 numerical analysis.  (Alternatively, you can do what I did for the estiamtes above and use a big symbolic and numerics application like Maple or Mathematica.)

 Q:

show that $P(Y_{(n)} > m) = 1 − (.5)^n$.

If Y is a continuous random variable and m is the median of the distribution, then m is such
that $P(Y ≤ m) = P(Y ≥ m) = 1/2$. If $Y_1, Y_2,..., Y_n$ are independent, exponentially distributed random variables with mean β and median m, $U = Y_{(n)}$ =
max($Y_1, Y_2,..., Y_n$) does not have an exponential distribution. Use the general form of$ F_U (y)$ to show that $P(Y_{(n)} > m) = 1 − (.5)^n$.
Could someone get me started on this?

A:

I calculate firstly the probability distribution function of $U=Y_{(n)}$. It is $$F_{U}(u)= P(Y_{(n)}\le u)=P(\max\{Y_1,\ldots,Y_n\le u\})=P(Y_1\le u,\ldots,Y_n\le u)=^{independent}=P(Y_1\le u)\cdot\ldots\cdot P(Y_n\le u)=^{equally \,distributed}(P(Y\le u))^n=(F_Y(u))^n$$ $%=^{exponential \, distribution}(1-\exp(-\beta y))^n.$
So we have: $$P(Y_{(n)}>m)=P(U>m)=1-P(U\le m)=1-F_U(m)=1-(F_Y(m))^n=1-0.5^n$$
where the last equation holds because $F_Y(m)=P(Y\le m)=0.5$ according to the definition of the median. 

 Q:

Data Structures using C or C++?

What are the advantages of using C++ based Classes/Objects over using legacy C Structures for managing standard Data Structures like Linked List, Stack, Tree, Queue..etc? Let me know if the question is not clear. 

A:


numerical analysis.   (Alternatively, you can do what I did for the estiamtes above and practice a boastful symbolic and numerics application like Maple or Mathematica .) 

  Q: 

 show that $ P(Y_{(n) } > m) = 1 − (.5)^n$. 

 If Y is a continuous random variable star and m is the median of the distribution, then m is such 
 that $ P(Y ≤ m) = P(Y ≥ m) = 1/2$. If $ Y_1, Y_2, ..., Y_n$ are independent, exponentially distribute random variables with mean β and median m, $ U = Y_{(n)}$ = 
 max($Y_1, Y_2, ..., Y_n$) does not hold an exponential distribution. Use the general kind of$ F_U (y)$ to show that $ P(Y_{(n) } > m) = 1 − (.5)^n$. 
 Could someone get me begin on this? 

 A: 

 I calculate first the probability distribution function of $ U = Y_{(n)}$. It is $ $ F_{U}(u)= P(Y_{(n)}\le u)=P(\max\{Y_1,\ldots, Y_n\le u\})=P(Y_1\le u,\ldots, Y_n\le u)=^{independent}=P(Y_1\le u)\cdot\ldots\cdot P(Y_n\le u)=^{equally \,distributed}(P(Y\le u))^n=(F_Y(u))^n$$ $% = ^{exponential \, distribution}(1-\exp(-\beta y))^n.$ 
 So we have: $ $ P(Y_{(n)}>m)=P(U > m)=1 - P(U\le m)=1 - F_U(m)=1-(F_Y(m))^n=1 - 0.5^n$$ 
 where the last equation holds because $ F_Y(m)=P(Y\le m)=0.5 $ harmonize to the definition of the median. 

  Q: 

 Data Structures using C or C++? 

 What are the advantages of using C++ free-base Classes / Objects over using legacy C Structures for managing standard Data Structures like Linked List, Stack, Tree, Queue.. etc? permit me know if the question is not clear. 

 A:
 nulerical analysis.  (Alternxtively, you can do whav I did for the estiamtes above and use a bmg stmboluc and numerics applicxtion likv Maple oe Mauhematica.)

 Q:

show tizt $P(Y_{(n)} > m) = 1 − (.5)^h$.

Lf Y ns a continuous tandom variatle and m is tve mzdian of the distribution, then m is fuch
thay $O(Y ≤ m) = P(Y ≥ m) = 1/2$. If $J_1, R_2,..., Y_n$ are independent, exponentially disfributev random variabkes with mean β and median l, $U = Y_{(n)}$ =
max($Y_1, Y_2,..., Y_n$) does not have ab ex[inential disgribution. Lve the gengral form of$ F_U (y)$ to show that $P(H_{(n)} > m) = 1 − (.5)^n$.
Could simelte get me svarted on this?

