(a) Area under the curve is
=∫13f(x)dx=∫1312dx=[x2]13=3−12=22=1= \int^3_1 f(x)dx = \int^3_1 \frac{1}{2} dx = [\frac{x}{2}]^3_1 \\ = \frac{3-1}{2} = \frac{2}{2} = 1=∫13f(x)dx=∫1321dx=[2x]13=23−1=22=1
(b) P(2<X<2.5)=∫22.512dxP(2<X<2.5) = \int^{2.5}_2 \frac{1}{2} dxP(2<X<2.5)=∫22.521dx
=2.5−22=0.52=0.25= \frac{2.5-2}{2} = \frac{0.5}{2} = 0.25=22.5−2=20.5=0.25
(c) P(X≤1.6)=∫11.6f(x)dx=∫11.612dxP(X≤1.6) = \int^{1.6}_1 f(x)dx = \int^{1.6}_1 \frac{1}{2} dxP(X≤1.6)=∫11.6f(x)dx=∫11.621dx
=12(1.6−1)=0.62=0.3= \frac{1}{2} (1.6 -1) = \frac{0.6}{2} = 0.3=21(1.6−1)=20.6=0.3
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