Question

The length of time it takes to find a parking space at 9 A.M. follows a normal distribution with a mean of 5 minutes and a standard deviation of 2 minutes. Seventy percent of the time, it takes more than how many minutes to find a parking space? (Round your answer to two decimal places.)

Answer 1

Answer:

$z=0.524$

And if we solve for a we got

$a=5 +0.524*2=6.048 \approx 6.05 min$

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:

$X \sim N(5,2)$  

Where $\mu=5$ and $\sigma=2$

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

$P(X>a)=0.3$   (a)

$P(X$   (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:

$P(X$  

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=0.524$

And if we solve for a we got

$a=5 +0.524*2=6.048 \approx 6.05 min$