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# Help with normal distribution question! watch

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1. Hi...I've been trying to do this question for ages and its not happening!

The first part is:
a) It is known that 60% of families in the UK have video recorders. If one takes a sample of 5 families what is the chance that at least 80% of them (i.e. 4 or 5 families) have a video recorder?

I've done this part and got 0.33696 using the binomial distribution.
n=5, p=0.6, r=0.4 and nCr pr qn-r

so P(4 in 5 successes) + P(5 in 5 successses) = 0.2592 + 0.07776 = 0.33696

Second part (part I can't do):
b) If the sample in part (a) was 1000 familes, why might you assume the number of families in the sample with video recorders is normally distributed? What is the mean and the variance of this normal distribution. What is the chance more than 800 (more than 80%) have video recorders?

Many Thanks

Emma
2. ok, well, i think that it is approximately normally distributed because you can approximate the binomial to normal if n>50, np>5 and nq>5. or it could have something to do with the central limit theorem, because it's a sample and n is large. actually, maybe the first reason. if thats the case then mean = np = 1000 * 0.6 = 600, variance = npq = 1000 * 0.6 * 0.4 = 240. can you do the rest?
3. if thats the case then mean = np = 1000 * 0.6 = 600, variance = npq = 1000 * 0.6 * 0.4 = 240. can you do the rest?[/QUOTE]

No...thats as far as I can get. Because then if you do (800-600)/√240, you get 12/12 which isnt on the normal distribution table!
4. which means that the probabily is roughly 0 then, although this is probably wrong. sorry, i hate distributions, lol. i think i forgot to divide the variance by 999 anyway but that wont help, just makes it worse. and also, you need a continuity correction but again that doesnt change it much . . .
5. [QUOTE=bob_54321]which means that the probabily is roughly 0 then, although this is probably wrong. sorry, i hate distributions, lol. i think i forgot to divide the variance by 999 anyway but that wont help, just makes it worse. and also, you need a continuity correction but again that doesnt change it much . . .[/QUOTE

ok, i think 0 is right! thanks!
6. sorry emma, workloads been crazy....

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Updated: February 18, 2006
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