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TriC-MathMOOC 21 Online
OpenStudy (jessicawade):

uniform distribution help? PLEASE D:

OpenStudy (jessicawade):

OpenStudy (jessicawade):

A.0.60 B.0.80 C.0.10 D.0.40

OpenStudy (jessicawade):

i dont know how to find the answer

OpenStudy (jessicawade):

do i subtract?

OpenStudy (anonymous):

You wan to find \(P(X<124.2)\). The uniform distribution is defined as \[f(x)=\begin{cases} \dfrac{1}{2}&\text{for }123<X<125\\\ 0&\text{elsewhere} \end{cases}\] The desired probability is given by the integral, \[P(X<124.2)=\int_{-\infty}^{124.2}f(x)~dx=\frac{1}{2}\int_{123}^{124.2}dx\] If you're NOT familiar with calculus: this means the same thing as the area under the curve between 123 and 124.2.

OpenStudy (jessicawade):

wow

OpenStudy (jessicawade):

i havnt taken calculus yet lol

OpenStudy (jessicawade):

do i subtract 124.2 from 123?

OpenStudy (anonymous):

Okay well if you understand what a probability density function is - which is \(f(x)\) in this case - then it's not too big a step to say that a portion of the area under the distribution represents the proportion or probability of \(X\) falling in some range of values. |dw:1410185924273:dw| Do you agree that the probability under the whole curve is 1? As in, the probability of getting any number between 123 and 125 is 100% ?

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