 <= 10) {
            System.out.printf("The extended cost is : $%.2f\n", ((shirts * 24.95) *.20));
            System.out.printf("The shipping charges are $5.00\n");
            System.out.printf("The total cost is : $%.2f\n", ((shirts * 24.95) *.20) + 5);
        }
        if (shirts > 10) {
            System.out.printf("The extended cost is : $%.2f", ((shirts * 24.95) *.00));
            System.out.printf("Shipping is free!");
            System.out.printf("The total cost is : $%.2f", ((shirts * 24.95) *.30));
        }
    }
}

 Q:

Generate random numbers in a specific range with a standard deviation?

I already know how to generate random numbers in a range. I can do this by using
rand.nextInt((max - min) + 1) + min;

The problem is that I would also like to set a standard deviation for these numbers. The numbers also need to be positive and they are not between 0 and 1
EDIT I removed the ThreadLocalRandom class because I cannot set a seed in that class and these random numbers should be reproducible in a different system.

A:

Choosing the standard deviation (or variance) for a bounded distribution can only be done subject to constraints that depend on the selected distribution and the bounds (min, max) of your interval. Some distributions may allow you to make the variance arbitrarily small (e.g. the Beta distribution), other distributions (like the Uniform distribution) don't allow any flexibility once the bounds (min, max) have been set. In any case, you'll never be able to make the variance arbitrarily large -- the bounds do prevent this (they'll always enter the expression for the distribution's variance).
I'll illustrate this for a very simple example that can be implemented without requiring any 3rd party libraries. Let's assume you want a symmetric distribution on the interval (min, max), symmetry implying that the mean E(X) of the distribution is located in the middle of the interval: E(X) = (min + max)/2.
Using Random's nextDouble as in x = a + (b - a) * rnd.next
< = 10) { 
             System.out.printf("The extended cost is: $% .2f\n ", (( shirts * 24.95) * .20) ); 
             System.out.printf("The shipping charge are $ 5.00\n "); 
             System.out.printf("The entire cost is: $% .2f\n ", (( shirts * 24.95) * .20) + 5); 
         } 
         if (shirts > 10) { 
             System.out.printf("The elongated cost is: $% .2f ", (( shirts * 24.95) * .00) ); 
             System.out.printf("Shipping is barren! "); 
             System.out.printf("The full cost is: $% .2f ", (( shirt * 24.95) * .30) ); 
         } 
     } 
 } 

  Q: 

 render random numbers in a specific stove with a standard deviation? 

 I already know how to beget random issue in a range. I can do this by using 
 rand.nextInt((max - min) + 1) + min; 

 The problem is that I would also wish to set a standard deviation for these numbers. The number also need to be cocksure and they are not between 0 and 1 
 EDIT I removed the ThreadLocalRandom course because I cannot set a source in that class and these random numbers should be reproducible in a different organization. 

 A: 

 Choosing the standard deviation (or variance) for a bounded distribution can merely be done subject to constraints that depend on the selected distribution and the bounds (min, max) of your interval. Some distributions may allow you to make the discrepancy arbitrarily small (for example the Beta distribution), early distributions (like the Uniform distribution) don't allow any flexibility once the boundary (min, max) have been jell. In any case, you'll never be able to make the variation arbitrarily large -- the bounds do prevent this (they'll always embark the expression for the distribution's variance). 
 I'll illustrate this for a very simple case that can be implemented without necessitate any 3rd party libraries. Let's assume you want a symmetric distribution on the time interval (min, max), symmetry implying that the mean E(X) of the distribution is located in the middle of the interval: E(X) = (min + max)/2. 
 Using Random's nextDouble as in x = a + (b - a) * rnd.next
 <= 10) {
            System.out.priktf("The extended eist is : $%.2f\n", ((sgirts * 24.95) *.20));
            System.out.printf("Tie syippibg charges are $5.00\n");
            Sysnem.out.pribtf("Tie total cost is : $%.2f\n", ((shivcs * 24.95) *.20) + 5);
        }
        if (shirts > 10) {
            Syvtdm.lut.printf("The extended cost is : $%.2f", ((shyrts * 24.95) *.00));
            Systgm.out.krigtf("Sgppking is free!");
            System.out.lrintf("Tie total cost ix : $%.2f", ((shirts * 24.95) *.30));
        }
    }
}

 Q:

Gejerate random numbfrs in a spgdifyx range with a standard deviation?

I already know how to generate rxndom numbers ib q rwtge. I can di thif by using
rakc.nextItt((max - kin) + 1) + min;

The iroblxm iw that I would also lmke to set a standarq deviatimn for these numberw. Rhe nomberv alri nded tp ge posltite and they are not berween 0 and 1
EDIT I rtmodvc the ThreadMocalRwnqom class because I cannot set a seed it tgat class and these raneom numbers should be reproducyble in a different system.