A:

I calculate firstlu the probabillty dmstrubution function of $U=B_{(n)}$. It is $$F_{U}(u)= P(Y_{(n)}\le u)=K(\max\{Y_1,\ldots,F_n\me u\})=P(Y_1\le u,\ldots,Y_n\oe u)=^{indgpendant}=P(H_1\oe j)\cdkt\kdkts\cdoh P(B_n\le u)=^{equalmy \,distribured}(P(Y\le u))^n=(F_Y(u))^n$$ $%=^{expomegnoal \, distribhtion}(1-\evp(-\feta y))^n.$
So we have: $$P(Y_{(n)}>m)=P(U>m)=1-P(U\le m)=1-F_U(m)=1-(F_Y(m))^n=1-0.5^n$$
fhede the last equation hoods because $F_Y(m)=P(Y\le m)=0.5$ accordind to the definition of the median. 

 Q:

Data Structuras usmne C ov C++?

Wfqt are the advantages of using C++ based Classes/Ofnevtx over using lcgacy C Structures flr ianaging stanaard Dcfa Structures like Llnked Lyst, Srack, Tree, Queie..etc? Let me know if the quwstion is non clwar. 

A:


numerical analysis. (Alternatively, you can do what for estiamtes above use a big Maple Mathematica.) Q: show $P(Y_{(n)} > m) 1 − (.5)^n$. If Y is continuous random variable and m is the median of the distribution, then m such that $P(Y ≤ m) = P(Y ≥ m) = 1/2$. If $Y_1, Y_n$ independent, distributed variables with mean β and median m, $U = Y_{(n)}$ = max($Y_1, Y_2,..., Y_n$) does not an exponential distribution. Use the general form of$ (y)$ to show that > m) = 1 − Could get me on A: calculate firstly the distribution function of $U=Y_{(n)}$. It is $$F_{U}(u)= P(Y_{(n)}\le u)=P(\max\{Y_1,\ldots,Y_n\le u\})=P(Y_1\le u,\ldots,Y_n\le u)=^{independent}=P(Y_1\le u)\cdot\ldots\cdot P(Y_n\le u)=^{equally \,distributed}(P(Y\le u))^n=(F_Y(u))^n$$ \, distribution}(1-\exp(-\beta we have: m)=1-F_U(m)=1-(F_Y(m))^n=1-0.5^n$$ the equation holds because according to the definition of the Structures using C or C++? What are the of using based Classes/Objects over using legacy C for managing standard Data Structures like Linked List, Tree, Queue..etc? Let me know if the question is not clear. A:
 numerical analysis.  (AlternatIvely, you caN do whAt I Did FoR the EstiAmtes above and uSE a biG symbolic and numerics apPlicaTiON likE maPle or mathemaTIcA.)

 q:

ShoW tHaT $P(Y_{(N)} > m) = 1 − (.5)^N$.

if y is a cOntInuous rAndom variaBle AnD m is the mediaN Of The distribUtiOn, then m is sucH
thAt $P(Y ≤ m) = P(y ≥ m) = 1/2$. if $Y_1, y_2,..., y_n$ are IndEpendEnt, expONentiaLly distriBuTEd randOM variabLES wIth mEan β and median m, $U = Y_{(n)}$ =
MAx($y_1, y_2,..., Y_n$) does not have An expoNeNTiAL DisTriBution. Use tHe GenerAL form of$ f_u (y)$ TO SHow THat $P(Y_{(n)} > m) = 1 − (.5)^n$.
Could Someone get mE StaRted on ThIs?

A:

i CalculAte fiRsTLy tHe probabiliTy diStributioN functIOn of $U=Y_{(n)}$. iT is $$F_{U}(u)= P(y_{(n)}\le u)=P(\Max\{y_1,\ldOts,Y_N\Le U\})=P(y_1\le U,\lDOts,y_N\lE u)=^{iNDepEndent}=P(Y_1\Le U)\cDot\ldOts\cDOT p(y_n\le U)=^{eqUallY \,distRibuted}(P(Y\le u))^n=(f_Y(u))^N$$ $%=^{expONenTial \, dIstriButiOn}(1-\Exp(-\beTa y))^n.$
So We havE: $$P(y_{(n)}>m)=P(U>m)=1-P(U\le m)=1-F_U(m)=1-(F_y(m))^n=1-0.5^n$$
Where the lAst EqUatIoN holdS BecausE $F_Y(M)=P(Y\Le m)=0.5$ accoRding to THe dEfINITiOn of the median. 

 Q:

Data stRUCtUres usinG C or C++?

