Question

The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of 6 minutes and a standard deviation of 3 minutes. Eighty percent of the time, it takes more than how many minutes to find a parking space

Answer 1

Answer:

$z=0.842<\frac{a-6}{3}$

And if we solve for a we got

$a=6 +0.842*3=8.526$

Eighty percent of the time, it takes more than 8.526 how many minutes to find a parking space

Step-by-step explanation:

Let X the random variable that represent the lenght of time it takes to find a parking space of a population, and for this case we know the distribution for X is given by:

$X \sim N(6,3)$  

Where $\mu=6$ and $\sigma=3$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.2$   (a)

$P(X<a)=0.8$   (b)

We want to find a value who accumulate 0.80 of the area on the left and 0.2 of the area on the right it's z=0.842.

If we use condition (b) from previous we have this:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.80$  

$P(z<\frac{a-\mu}{\sigma})=0.80$

Using this formula we have:

$z=0.842<\frac{a-6}{3}$

And if we solve for a we got

$a=6 +0.842*3=8.526$

Eighty percent of the time, it takes more than 8.526 how many minutes to find a parking space