Respuesta :
Answer:
[tex]P(X>55)=P(\frac{X-\mu}{\sigma}>\frac{55-\mu}{\sigma})=P(Z>\frac{55-60}{5.5})=P(z>-0.909)[/tex]
And we can find this probability with the complement rule:
[tex]P(z>-0.909)=1-P(z<-0.909)[/tex]
And we can use excel or the normal standard table and we got:
[tex]P(z>-0.909)=1-P(z<-0.909)=1-0.182= 0.818[/tex]
So then we expect about 81.8% of students that will be promoted
Step-by-step explanation:
Let X the random variable that represent the scores of promotion of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(60,5.5)[/tex]
Where [tex]\mu=60[/tex] and [tex]\sigma=5.5[/tex]
We want to find the following probability:
[tex]P(X>55)[/tex]
And we can use the z score formula given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Using the z score formula we got:
[tex]P(X>55)=P(\frac{X-\mu}{\sigma}>\frac{55-\mu}{\sigma})=P(Z>\frac{55-60}{5.5})=P(z>-0.909)[/tex]
And we can find this probability with the complement rule:
[tex]P(z>-0.909)=1-P(z<-0.909)[/tex]
And we can use excel or the normal standard table and we got:
[tex]P(z>-0.909)=1-P(z<-0.909)=1-0.182= 0.818[/tex]
So then we expect about 81.8% of students that will be promoted
