, P(45<X<81) = P(-3<z<3)=0.9974 or 99.74%
