Respuesta :
Answer:
a) [tex]P(148<X<175)=P(\frac{148-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{175-\mu}{\sigma})=P(\frac{148-160}{8}<Z<\frac{175-160}{8})=P(-1.5<z<1.875)[/tex]
[tex]P(-1.5<z<1.875)=P(z<1.875)-P(z<-1.5)= 0.970-0.0668= 0.9032[/tex]
b) [tex]P(X>164)=P(\frac{X-\mu}{\sigma}>\frac{164-\mu}{\sigma})=P(Z>\frac{164-160}{8})=P(z>0.5)[/tex]
[tex]P(z>0.5)=1-P(z<0.5)= 1-0.691= 0.309[/tex]
c) [tex]P(X<179)=P(\frac{X-\mu}{\sigma}<\frac{179-\mu}{\sigma})=P(Z<\frac{179-160}{8})=P(z<2.375)[/tex]
[tex]P(z<2.375)= 0.991[/tex]
Step-by-step explanation:
Let X the random variable that represent the heights of Guinean travels , and for this case we know the distribution for X is given by:
[tex]X \sim N(160,8)[/tex] Â
Where [tex]\mu=160[/tex] and [tex]\sigma=8[/tex]
Part a
We are interested on this probability
[tex]P(148<X<175)[/tex]
We can ue the z score formula given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Using this formula we got:
[tex]P(148<X<175)=P(\frac{148-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{175-\mu}{\sigma})=P(\frac{148-160}{8}<Z<\frac{175-160}{8})=P(-1.5<z<1.875)[/tex]
And we can find this probability with this difference using the normal standard distribution or excel:
[tex]P(-1.5<z<1.875)=P(z<1.875)-P(z<-1.5)= 0.970-0.0668= 0.9032[/tex]
Part b
[tex]P(X>164)=P(\frac{X-\mu}{\sigma}>\frac{164-\mu}{\sigma})=P(Z>\frac{164-160}{8})=P(z>0.5)[/tex]
And we can find this probability with this difference using the normal standard distribution or excel:
[tex]P(z>0.5)=1-P(z<0.5)= 1-0.691= 0.309[/tex]
Part c
[tex]P(X<179)=P(\frac{X-\mu}{\sigma}<\frac{179-\mu}{\sigma})=P(Z<\frac{179-160}{8})=P(z<2.375)[/tex]
And we can find this probability with this difference using the normal standard distribution or excel:
[tex]P(z<2.375)= 0.991[/tex]