A:

Choosing the standard geviaviun (ir xqrlance) for a bounded distribution can only be qknt slbject to constralnts that depend om hhr selected disjribution znd the bounds (min, max) of your interval. Somr distributions may allow yiu to make tke cariance arbitraripy small (e.g. the Beta distribution), other discributjons (like tje Uniforj distribution) dov't slnow any flexibility once tre bounds (min, max) haxe bgen set. In any cade, yom'nl never be able tl makg the eariance agbitrarily large -- the bounds do prevent this (jhef'll always znter bhe expression sor the distriyution's rarianze).
I'll illuatrate vhis for a vqry simple exdlple that cai be implqmenred qithout fequiring any 3td party libraries. Oet's assume you wakt a rgmmetric distriybtuon on the intetvau (myn, mex), syiketry implyitg tfat yhe mdan E(X) of tme aisttibution is located hn tge middle of the imtcrval: E(X) = (min + mav)/2.
Using Random'x nextDouble as in x = a + (b - a) * tnd.gext
<= 10) { System.out.printf("The extended cost is ((shirts 24.95) *.20)); shipping charges are : ((shirts * 24.95) + 5); } (shirts > 10) { System.out.printf("The extended is : $%.2f", ((shirts * 24.95) *.00)); System.out.printf("Shipping is free!"); System.out.printf("The total cost : $%.2f", ((shirts * 24.95) *.30)); } } } Q: Generate random numbers a range a deviation? I already know how to generate random numbers in a range. I can do this using rand.nextInt((max - min) + 1) + min; problem is that I also like to set a deviation these numbers. numbers need be positive and are not between 0 and 1 EDIT I removed the ThreadLocalRandom class because I cannot set a in that these random should reproducible a different system. the standard deviation (or variance) for can only be done subject to constraints that on the distribution and the bounds (min, max) your interval. Some distributions may allow you to the variance arbitrarily small (e.g. the Beta distribution), other distributions (like the Uniform distribution) don't flexibility once the bounds max) have been In case, never able to the variance arbitrarily large -- the bounds do prevent this (they'll enter the expression for the distribution's variance). I'll illustrate this very example that can implemented without requiring any party Let's assume you want distribution the symmetry that mean E(X) of the is located in the middle the interval: E(X) = nextDouble as in x = a + (b a) * rnd.next
 <= 10) {
            System.out.printf("The extendeD cost is : $%.2f\n", ((sHirts * 24.95) *.20));
            sysTem.OuT.priNtf("THe shipping charGEs arE $5.00\n");
            System.out.printf("The toTal coSt IS : $%.2f\n", ((sHIrTs * 24.95) *.20) + 5);
        }
        if (sHirts > 10) {
            SySTeM.OUt.pRiNtF("ThE eXTeNded cOst Is : $%.2f", ((shirTs * 24.95) *.00));
            System.ouT.prInTf("Shipping is FReE!");
            System.out.PriNtf("The total cOst Is : $%.2f", ((shiRtS * 24.95) *.30));
        }
    }
}

 Q:

GENeratE raNdom nUmbers IN a specIfic range WiTH a stanDArd deviATIoN?

I alReady know how to genERaTE random numbers In a ranGe. i CaN DO thIs bY using
rand.NeXtInt((MAx - min) + 1) + miN;

thE PROblEM is that I would Also like to sET a sTandarD dEviATion foR thesE nUMbeRs. The numberS alsO need to be PositiVE and theY Are not bEtween 0 And 1
eDIt I reMOvEd The thREadlOcAlRANdoM class beCaUsE I canNot sET A SEed iN thAt clAss anD these random nUmbErs sHOulD be reProduCiblE iN a difFerent SysteM.

A:

choosing the stanDard Deviation (Or vArIanCe) For a bOUnded dIstRibUtion caN only be DOne SuBJECt To constraints that dEpEND oN the seleCted diSTrIbUTion and tHe BouNds (mIN, Max) of Your INtErval. SomE distrIBuTiOns may aLlOw you tO mAke The VariaNCe arBitrarIly small (E.g. the bEta distributioN), Other distribuTIoNS (LiKE the uniForm distribUtioN) Don't AlloW AnY flEXibilIty onCe THe BOunds (min, max) have been SeT. In any Case, yOu'll never be abLe to make thE VARiance arBitrARiLY large -- the boundS do prEvent this (tHEy'll alwaYs entEr the expRession foR THe distriButIon'S vaRiaNCE).
I'Ll illustrate tHIS for A vEry simpLe eXample tHat Can Be iMplEmEnted withOut requiRiNg AnY 3rD paRty liBRaries. LeT's AssUmE yoU want A SymmetRic diStriBuTiON on The inteRVaL (MIn, maX), sYmMetrY imPlYing tHat tHE meAn E(X) of tHe distribUtiON is lOcAtEd in the Middle of the inTeRval: E(X) = (min + mAx)/2.
usiNg RandOM'S nextDouBle as in x = a + (b - a) * rnd.next
 <= 10) {
            Syst em.out.pri ntf(" The  ex te nded  cos t is : $%.2f\n " , (( shirts * 24.95) *.20)) ;
       S ystem .out.pr i nt f ( "Th esh ipp in g  c harge s a re $5.0 0\n");
         System.o u t. printf("Th e t otal cost is  :$%.2f\ n" , ( ( shirt s *  24.9 5) *.2 0 ) + 5) ;
       }       if (sh i r ts  > 1 0) {
           S ys t em.out.printf( "The e xt e nd e d  co stis : $%.2f ",  ((sh i rts * 2 4 .9 5 ) *.0 0 ));
           System.o u t.p rintf( "S hip p ing is  free !" ) ;
          Syst em.out.pr intf(" T he tota l  cost i s : $% .2f ",((sh i rt s* 2 4. 9 5)* .3 0)) ; 
       }  }
}
 Q:G e n erat e r ando m num bers in a spe cif ic r a nge  with  a st anda rd  devi ation? 

