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Mathematics 5 Online
OpenStudy (anonymous):

You have a normal distribution of hours per week that music students practice. The mean of the values is 8 and the standard deviation of the values is 4. According to the normal distribution model that corresponds to this population, what percentage of the students practice between 6 to 10 hours per week?

OpenStudy (anonymous):

first you find the z scores...

OpenStudy (anonymous):

\[ z = \frac{x-\mu}{\sigma} \]So \(\mu\) is average, \(\sigma\) is the standard deviation, and \(x\) is the value.

OpenStudy (anonymous):

so you need to find \(z\) that corresponds to 6 and the \(z\) that corresponds to 10.

OpenStudy (anonymous):

Would u be 16?

OpenStudy (anonymous):

I think the answer is 34?

OpenStudy (anonymous):

wait, have you found the z scores yet?

OpenStudy (anonymous):

34%

OpenStudy (anonymous):

where are you stuck? how did you get the answers?

OpenStudy (anonymous):

Yeah 16 right? I'm estimating

OpenStudy (anonymous):

Okay I think I've figure it out thanks for the help

OpenStudy (anonymous):

oh, you are guessing because it is withing a standard deviation?

OpenStudy (anonymous):

Yeah

OpenStudy (anonymous):

okay you wanna do this: http://www.wolframalpha.com/input/?i=within+half+standard+deviation

OpenStudy (anonymous):

it should be 'confidence level'

OpenStudy (anonymous):

Oh I see thanks!

OpenStudy (anonymous):

So the answer is 38%

OpenStudy (anonymous):

Yeah, I think so.

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