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

The cdf a certain college library checkout duration X is as follows
$F\left(x\right)=\left\{\begin{array}{ll}0& x<0\\ \frac{4{x}^{2}}{49}& 0\le x<3.5\\ 1& 3.5\le x\end{array}$
Use this to compute th efollowing.
a) $P\left(x\le 1\right)$
b) $P\left(0.5\le x\le 1\right)$
c) $P\left(X>0.5\right)$
d) The median checkout duration
e) ${F}^{\prime }\left(x\right)$ to obtain the density function f(x)
f) $E\left(X\right)$
g) $V\left(X\right)$
${\sigma }_{x}$
h) If the borrower is charged an amount $h\left(X\right)={x}^{2}$ when checkout duration is X, compute the expected charge $E\left[h\left(X\right)\right]$

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yotaniwc

Step 1
Given:
$F\left(x\right)=\left\{\begin{array}{ll}0& x<0\\ \frac{4{x}^{2}}{49}& 0\le x<3.5\\ 1& 3.5\le x\end{array}$
$F\left(x\right)=P\left(x\le x\right)$
a) $P\left(x\le 1\right)=F\left(1\right)=\frac{4{\left(1\right)}^{2}}{49}=\frac{4}{49}$
Step 3
b) $P\left(0.5\le x\le 1\right)=P\left(x\le 1\right)-P\left(x\le 0.5\right)$
$=F\left(1\right)-F\left(0.5\right)$
$=\frac{4}{49}-\frac{4×{\left(0.5\right)}^{2}}{49}=\frac{4}{49}\left(1-\frac{1}{4}\right)=\frac{3}{49}$
Step 4
c) $P\left(x>0.5\right)=1-P\left(x\le 0.5\right)=1-F\left(0.5\right)$
$=1-\frac{4x{\left(0.5\right)}^{2}}{49}=\frac{48}{49}$

###### Not exactly what you’re looking for?
sirpsta3u

Step 1
d) at median the Cumulative density function will be 0.5, so
${\int }_{0}^{x}\frac{4{x}^{2}}{49}=0.5$
$\frac{4{x}^{3}}{3×49}=0.5$
$x=2.6388$
e) we know that , so
$f\left(x\right)=\frac{d\left(\frac{4{x}^{2}}{49}\right)}{dx}$
$=\frac{8x}{49}$
Step 2
f) from the Cumulative density function we see that the at and at , means the probability mass function is in between 0 and 3.5.
Expectation is given as
$E\left(X\right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x\frac{8x}{49}dx$
$={\int }_{0}^{3.5}\frac{8{x}^{2}}{49}dx$
$=2.333$
g) Variance is given as
$V\left(X\right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^{2}f\left(x\right)dx-{\left[{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}xf\left(x\right)dx\right]}^{2}$, so
$V\left(X\right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^{2}f\left(x\right)dx-{\left[{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}xf\left(x\right)dx\right]}^{2}$
$={\int }_{0}^{3.5}\frac{8{x}^{3}}{49}dx-{\left[{\int }_{0}^{3.5}\frac{8{x}^{2}}{49}dx\right]}^{2}$
$=6.125-5.442$
$=0.683$
${\sigma }_{x}=\sqrt{V\left(X\right)}$
$=0.8264$
Step 3
h) The required expectation can be written as
$E\left(h\left(x\right)\right)=E\left({X}^{2}\right)$
$={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^{2}f\left(x\right)dx$
$={\int }_{0}^{3.5}\frac{8{x}^{3}}{49}dx$

###### Not exactly what you’re looking for?
Jeffrey Jordon

Step 1
a) $P\left(x\le 1\right)=F\left(1\right)$
Reason: because CDF gives us the area left to the data point.
$P\left(x\le 1\right)=F\left(1\right)=\frac{4{x}^{2}}{49}=\frac{4\left(1{\right)}^{2}}{49}=\frac{4}{49}$
b) $P\left[\left(0.5\right)\le x\le 1\right]=F\left(1\right)-F\left(0.5\right)$
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
$P\left(0.5\le x\le 1\right)=\frac{4\left(1{\right)}^{2}}{49}-\frac{4\left(0.5{\right)}^{2}}{49}$
$=\frac{4}{49}-\frac{1}{49}=\frac{3}{49}$
c) $P\left(x>0.5\right)=1-P\left(x\le 0.5\right)$
$=1-F\left(0.5\right)$
$=1-\frac{4\left(0.5{\right)}^{2}}{49}$
$=\frac{48}{49}$
d) median $F\left(\stackrel{~}{x}\right)=0.5$
$⇒\frac{4{\stackrel{~}{x}}^{2}}{49}=0.5$
$4{\stackrel{~}{x}}^{2}=24.5$
${\stackrel{~}{x}}^{2}=6.125$
$\stackrel{~}{x}=2.475$
Step 2
e) $f\left(x\right)={F}^{\prime }\left(x\right)=\frac{d}{dx}\left(\frac{4{x}^{2}}{49}\right)$
$=\frac{8x}{49}$ for $0\le x\le 3.5$
Here wise $f\left(x\right)=0$
$\therefore {F}^{\prime }\left(x\right)=0$
for c=constant
f) $E\left(r\right)={\int }_{0}^{3.5}x×f\left(x\right)dx={\int }_{0}^{3.5}x×\frac{8x}{49}dx$
NSK
Note: limit is from
$=\frac{8}{49}\left[\frac{{x}^{3}}{3}{\right]}_{0}^{3.5}=2.333$
0 to 3.5 because f(x) is zero every where exup his interval
Step 3
g) $v\left(x\right)={\int }_{0}^{3.5}{x}^{2}×f\left(x\right)dx$
$={\int }_{0}^{3.5}\frac{8}{49}{x}^{3}dx$
$=\frac{8}{49}\left[\frac{{x}^{4}}{4}{\right]}_{0}^{3.5}$
$=6.125$
$=\sqrt{v\left(x\right)}=\sqrt{6.125}$
$=2.475$