cheekbones3: (Default)
cheekbones3 ([personal profile] cheekbones3) wrote2003-11-03 01:52 pm

(no subject)

Being in a bad mood actually does make me want to post more! Anyone got any crap jobs they need doing? Collation of papers, file renaming that sort of thing. £5 an hour is my rate! In fact, I could do some tutoring actually. Maths, physics, or any science up to 16 year old level, plus geography and related I suppose. I really am that poor. Piss!
ext_52479: (Default)

[identity profile] nickys.livejournal.com 2003-11-03 06:15 am (UTC)(link)
Well, I still haven't organised a new cleaner, so that job's open... although [livejournal.com profile] lilitufire made some interesting suggestions on the subject :o)

[identity profile] cheekbones3.livejournal.com 2003-11-05 05:01 am (UTC)(link)
Okay - maybe next week? I'll see how I feel! And I think I'd just be doing it in conventional style - my submissive tendencies aren't that well developed!
ext_52479: (flower)

[identity profile] nickys.livejournal.com 2003-11-05 06:26 am (UTC)(link)
That's okay. I don't think I could avoid laughing if I tried to be a domme... :o)

Are either Wednesday or Friday feasible for you?

[identity profile] cheekbones3.livejournal.com 2003-11-06 08:09 am (UTC)(link)
It looks like I can get the odd pub shift at the moment, so I'm quite happy to put it off for a couple of weeks if you can cope for that long!

[identity profile] sharmaya.livejournal.com 2003-11-03 06:24 am (UTC)(link)
Love your choice of 'Current Mood'. Stick in there. Put some postcards in shop windows near to local schools, offering tutoring (I'd say more than £5 per hour for that, though!!) if you want to go that way. And keep on posting in LJ today, I keep hitting F5 and you're the only one posting anything!! *another hug, for good luck*

[identity profile] cheekbones3.livejournal.com 2003-11-05 05:01 am (UTC)(link)
Thanks :O)

[identity profile] toffeecat.livejournal.com 2003-11-03 02:37 pm (UTC)(link)
Up to a 16yr old level! Damn! I need help in Business Stats......its ass raping me with a 12inch dildo :(


[identity profile] cheekbones3.livejournal.com 2003-11-03 05:52 pm (UTC)(link)
Well I can do degree level too if you really want! What are you struggling with?

[identity profile] toffeecat.livejournal.com 2003-11-03 06:01 pm (UTC)(link)
Continuous Probability Distributions
Uniform Distribution
Binomial Distribution

Changing things to Z-Scores


:-/ I am going to see my Math prof. on Weds hopfully he can help me too

[identity profile] cheekbones3.livejournal.com 2003-11-03 06:02 pm (UTC)(link)
Easy easy easy! Ask me and I can tell :O)

[identity profile] toffeecat.livejournal.com 2003-11-03 10:39 pm (UTC)(link)
What is the point on Z-scores......what do they measure.......erm, i'll think of more questions tomorrow. Its 1:30am so i am sleepy.....but tomorrow i will ask more

[identity profile] cheekbones3.livejournal.com 2003-11-04 02:20 am (UTC)(link)
Z is a simple way to represent probabilities on the normal distribution. It comes from X=rho.Z+mu, where X is a general normal dist., and gives a simple way to measure the probabilty of finding a point within a certain range of this distribution.

Z expresses x (a value from a given distribution) in the number of standard deviations from the mean. A very useful fact is that there is virtually 95% probabilty of finding a value in the range -2<Z<2, since 95% is the most common test for statistical significance. It's easy to grasp the uses of Z with examples - the examples should really easy. Symmetry is also very useful when using Z.

[identity profile] toffeecat.livejournal.com 2003-11-04 10:43 am (UTC)(link)
One of the questions is

Suppose that x has a binomial distribution with n=200 and p=0.4

a) Show that the normal approximation to the binomial can appropriately to used to calculate probablities about x.

b)Make continuity corrections for each of the following, and then use the normal aprroximation to the binomial to find each probability
1) P(x=80)
2) P(x<65)
3) p(x>100)


[identity profile] cheekbones3.livejournal.com 2003-11-04 01:57 pm (UTC)(link)
Erm I never did this, but you can say for a) that the central limit theorem can be used to sum all the independent random variables that make up the binomial r.v., which means that the normal can be used when n is big enough, so mean =n.p and SD=sqroot(n.p.(1-p)).

b)1) Mean=80, SD=6.93, Z=see a table for the mid value of n=200 I think...
2) 0.0125
3) 0.002

I find that I'm actually bloody rusty, so this is helpful work for me as well - that's what I make those to be - could you tell me if I'm right? Lots of practise is the key...