I guess I was hoping if anybody could check my work? This is a 4 part question, I hope I am not breaking rules by posting so much?? I took most of them as far as I could, a couple I am completely clueless on....!

Suppose that for a school, distances between students' homes and school observe a normal distribution, with mean 4.76 miles and standard deviation being 1.76 miles.

a) What percentage of students live farther than 6.78 miles from the school?

P(X>6.78 miles)

Z=X-mean/std --> 6.78-4.76/1.74= 1.16

P(0<z<1.16)= 0.3770

0.5- 0.3770= 0.123

Thus, 12.3% of students live farther than 6.78 miles from the school.

(yes? no? i think i may have got the right answer for that one)

b) A survey shows that 8% of students who live closest to the school chose to walk to the school. What is the maximum walking distance of these 8% of students? In other words, what is the distance below which these 8% of students walk to school?

(I really don't know how to start on this one, could anybody set me in the right direction?)

c) Suppose there is a new policy that allows students living beyond 4.50 miles to take a bus to school. There are 3,567 students enrolled. How many students are not eligible to take school buses?

Z=4.5-4.76/ 1.74= -0.15

P(-0.15<z<0)= 0.0596

0.50 + 0.0596 = 0.5596 or 55.96%

1-0.5596= 0.4404

0.4404 * 3567= 1570.91 = 1571

Thus, approximately 1,571 students are not eligible to take school buses.

d) Suppose all samples of size 12 are taken. What percentage of sample means has a value larger than 6.78 miles?

*Since this one is talking about sample means, to find the z-score is:

Z=Mean-Mean of sample means/ std of sample means, so

Z=6.78-4.76/ (1.74/[sqrt]12)= 4.02

I get lost here, because how am I supposed to find a z-score of 4.02 in my table? *bangs head* Anybody?