The cdf a certain college library checkout duration X is as followsF(x)={0x&lt;04x2490≤x&lt;3.513.5≤xUse this to compute th efollowing.a) P(x≤1)b) P(0.5≤x≤1)c) P(X&gt;0.5)d) The median checkout duration μ~ [solve, S=F(μ)]e) F′(x) to obtain the density function f(x)f) E(X)g) V(X)σxh) If the borrower is charged an amount h(X)=x2 when checkout duration is X, compute the expected charge E[h(X)]

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The cdf a certain college library checkout duration X is as followsF(x)={0x&amp;lt;04x2490≤x&amp;lt;3.513.5≤xUse this to compute th efollowing.a) P(x≤1)b) P(0.5≤x≤1)c) P(X&amp;gt;0.5)d) The median checkout duration μ~ [solve, S=F(μ)]e) F′(x) to obtain the density function f(x)f) E(X)g) V(X)σxh) If the borrower is charged an amount h(X)=x2 when checkout duration is X, compute the expected charge E[h(X)]

yotaniwc · Accepted Answer

Step 1 Given: F(x)={0x&amp;lt;04x2490≤x&amp;lt;3.513.5≤x F(x)=P(x≤x) a) P(x≤1)=F(1)=4(1)249=449 Step 3 b) P(0.5≤x≤1)=P(x≤1)−P(x≤0.5) =F(1)−F(0.5) =449−4×(0.5)249=449(1−14)=349 Step 4 c) P(x&amp;gt;0.5)=1−P(x≤0.5)=1−F(0.5) =1−4x(0.5)249=4849

sirpsta3u · Accepted Answer

Step 1 d) at median the Cumulative density function will be 0.5, so ∫0x4x249=0.5 4x33×49=0.5 x=2.6388 e) we know that f(x)=d F(x)dx, so f(x)=d(4x249)dx =8x49 Step 2 f) from the Cumulative density function we see that the at x=0, F(x)=0 and at x=3.5, F(X)=1, means the probability mass function is in between 0 and 3.5. Expectation is given as E(X)=∫−∞∞x8x49dx =∫03.58x249dx =2.333 g) Variance is given as V(X)=∫−∞∞x2f(x)dx−[∫−∞∞xf(x)dx]2, so V(X)=∫−∞∞x2f(x)dx−[∫−∞∞xf(x)dx]2 =∫03.58x349dx−[∫03.58x249dx]2 =6.125−5.442 =0.683 σx=V(X) =0.8264 Step 3 h) The required expectation can be written as E(h(x))=E(X2) =∫−∞∞x2f(x)dx =∫03.58x349dx

Jeffrey Jordon · Accepted Answer

Step 1a) P(x≤1)=F(1)Reason: because CDF gives us the area left to the data point.P(x≤1)=F(1)=4x249=4(1)249=449b) P[(0.5)≤x≤1]=F(1)−F(0.5)Here F(1) gives area left of point (1) and F(0.5) gives area left of point (0.5)⇒ difference gives the probability requiredP(0.5≤x≤1)=4(1)249−4(0.5)249=449−149=349c) P(x&amp;gt;0.5)=1−P(x≤0.5)=1−F(0.5)=1−4(0.5)249=4849d) median F(x~)=0.5⇒4x~249=0.54x~2=24.5x~2=6.125x~=2.475Step 2e) f(x)=F′(x)=ddx(4x249)=8x49 for 0≤x≤3.5Here wise f(x)=0∴F′(x)=0for c=constantf) E(r)=∫03.5x×f(x)dx=∫03.5x×8x49dxNSKNote: limit is from=849[x33]03.5=2.3330 to 3.5 because f(x) is zero every where exup his intervalStep 3g) v(x)=∫03.5x2×f(x)dx=∫03.5849x3dx=849[x44]03.5=6.125=v(x)=6.125=2.475

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