Question

Roll a fair die 100 times and let T be the sum of the 100 results.

(a) Find a very good approximation for the value of P( T > 315 )

(b). Find P( T = 315 ), approximately

Answer 1

here for probability of each number on die =P(x)=1/6

therefore mean on one roll E(X)=xP(x)=(1/6)*(1+2+3+4+5+6)=3.5

and E(X²)=x²P(x)=(1/6)*(1²+2²+3²+4²+5²+6²)=15.1667

hence Var(X)=E(X²)-(l E(X)))² =2.9167

here for 100 rolls expected sum =100*3.5=350

and std deviation =(Var(x)*n)^1/2 =(2.9167*100)^1/2 =17.078

a)

from normal approximation and continuity correction

as z score =(X-mean)/std deviatio

P(T>315)=1-P(T<=314)=1-P(Z<(314.5-350)/17.078)=1-P(Z<-2.08)=1-0.0188 =0.9812

b)

\P(T=315)=P(314.5<X<315.5)=P((314.5-350)/17.078<Z<(315.5-350)/17.078)=P(-2.08<Z<-2.02)=0.0217-0.0188

=0.0029