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Mathematics 14 Online
OpenStudy (mrhoola):

Suppose X has a continuous uniform distribution over the interval [2.4, 5.4]. What is P(X < 4.2)

OpenStudy (anonymous):

Since it is uniform, f(x) = k Solve: $$ \Large \int_{2.4}^{5.4} k ~dx= 1 $$

OpenStudy (anonymous):

the area under the curve must equal to 1 , over the domain

OpenStudy (mrhoola):

makes sense, yes.

OpenStudy (anonymous):

In general : For a uniform distribution f(x) = k defined over interval [a,b] $$ \Large{ \int_{a}^{b} k ~dx= 1 \\ ~\\ \left [~kx~ \right ]_{b}^{a}= 1 \\ ~\\ k(b-a) = 1 \\ ~\\ k = \frac{1}{b-a} } $$

OpenStudy (anonymous):

therefore f(x) = 1 / ( 5.4 - 2.4) = 1/3

OpenStudy (anonymous):

$$ \Large { P(X<4.2) = \int_{2.4}^{4.2} \frac1 3 ~dx } $$

OpenStudy (mrhoola):

1/3(4.2-2.4) = 1/3 (1.8) =0.6

OpenStudy (mrhoola):

Thank you very much !!!!!!!! :D

OpenStudy (anonymous):

theres a small typo above

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