Respuesta :
Answer:
[tex]z=0.524<\frac{a-5}{2}[/tex]
And if we solve for a we got
[tex]a=5 +0.524*2=6.048 \approx 6.05 min[/tex]
Step-by-step explanation:
Let X the random variable that represent the lenght time it takes to find a parking space at 9AM of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(5,2)[/tex]
Where [tex]\mu=5[/tex] and [tex]\sigma=2[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.3[/tex] (a)
[tex]P(X<a)=0.7[/tex] (b)
As we can see on the figure attached the z value that satisfy the condition with 0.7 of the area on the left and 0.3 of the area on the right it's z=0.524
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.7[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.7[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=0.524<\frac{a-5}{2}[/tex]
And if we solve for a we got
[tex]a=5 +0.524*2=6.048 \approx 6.05 min[/tex]
