1. ## Probability and proportions?

I can't work out how to do this question. It doesn't look like anything we've done before, but I don't know if I'm just misreading it. I'm assuming it has something to do with binomial distributions, but I'm not sure. Any help is appreciated!

A road is constructed so that the right-turn lane at an intersection has a capacity of 3 cars. Suppose that 30% of cars approaching the intersection want to turn right, and they do so independently.

a) If 15 cars approach the intersection, what is the probability that the right-turn lane will not hold all the cars wanting to turn right?

b) Over a week, 1428 cars used the intersection. What is the distribution of the sample proportion of cars turning right at this intersection, in random samples of this size?

2. Originally Posted by mightysparks
I can't work out how to do this question. It doesn't look like anything we've done before, but I don't know if I'm just misreading it. I'm assuming it has something to do with binomial distributions, but I'm not sure. Any help is appreciated!

A road is constructed so that the right-turn lane at an intersection has a capacity of 3 cars. Suppose that 30% of cars approaching the intersection want to turn right, and they do so independently.

a) If 15 cars approach the intersection, what is the probability that the right-turn lane will not hold all the cars wanting to turn right?

b) Over a week, 1428 cars used the intersection. What is the distribution of the sample proportion of cars turning right at this intersection, in random samples of this size?
a) Binomial(n=15, p=3/10). You want P(X>3).
b) Binomial(n=1428, p=3/10)

3. So a) is just P(X>3) = .2969?

and for b) is B(1428, 3/10) all I have to write? These answers seem too simple, which is why I questioned it in the first place.

Thanks.

4. Or is it 1-P(X<3) = 1 -.2969 = .7031?

5. it is the 0.7031

b) i) Binomial(n=1428, p=3/10)
that is not the case, it is not a binomial distribution it is a normal distribution
ii) you need to work out the SD=square root of variance; the mean=np
as well as the Z score

i just have a quick question for you....
b) i) how do you expect the sample mean brain weight to be distributed for random sample of size 20 from this population?
(i think that i need to use the central limit theorem for this answer, i just do not know how to word it)

ii) in such samples, how unusual would it be to find a mean brain weight in excess of 1470g?

can you help?