Respuesta :
Answer:
[tex]P(11.5<X<15.5)=P(\frac{11.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{13.5-\mu}{\sigma})=P(\frac{11.5-13.5}{2}<Z<\frac{15.5-13.5}{2})=P(-1<z<1)[/tex]
And we can find the probability with this difference
[tex]P(-1<z<1)=P(z<1)-P(z<-1)[/tex]
And we can use the normal standard distribution or excel and we got:
[tex]P(-1<z<1)=P(z<1)-P(z<-1)=0.841-0.159=0.682[/tex]
So then we expect a proportion of 0.682 between 11.5 and 13.5
Step-by-step explanation:
Let X the random variable that represent the price earning ratios of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(13.5,2)[/tex]
Where [tex]\mu=13.5[/tex] and [tex]\sigma=2[/tex]
We want to find the following probability
[tex]P(11.5<X<15.5)[/tex]
And we can use the z score formula given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Using this formula we got:
[tex]P(11.5<X<15.5)=P(\frac{11.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{13.5-\mu}{\sigma})=P(\frac{11.5-13.5}{2}<Z<\frac{15.5-13.5}{2})=P(-1<z<1)[/tex]
And we can find the probability with this difference
[tex]P(-1<z<1)=P(z<1)-P(z<-1)[/tex]
And we can use the normal standard distribution or excel and we got:
[tex]P(-1<z<1)=P(z<1)-P(z<-1)=0.841-0.159=0.682[/tex]
So then we expect a proportion of 0.682 between 11.5 and 13.5
