Statistic Probability, Normal distribution!!!!!!help?

Question

An Executive at Westinghouse Canada drives from his home in the suburbs near Toronto to his office in the Center of the city. His driving times can be approximated by the normal distribution with a mean of 45 minutes and a standard deviation of 8 minutes.
a) Somedays there will be accidents or other delays, so the trip will take longer than unusual. How long will the longest 10% of this trip???
Please anyone answer this question with explanation, I cannot figure the answer out. Thanks!!!!

Answer 1

So far I have come up with 55.24 minutes. I would check this answer though.

Normal distribution given in the problem:

mean = 45 minutes standard deviation = 8 minutes


Remember

*Standard Normal Distribution

mean = 0 standard deviation = 1

I think you are looking for the 90th percentile.

You need to compare the values of the normal distrubtion given in the problem to a standard normal distribution.

***Update***
How I obtained the answer
***What the formula below says is to use the z value you would get for the 90th percentile in the standard normal distribution.
Multiply to the standard deviation of the normal distribution in the problem. Then had this to the mean of the normal distribution given in the problem. This is to get the z value of
the normal distribution in your problem

This formula compares an arbitrary normal distribution to
a standard normal distribution

100pth percentile for = mean + (100p)th percentile for s * std
normal distribution


""""""""""""""""""""""""""""
s = standard normal distribution
std = standard deviation
p: is the percent
""""""""""""""""""""""""""""""

"""""""""""""""""""""""""""""""
for example p = .9 for the 90th percentile

using the above formula I got

The 90th percentile of normal = 45+ (1.28)*(8)
= 55.24 minutes

* Note: The z-critical value for the 90th percentile (1.28) for the standard normal distribution I got from using a z critical value with an alpha of .1.

Finding values with a table

On this site step 3 of question 1 gives an example as to
find values using a standard normal distribution table.
http://www.oswego.edu/~srp/stats/z.htm
* If you have a table of values for the Standard Normal curve
you can find the z-critical value. My table has the tenths (.1)th place of the z value(it should have negative and postive values) in the rows. The hundrenths and
thousandths place (you should see numbers like .00, .01,.02,.03,.04 up to and including .09) in the columns. The entries of the table are
percentiles of the standard normal curve. The percentiles
are not exact (example .8997 would be the 90th percentile .7389 would be the 74th percentile). So if you were looking for the 85th percentile you would see the value
.8508 look at the places of the tenths place of the z value in the row (which in this case is 1.0) and then add the value in
the column (representing the hundrenths and thousandths place) in this case which is .04. Thus you would get a
z-critical value of 1.04 for the 85th percentile of a standard normal distribution.

** Update Saturday June 16th

I got a similar answer with a TI-83 Calculator

1. Press 2nd vars
2. Scroll down to choice 3 : invNorm
3. Type invNorm(.9,45,8)
4. It should say 55.25

*Note in step 3 (.9 refers to the 90th percentile), 45 is the
mean of the normal distribution in the problem , and the 8 is
the mean in the normal distribution in the problem.
This function compares the numbers to a standard normal
distribution.