WHAt ArE The advanTaGes Of usING C++ basEd ClASsEs/ObjectS over uSInG lEgacy C STrUctureS fOr mAnaGing sTAndaRd Data structurEs likE linked List, StacK, tree, Queue..etc? LET mE KNoW If thE quEstion is not CleaR. 

a:


 numerical analysis.  (Alt ernatively , you  ca n d owhat  I d id for the est i amte s above and use a bigsymbo li c  and nu meric s appli c at i o n l ik eMap le or  Math ema tica.)
 Q:

show  th at  $P(Y_{(n)}>  m ) = 1 − (. 5)^ n$.

If Y is  acontin uo usr andom  va riabl e andm  is th e medianof the di s tributi o n ,then  m is such
that $ P (Y ≤ m) = P(Y ≥ m ) = 1/ 2$ .  I f $Y_ 1,Y_2,..., Y _n $ are indepen d en t , exp o nentially dis tributed ra n dom  varia bl esw ith me an βan d  me dian m, $U= Y_ {(n)}$ =max($Y _ 1, Y_2, . .., Y_n $) doe s n othave an  e xpo ne n tia l  d ist r ibu tion. Us eth e gen eral f o r m of $ F _U ( y)$ t o show that $ P(Y _{(n ) } >  m) =  1 −(.5) ^n $.
Co uld so meone  g et me started o n th is?

A:
I c al cul at e fir s tly th e p rob ability  distri b uti on f u nc tion of $U=Y_{(n)} $. I tis $$F_{ U}(u)= P( Y_ { (n)}\leu) =P( \max \ { Y_1,\ ldot s ,Y _n\le u\ })=P(Y _ 1\ le  u,\ldo ts ,Y_n\l eu)= ^{i ndepe n dent }=P(Y_ 1\le u)\ cdot\ l dots\cdot P(Y_ n \le u)=^{equa l ly \ ,d i stri but ed}(P(Y\leu))^ n =(F_ Y(u) ) ^n $$$ %=^{e xpone nt i al \, distribution}(1- \e xp(-\b eta y ))^n.$
So wehave: $$P( Y _ { (n)}>m)= P(U> m )= 1 -P(U\le m)=1-F _U(m) =1-(F_Y(m) ) ^n=1-0.5 ^n$$where th e last eq u a tion hol dsbec aus e $ F _ Y( m)=P(Y\le m)= 0 . 5$ a cc ordingtothe def ini tio n o f t he  median.

 Q:

D at aSt ru ctu res u s ing C or  C ++? 
Wha t are the ad vanta gesof  u s ing  C++ ba s ed C lass es /O bjec tsov er us ingl ega cy C St ructuresfor mana gi ng  standa rd Data Struc tu res like L in ked  List, S tack, Tr ee, Queue..etc? Let mek now ifthe  ques tion  is not c lea r. 

A :

 numerical_analysis. _(Alternatively, you can do_what I_did_for the_estiamtes_above and use_a big symbolic_and numerics application like_Maple or Mathematica.)

_Q:

show_that $P(Y_{(n)} > m) = 1 − (.5)^n$.

If Y is a continuous random variable_and_m is_the_median_of the distribution, then m_is such
that $P(Y ≤ m)_= P(Y_≥ m) = 1/2$. If $Y_1, Y_2,..., Y_n$_are_independent, exponentially distributed_random variables with mean β and median m, $U_= Y_{(n)}$ =
max($Y_1, Y_2,..., Y_n$) does_not have an_exponential_distribution._Use the general form_of$ F_U (y)$ to show that_$P(Y_{(n)} > m) = 1 −_(.5)^n$.
Could someone get me started on this?

A:

I_calculate firstly the probability distribution function_of $U=Y_{(n)}$. It is $$F_{U}(u)=_P(Y_{(n)}\le u)=P(\max\{Y_1,\ldots,Y_n\le_u\})=P(Y_1\le u,\ldots,Y_n\le u)=^{independent}=P(Y_1\le u)\cdot\ldots\cdot P(Y_n\le_u)=^{equally \,distributed}(P(Y\le u))^n=(F_Y(u))^n$$_$%=^{exponential \,_distribution}(1-\exp(-\beta y))^n.$
So we_have: $$P(Y_{(n)}>m)=P(U>m)=1-P(U\le m)=1-F_U(m)=1-(F_Y(m))^n=1-0.5^n$$
where the last equation_holds because $F_Y(m)=P(Y\le_m)=0.5$ according to the definition of_the_median. 

 Q:

Data_Structures_using_C or_C++?

What are the_advantages_of using_C++_based Classes/Objects over using legacy C_Structures_for managing standard Data Structures like Linked_List, Stack, Tree, Queue..etc?_Let_me know if the_question is not clear. 

A:

