Hi Bryan my name's Matt.
This is the first time I've posted anything on here and I have to say I
probably (no pun intended) wouldn't have bothered if it weren't for all the
hilarious replies you were getting!! I was in fits of laughter (especially
from you Peter Butler, fantastic!).
Anyway, just for your (plural) interest, I'm a student myself in my final
year at university. I completely understand how you must be feeling because
when I was doing my A-levels I was permanetly stuck with my Physics. My
parents even went so far as to hire a personal tutor, who was little help.
I just don't think I am cut out for Physics, but I am wandering off the
point now, sorry, so here we go...
Question 1...
The question describes a normal distribution curve with a mean of 54 and st.
dev. of 9.5. This is absolutely no help in calculating anything, so we need
to convert it to a curve with a mean of 0 and a st. dev. of 1. For this, we
use the equation...
_
z = ( x - x ) / sigma
= 48 - 54 / 9.5
= -0.63
What this tells us is that on a normal curve of mean 0, st. dev. 1 the
48km/h equals -0.63
We want to calculate the percentage above this so you have to use those
normal tables that i'm sure you've seen before. (it's best if you draw this
all out, makes a lot more sense that way!!). I don't actually have a table
to hand (I'm sure my dad's got millions somewhere, being a statistician, but
i can't be bothered to find them). Anyway, that's part a done!! (simple
enough eh?!)
b. we now want to find 72 on our N(0, 1) curve. So use the same formula as
before...
z = 72 - 54 / 9.5 = 1.89
Find the area (the percentage, or probability) in the tables where z > 1.89
c. this part is the trickiest, but not by much... we need to work
backwards!! we want to find the area in the middle that takes up 50% of the
curve. this is tricky to find in the table but thinking about it, that
leaves 25% on either side of the "shaded area" of the curve that i'm sure
you have already drawn. so look in the table for an area of .25 and that is
the upper limit of the 50%. and then of course, stick a negative sign in
front to get the lower limit!!!
all you need to do then is re-arrange that equation from above to get them
back into km/h!
Question 2..
(this is a bit of a bitch compared to question 1, but then i always hated
these things and think it maybe more psychological on my part than anything
else, so here we go...)
n = 56
p = 0.72
st. dev = SQRT[ (p(1-p)) / n ]
= SQRT[ (0.72 x 0.28) / 56 ]
= 0.06
All that above means the standard deviation of the sample given is 0.06
Draw the normal curve again with mean = 0.72
The 95% confidence limit that we're after lies between +/- 1.96 of 0.06
(does that make sense?). i know what you're thinking - where did that 1.96
come from right? that's just like a fact. 95% of the normal curve lies
between +/- 1.96 standard devations from the mean (trust me ok?).
0.06 x 1.96 = 0.1176
so the ranges are between 0.72 + 0.1176 and 0.72 - 0.1176 !!!
part b.
we want 1.96 x st. dev to equal 0.03 with some sample size (that's bigger
than 56). so rearrange the equation
st. dev = SQRT[ (p(1-p)) / n ]
for n, where st. dev = 0.03/1.96, p is still .72
I hope that makes sense (to everyone).
Best wishes for the future Bryan.
Matt
P.S. I have a hunch that no-one else answered you because they were all
afraid they wouldn't be able to and didn't want to look foolish in front of
all their little friends!! But I don't know anyone on here (although my dad
probably does) so I'm not overly bothered.
|