SOLUTION: more normal distribution that i need help understanding. and learning how to use technology to help. thank you. Suppose that Denver workers have a mean commute of 22 minutes and

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Question 1195499: more normal distribution that i need help understanding. and learning how to use technology to help. thank you.
Suppose that Denver workers have a mean commute of 22 minutes and commute times are normally distributed with a standard deviation of 15 minutes. Let X represent the commute time for a randomly selected Denver worker.
Find the 85th percentile for the commute time of Denver workers. Round your answer to 1 decimal place.
The 85th percentile is minutes.
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
the mean is 22 minutes.
the standard deviation is 15 minutes.
85th percentile means that 85% have a commute time less than 22 minutes.
this means that 15% have a commute time greater than 22 minutes.
z-score is equal to (x - m) / s
x is the raw score
m is the mean
s is the standard deviation, in this case.
you are looking for a z-score that has 85% of all z-scores less than that.
using a statistical calculator, you find that the z-score that has 85% of the commute time less than it would be equal to 1.03643338.
use the z-score formula to find the raw score.
z = (x - m) / s becomes 1.03643338 = (x - 22) / 15.
solve for x to get:
x = 15 * 1.03643338 + 22 = 37.5465007 minutes.
that's the commute time that is in the 85th percentile, meaning that 85% of all commute times with a mean of 22 minutes and a standard deviation of 15 minutes will have a commute time less than that.
you can use a z-score calculator online to confirm that.
the online calculator that i used is at https://www.calculator.net/z-score-calculator.html
the results from using that calculator are shown below:

let me know if you have any questions.
theo