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# Statistics watch

1. hi,
can someone help with the following....

the mean number of faulty bolts produced by a machine per hour is 0.3. the machine runs non stop for 8hrs. wat is the probability that more than 4 bolts are faulty?

i came up with p = 3/5 but not sure if its the right answer.
2. In 8 hrs, the mean number of faulty bolts is 8*0.3 = 2.4

X ~ Po(2.4)

P(X>4) = 1 - P(X<4) = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)]

= 1 - e(-2.4)[1 + 2.4 + (2.42/2) + (2.43/6) + (2.44/24)]

= 0.096

I'm not very good with some of my stats, so I wouldn't take my word on it. How did you do it?
3. I'm probably wrong as I'm pants at stats but this sounds like binomial prob so therefore X- Bin(8, p) and mean = 8*p so p= 0.0375.

Then you'll have to work out P(X>4).

Hang on what stats is this?
4. (Original post by fabz)
I'm probably wrong as I'm pants at stats but this sounds like binomial prob so therefore X- Bin(8, p) and mean = 8*p so p= 0.0375.

Then you'll have to work out P(X>4).

Hang on what stats is this?
It's Poisson not binomial because there can be more than one faulty bolt in an hour.
5. (Original post by Jonny W)
It's Poisson not binomial because there can be more than one faulty bolt in an hour.
Poo, I should be shot. No A for me.

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