Respuesta :
Answer:
[tex]P(49.5<X<67.2)=P(\frac{49.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{67.2-\mu}{\sigma})=P(\frac{49.5-55.7}{5.2}<Z<\frac{67.2-55.7}{5.2})=P(-1.192<z<2.212)[/tex]
[tex]P(-1.192<z<2.212)=P(z<2.212)-P(z<-1.192)[/tex]
[tex]P(-1.192<z<2.212)=P(z<2.212)-P(z<-1.192)=0.987-0.117=0.870[/tex]
Step-by-step explanation:
We define X the random variable that represent the heights of a population for ten year old children, and for this case we know the distribution for X is given by:
[tex]X \sim N(55.7,5.2)[/tex]
Where [tex]\mu=55.7[/tex] and [tex]\sigma=5.2[/tex]
We want to find this probability:
[tex]P(49.5<X<67.2)[/tex]
We can use the z score formula to solve this problem given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Using this formula we have:
[tex]P(49.5<X<67.2)=P(\frac{49.5-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{67.2-\mu}{\sigma})=P(\frac{49.5-55.7}{5.2}<Z<\frac{67.2-55.7}{5.2})=P(-1.192<z<2.212)[/tex]
And we can find this probability with this difference
[tex]P(-1.192<z<2.212)=P(z<2.212)-P(z<-1.192)[/tex]
We can use tables for the normal standard distribution, excel or a calculator and we got this
[tex]P(-1.192<z<2.212)=P(z<2.212)-P(z<-1.192)=0.987-0.117=0.870[/tex]
