Respuesta :
Answer:
[tex]z=-1.28<\frac{a-8.54}{4.91}[/tex]
And if we solve for a we got
[tex]a=8.54 -1.28*4.91=2.2552[/tex]
And for this case we can conclude that the time required needs to be 2.2552 minutes or less in order to be sponsored
Step-by-step explanation:
Let X the random variable that represent the time for goals in hockey games of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(8.54,4.91)[/tex]
Where [tex]\mu=8.54[/tex] and [tex]\sigma=4.91[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X<a)=0.10[/tex] (a)
[tex]P(X>a)=0.90[/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=-1.28.
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 do the following
[tex]z=-1.28<\frac{a-8.54}{4.91}[/tex]
And if we solve for a we got
[tex]a=8.54 -1.28*4.91=2.2552[/tex]
And for this case we can conclude that the time required needs to be 2.2552 minutes or less in order to be sponsored
