so does anyone know how to answer this problem or can tell me what im doing wrong?

The Sullivan Advertising Agency has determined that the average cost to develop a 30-second commercial is 20000. The standard deviation is 4000. Suppose a random sample of 50 commercials is selected and the average is 22800. Carry out your answers to 4 decimal places.
(a) What are the chances of finding a sample mean this high or higher?

so i figure x-bar= 22800, mean= 20000, stdev= 4000, n= 50

i intended to solve for z, by taking x-bar minus mean/ (stdev/(square root of n)). my answer is 4.94, but thats way too high for a z-score....

PLEASE HELP!

10/15/2006 4:01:26 PM

the only thing that i could think of is you get a z-score of .7 which corresponds to .7580, but since you want the higher side, you do 1 minus .7580 to get .242 or 24.2 %. but that could be wrong, i would of initially gone with the same formula you did.

[Edited on October 15, 2006 at 4:51 PM. Reason : trying to remember]

10/15/2006 4:45:29 PM

it doesn't really matter if it seems the z-score is too high. It seems like you got the correct z-value so the P(Z > 4.94) = 1 - P(Z < 4.94) which is approximately 1 - 1 = 0. So it's a pretty unlikely probability that you'll find a sample mean this high or higher.

10/15/2006 5:30:45 PM

yeah, you two are right. if you think about it, 22800 is not even close to 1 standard deviation away from the mean, and 3 standard deviations away includes 99.7 % for a normal distribution, so the odds are extremely high that you get a number slightly above the mean (even though it is 2800) in terms of Standard Deviation, it is nothing.

10/15/2006 5:39:39 PM

hmm.... what am i missing? ^ im a little confused... 22800 falls between one standard deviation to the right of 20,000 with a stdev of 4000. i tried the .7505 and didnt get the right answer.

10/15/2006 6:14:23 PM

disregard what i said in my first post, i was just trying to think of something else that you could do if what you had wasnt the right answer. But im pretty sure you got the right answer. It should be pretty much an 100% chance of finding a sample mean that high or higher. It is easier to think about it like this. Say you have taken 300 tests in your life, and you have an average of like 80 and standard deviation 10. Now out of those 300 tests...you take random samples from it of 10 tests (which would give you a total of 29 other sets of data of 10 tests). Now the question is, what are the odds that out of those 30 total sets of 10, that your test average was above about 1/2 a standard deviation about your mean of 80 (1/2 a standard deviation would be 85). Your odds are pretty much 100 percent that you would get that in one of those. That is what your question is essentially saying. Hope that makes more sense

10/15/2006 6:34:47 PM