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Let the random variable $X$ have the density $f(x)=k x$ if $0 \leq x \leq 3$. Find $k$. Find $x_{1}$ and $x_{2}$ such that $P\left(X \leq x_{1}\right)=0.1$ and $P\left(X \leq x_{2}\right)=0.95 .$ Find $P(|X-1.8|<0.6)$.
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To make $\int_{0}^{3} f(x) d x=1$ we must take $k=2 / 9$. For any $c$ in $(0,3), P(X \leq c)=c^{2} / 9$ so $x_{1}=\sqrt{0.9}=$ $0.9487$ and $x_{2}=\sqrt{8.55}=2.9240 . P(|X-1.8|<0.6)=P(1.2<X<2.4)=\left(2.4^{2}-1.2^{2}\right) / 9=0.48$.
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