Respuesta :
Answer:
[tex]z=-2.33<\frac{a-8.9}{1.9}[/tex]
And if we solve for a we got
[tex]a=8.9 -2.33*1.9=4.47[/tex]
And the best answer for this case would be:
A) 4.47 lb
Step-by-step explanation:
Let X the random variable that represent the weights of juvenile turkeys, and for this case we know the distribution for X is given by:
[tex]X \sim N(8.9,1.9)[/tex]
Where [tex]\mu=8.9[/tex] and [tex]\sigma=1.9[/tex]
The z score formula very useful for this case is given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.99[/tex] (a)
[tex]P(X<a)=0.01[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.01 of the area on the left and 0.99 of the area on the right it's z=-2.33. On this case P(Z<-2.33)=0.01 and P(z>-2.33)=0.99
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.01[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.01[/tex]
So we have this relation
[tex]z=-2.33<\frac{a-8.9}{1.9}[/tex]
And if we solve for a we got
[tex]a=8.9 -2.33*1.9=4.47[/tex]
And the best answer for this case would be:
A) 4.47 lb