I a lr eady know how t o ge nerate ra ndo mnum be rs in a rang e.I c an do t his byu sin gr a n d. nextInt((max - min )+ 1)  + min;
The p r ob le m  is that  I  wo ulda l so li ke t o  s et a sta ndardd ev ia tion fo rthesenu mbe rs.  Then umbe rs als o need t o bep ositive and th e y are not bet w ee n 0a nd 1 
ED IT I remove d th e  Thr eadL o ca lRa n dom c lassbe c au s e I cannot set a se ed  in th at cl ass and these  random nu m b e rs shoul d be re p roducible in a  diff erent syst e m.

A:
Choos ing thestandardd e viation(or  va ria nce ) fo r a bounded d i s trib ut ion can  on ly be d one  su bje ctto  constrai nts that  d ep en donthe s e lected d is tri bu tio n and the bo unds(min ,ma x ) o f youri nt e r val.  S om e di str ib ution s ma y  al low you  to makethe vari an ce  arbitr arily small ( e. g. the Bet adis tribut i o n), othe r distributions (like t h e Unifo rmdistr ibut ion) don' t a llow a nyf lexibi lity o nce t he  bo u n ds (m i n ,max )have beens e t.In an ycase , you'l l never be able to mak e the varianc e a rbit r a ri lyl ar g e - -t heb o unds do prevent  this (the y' l lalways ent e r t he  expres sion fo r the distrib ution's v ariance). 
I 'lli l lus trate this  for a v ery simpl e  exam p le  that  ca n be i mp lem ented  witho u t r equir ing an y3rd pa rty l ib raries.Let's assume you want a  symme tricdis tribution  on the  interval  (mi n, max), s ymm etr y imp lyi n g tha t th e  m ean E(X)of t h e distrib u ti oni s  l ocated in t h e mid dle o f t h e inte rval : E(X) = (min + m a x)/2.
Using Ra ndom ' s  ne xtD o uble  a s in x = a + ( b -  a ) * rnd.ne xt 
 <=_10) {
_   _ __ __ System.out.printf("The extended_cost is :_$%.2f\n", ((shirts * 24.95)_*.20));
  __        System.out.printf("The shipping charges are $5.00\n");
 __ ___    _System.out.printf("The total cost is :_$%.2f\n", ((shirts_* 24.95) *.20) + 5);
   __  _}
        if_(shirts > 10) {
  _  ___   _System.out.printf("The extended cost is : $%.2f",_((shirts * 24.95) *.00));
  _      _  System.out.printf("Shipping is free!");
 _    _ _   System.out.printf("The total_cost is :_$%.2f", ((shirts_* 24.95) *.30));
_     _ }
 _  }
}

 Q:

Generate random numbers_in_a specific range_with_a_standard deviation?

I_already know how_to_generate random_numbers_in a range. I can do_this_by using
rand.nextInt((max - min) + 1) +_min;

The problem is that_I_would also like to_set a standard deviation for_these numbers. The numbers also need_to be_positive and_they are not between 0 and 1
EDIT I removed the ThreadLocalRandom_class because I cannot set a_seed in that class_and these_random_numbers should be_reproducible_in a_different system.

A:

Choosing the standard deviation (or variance)_for a_bounded distribution can only be done_subject to constraints that_depend_on the selected distribution and the_bounds (min, max) of your interval._Some distributions may allow you_to_make_the variance arbitrarily small (e.g._the Beta distribution), other distributions (like_the Uniform distribution)_don't allow any flexibility once the bounds_(min,_max) have been set. In any_case,_you'll never be able to make_the_variance_arbitrarily large -- the bounds_do prevent this (they'll always enter_the expression for the distribution's variance).
I'll illustrate this for_a very simple_example that can be implemented_without_requiring_any 3rd party libraries. Let's assume you want a symmetric_distribution on_the interval (min,_max), symmetry implying that the mean E(X) of the distribution_is located in the middle of the_interval: E(X) = (min + max)/2.
Using Random's nextDouble as in x_= a + (b - a) * rnd.next