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Mathematics 7 Online
OpenStudy (tomfoolery1):

x P(x) 0 0.26 1 0.11 2 0.16 3 0.05 4 0.42 find mu 2.16 2.42 2.52 2.26

OpenStudy (tomfoolery1):

the last numbers are the choices

OpenStudy (mrm):

mu is your mean value, or average. Since each x has a different probability P(x), then you must do a Weighted Average. A weighted average is calculated as follows: \[average = \frac{ x_nP(x_n) }{ n }\] where n is the number of rows

OpenStudy (mrm):

I didnt exactly write that equation out well... let me expand it

OpenStudy (mrm):

\[\mu = \frac{x_1P(x_1)+x_2P(x_2)+...}{n}\]

OpenStudy (tomfoolery1):

what is x? and i also am not sure what P is

OpenStudy (mrm):

I'm sorry... I told you wrong.... I apologize.

OpenStudy (mrm):

You have a column of numbers x, and a column of numbers P(x). x is your value, P(x) is the probability of that value occurring. \[\mu = x_1P(x_1)+x_2P(x_2)+...\] Multiply each x by P and then all all the results together.

OpenStudy (mrm):

For example... (0)(0.26)+(1)(0.11)+...

OpenStudy (tomfoolery1):

ok i got 2.26!

OpenStudy (tomfoolery1):

is that close?

OpenStudy (mrm):

That's right! Good job.

OpenStudy (tomfoolery1):

Ahh thank you so much!! super helpful!

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