Respuesta :
Answer:
a) [tex] P(0<z<1.67)= P(z<1.67) -P(Z<0) = 0.953 -0.5=0.453[/tex]
b) [tex] P(Z<-0.33) = 0.371[/tex]
c) [tex] P(Z>0.33) =1-P(Z<0.33) = 1-0.629= 0.371[/tex]
d) [tex] P(2<z<2.67)= P(z<2.67) -P(Z<2) = 0.996 -0.977=0.019[/tex]
Step-by-step explanation:
Assuming the following questions:
a. Between 20 and 25 gallons
Let X the random variable that represent the sports drink consumption of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(20,3)[/tex]
Where [tex]\mu=20[/tex] and [tex]\sigma=3[/tex]
We are interested on this probability
[tex]P(20<X<25)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
And if we find the z score for each limit we got:
[tex] z = \frac{20-20}{3}=0[/tex]
[tex] z = \frac{25-20}{3}=1.67[/tex]
And we can use the normal standard distirbution table and we got:
[tex] P(0<z<1.67)= P(z<1.67) -P(Z<0) = 0.953 -0.5=0.453[/tex]
b. Less than 19 gallons
[tex] z = \frac{19-20}{3}=-0.33[/tex]
And using the normal table we got:
[tex] P(Z<-0.33) = 0.371[/tex]
c. More than 21 gallons
[tex] z = \frac{21-20}{3}=0.33[/tex]
And using the normal table and the complement rule we got:
[tex] P(Z>0.33) =1-P(Z<0.33) = 1-0.629= 0.371[/tex]
d. Between 26 and 28 gallons
[tex] z = \frac{26-20}{3}=2[/tex]
[tex] z = \frac{28-20}{3}=2.67/tex]
And we can use the normal standard distirbution table and we got:
[tex] P(2<z<2.67)= P(z<2.67) -P(Z<2) = 0.996 -0.977=0.019[/tex]
