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szamy004 · Aug 30, 2009, 06:37 PM

I dot understand what to do in this question? For the SD I know how to find that using a computer but is there a manual way to do it?

Question

An iPod had about 10,000 songs. Assume the standard deviation for the population is 300 seconds.
What is the standard deviation of the average time when you take a simple random sample of 36 songs from this population?
How many songs would you need to sample if the standard deviation of x to be 30 seconds?
Suppose the true mean duration of the play time for the song in the iPod is 350 seconds. What is the probabiltiy that the sample mean differs from the population mean by more then 19 seconds when 30 songs are chosen?

Help please??

Thanks in advance :-)

morgaine300 · Aug 30, 2009, 10:08 PM

As to standard deviation, do you mean the standard deviation for the sample mean? That is, the population one is the 300, and you can also figure out the s.d. of the sample itself, except that it doesn't provide the information to do so. That's an annoying tedious process anyway.

But then there's the s.d. of the sample means, which is what you need to work with this. That equation is:
$\frac{\sigma}{sqrt{n}}$

That is
$\frac{300}{sqrt{36}}$

To get how many would be needed, plug the answer of 30 back into the above equation and then solve for n.

${\frac{300}{sqrt{n}}=30}$

As for that last part, either someone else can grab that or I might if I have time later. Right now my brain's getting to be a fog. :D

szamy004 · Aug 31, 2009, 03:10 AM

by plug 30 back into the equation do you mean, 300/30?
sorry I'm starting to get confused, thanks for your help though, its really helping me out.

morgaine300 · Sep 1, 2009, 01:17 AM

I already wrote it out for you at the bottom of the post. 30 is the answer, not the denominator.

szamy004 · Sep 1, 2009, 01:40 AM

Oh sorry, thank you so much for your help :-) I'm trying to the other question now god my lecturer is so bad. Again thank you