[SPLIT] prob. that fee exceeds $29 / find mean, st. dev.

[Pensuave]

I"m still not understanding this process. where does the answer come in for these problems? This isn't really homework, as much as it is me trying to grasp these concepts for an upcoming test. Here is another practice problem that I am confused with...

1) The amount that households pay service providers for access to the Internet varies quite a bit, but the mean monthly fee is $28 and the standard deviation is $10. The distribution is not Normal: many households pay about $10 for limited dial-up access or about $25 for unlimited dial-up access, but some pay much more for faster connections. A sample survey asks an SRS of 500 households with Internet access how much they pay. What is the probability that the average fee paid by the sample households exceeds $29?

So far I have done the following:

. . .10 / √500
. . . . .= 10 / 22.36067977
. . . . .= 0.4472135956

I am stuck here. I know the answer is 0.0125 but i can't figure out what to do with these numbers.

Another practice problem, which seems to tie into the above, is:

2) An automatic girnding machine in an auto parts plant prepares axles with a target diameter mu = 40.125 millimeters. The machine has some variability, so the standard deviation of the diameters is sigma = 0.002 millimeters. A sample of 4 axles is inspected each hour for process control purposes, and records are kept of the sample mean diameter. What will be the mean and standard deviation of the numbers recorded?

I think that is tied into the first, but I'm really confused by this. Our teacher skipped 6 chapters in our book and everyone in our class now seems to be lost.

 

[galactus]

1) The amount that households pay service providers for access to the Internet varies quite a bit, but the mean monthly fee is $28 and the standard deviation is $10. The distribution is not Normal: many households pay about $10 for limited dial-up access or about $25 for unlimited dial-up access, but some pay much more for faster connections. A sample survey asks an SRS of 500 households with Internet access how much they pay. What is the probability that the average fee paid by the sample households exceeds $29?[/color]

So far I have done the following:

. . .10 / √500
. . . . .= 10 / 22.36067977
. . . . .= 0.4472135956

I am stuck here. I know the answer is 0.0125 but i can't figure out what to do with these numbers.

I see the distribtion is not normal, but:

\(\displaystyle \L\\z=\frac{x-x_{1}}{\frac{s}{\sqrt{n}}}\)

\(\displaystyle \L\\\frac{29-28}{\frac{10}{\sqrt{500}}}=\sqrt{5}\approx{2.236}\)

Find 2.236 in the table and we find 0.9873.

Since they're asking for the probability > than, subtract from 1:

1-0.9873=0.0127. That's mighty close to the answer you mentioned.