The cdf a certain college library checkout duration X is as followsNS

Deragz 2021-12-15 Answered

The cdf a certain college library checkout duration X is as follows
F(x)={0x<04x2490x<3.513.5x
Use this to compute th efollowing.
a) P(x1)
b) P(0.5x1)
c) P(X>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)
σx
h) 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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Expert Answer

yotaniwc
Answered 2021-12-16 Author has 34 answers

Step 1
Given:
F(x)={0x<04x2490x<3.513.5x
F(x)=P(xx)
a) P(x1)=F(1)=4(1)249=449
Step 3
b) P(0.5x1)=P(x1)P(x0.5)
=F(1)F(0.5)
=4494×(0.5)249=449(114)=349
Step 4
c) P(x>0.5)=1P(x0.5)=1F(0.5)
=14x(0.5)249=4849

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sirpsta3u
Answered 2021-12-17 Author has 42 answers

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
=6.1255.442
=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

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Jeffrey Jordon
Answered 2021-12-26 Author has 2495 answers

Step 1
a) P(x1)=F(1)
Reason: because CDF gives us the area left to the data point.
P(x1)=F(1)=4x249=4(1)249=449
b) P[(0.5)x1]=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 required
P(0.5x1)=4(1)2494(0.5)249
=449149=349
c) P(x>0.5)=1P(x0.5)
=1F(0.5)
=14(0.5)249
=4849
d) median F(x~)=0.5
4x~249=0.5
4x~2=24.5
x~2=6.125
x~=2.475
Step 2
e) f(x)=F(x)=ddx(4x249)
=8x49 for 0x3.5
Here wise f(x)=0
F(x)=0
for c=constant
f) E(r)=03.5x×f(x)dx=03.5x×8x49dx
NSK
Note: limit is from
=849[x33]03.5=2.333
0 to 3.5 because f(x) is zero every where exup his interval
Step 3
g) 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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