as walpole doesn't gives even number answers, pls advise if my solution is correct below.

walpole

8.22 b)

P (Z<0.94) - P(Z<-1.49)

=0.7583

no. of sample mean that falls betw 172.5 and 175.8 inclusive = 200 x 0.7583-151.66, approx 152.

c)p(X bar < 172.0 ) = P(Z< -1.81)

=0.351

no of sample mean that falls below 172 = 200 x 0.0351 = 7.02 = 7 (round off).

8.28

P ((3.4-5.0) / 1.118) < Z < ((5.9-5.0)/1.118)

= P (Z< 0.805) - P (Z<-1.43)

8.51

T= (24-20 )/(4.1/3)

=2.93

so what % of confidence interval to use ?

if i use 95%, t at 0.025, with degrees of freedom =8 -> 2.306.

so since 2.306 falls out the range of 2.93 and -2.93, it is unlikely to obtain a sample size of size 9 form this population with mean of 24 and std dev of 4.1.

Am i right ?

thks.