Respuesta :
Answer:
[tex]P(z<\frac{a-\mu}{\sigma})=0.10[/tex]
and we can set up the following equation
tex]z=-1.282<\frac{a-104.5}{23.62}[/tex]
And if we solve for a we got
[tex]a=104.5 -1.282*23.62=74.22[/tex]
And the best answer for this case would be:
b) $74.23
Step-by-step explanation:
Let X the random variable that represent the stocks price of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(104.5,23.62)[/tex]
Where [tex]\mu=104.5[/tex] and [tex]\sigma=23.62[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.90[/tex] (a)
[tex]P(X<a)=0.10[/tex] (b)
As we can see on the figure attached the z value that satisfy the condition with 0.10 of the area on the left and 0.90 of the area on the right it's z=-1.282. On this case P(Z<-1.282)=0.10 and P(z>-1.282)=0.90
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.10[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.10[/tex]
and we can set up the following equation
tex]z=-1.282<\frac{a-104.5}{23.62}[/tex]
And if we solve for a we got
[tex]a=104.5 -1.282*23.62=74.22[/tex]
And the best answer for this case would be:
b) $74.23
