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Question 82 of 88

Q.For the following probability density function of a random variable X, find

(a) P(X<1)P(X<1) and
(b) P(∣X∣<1)P(|X|<1). f(x)=x+218f(x) = \dfrac{x+2}{18}; for −2<x<4-2<x<4; =0= 0, otherwise
Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2023Subjective· 3mImportance★★★★★
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Integrate the given pdf over the required intervals.

f(x)=x+218f(x)=\dfrac{x+2}{18} on (−2,4)(-2,4).

(a) P(X<1)=∫−21x+218 dx=118[x22+2x]−21=118(2.5−(−2))=4.518=14P(X<1)=\displaystyle\int_{-2}^{1}\dfrac{x+2}{18}\,dx=\dfrac1{18}\left[\dfrac{x^2}{2}+2x\right]_{-2}^{1}=\dfrac1{18}\big(2.5-(-2)\big)=\dfrac{4.5}{18}=\dfrac14

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