Begin2.DVI

I11-5. (xj− ̄x)^2 =x^2 j− 2 xj ̄x+ ̄x^2 so that

s^2 =^1

n− 1







∑m

j=1

x^2 jnfj−2 ̄x

∑m

j=1

xjnfj+ ̄x^2 n

∑m

j=1

fj







But

∑m

j=1fj= 1andn

∑m

j=1xjfj= ̄xnso that

s^2 =

1

n− 1







∑m

j=1

x^2 jnfj− 2 nx ̄^2 +nx ̄^2







=

1

n− 1







∑m

j=1

x^2 jnfj−n ̄x^2







=

1

n− 1







∑m

j=1

x^2 jnfj−n

(

1

n^2

)



∑m

j=1

xjfjn





2 



I11-6. P(2) =P(12) = 1/ 36 , P(3) =P(11) = 2/ 16 , P(4) =P(10) = 3/ 36 ,

P(5) =P(9) = 4/ 36 , P(6) =P(8) = 5/ 36 , P(7) = 6/ 36

I11-8. Binomial distribution(p+q)n=pn+npn−^1 q+···wherep= 1/ 2 ,q= 1/ 2 and

n= 10gives(1/2)^10 = 0. 00097956

I11-9. (a)

xj f ̃j fj F(x) xjf ̃j x^2 jf ̃j

0.725 2 0 .0333 = 2/ 60 0 .0333 = 2/ 60 1.450 1.0513

0.728 7 0 .1167 = 7/ 60 0 .1500 = 9/ 60 5.096 3.7099

0.731 11 0 .1833 = 11/ 60 0 .3333 = 20/ 60 8.041 5.878

0.734 14 0 .2333 = 14/ 60 0 .5666 = 34/ 60 10.276 7.5426

0.737 13 0 .2167 = 13/ 60 0 .7833 = 47/ 60 9.581 7.0612

0.740 7 0 .1167 = 7/ 60 0 .9000 = 54/ 60 5.180 3.8332

0.743 3 0 .05 = 3/ 60 0 .95 = 57/ 60 2.229 1.6561

0.746 3 0 .05 = 3/ 60 1 .00 = 60/ 60 2.238 1.6695

(c)mean= 0. 73488 , variance= 0. 00002461 , Standard deviation= 0. 00496

(e)(i)P(X≤ 0 .737) = 0. 783 ,

(ii)P(0. 728 < X < 0 .734) =P(X≤ 0 .734)−P(X≤ 0 .728) = 0. 42 ,

(iii)P(X > 0 .734) = 1−P(X≤ 0 .734) = 0. 4334

Solutions Chapter 11