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.)

And if we solve for a we got

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:

Where and

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

  (a)

  (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:

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

And if we solve for a we got

5

step-by-step explanation:

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200.   after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes           1800 - 6400 which gets you 11,600.